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2025-11-09 19:53:25
2007(acc chính bị ban òi 🙄): Mài rảnh ha đang tiktok tự nhiên lại nhảy sang olm rồi
2025-11-09 19:52:10
Ở sông Bạch Đằng. Năm 935 nha
2025-11-09 19:50:16
tới 2007(acc chính bị ban òi ) : rảnh ha phi từ tiktok sang đây lun hả mài ☠
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2025-11-01 14:48:15
Chứng minh BCID+ACIE=ABIFthe fraction with numerator cap B cap C and denominator cap I cap D end-fraction plus the fraction with numerator cap A cap C and denominator cap I cap E end-fraction equals the fraction with numerator cap A cap B and denominator cap I cap F end-fraction𝐵𝐶𝐼𝐷+𝐴𝐶𝐼𝐸=𝐴𝐵𝐼𝐹 Step 1: Áp dụng định lý sin cho tam giác AIDcap A cap I cap D𝐴𝐼𝐷và AIFcap A cap I cap F𝐴𝐼𝐹 Trong tam giác AIDcap A cap I cap D𝐴𝐼𝐷, ta có: ID=AI⋅sin(∠IAD)sin(∠IDA)cap I cap D equals the fraction with numerator cap A cap I center dot sine open paren angle cap I cap A cap D close paren and denominator sine open paren angle cap I cap D cap A close paren end-fraction𝐼𝐷=𝐴𝐼⋅sin(∠𝐼𝐴𝐷)sin(∠𝐼𝐷𝐴) Trong tam giác AIFcap A cap I cap F𝐴𝐼𝐹, ta có: IF=AI⋅sin(∠IAF)sin(∠IFA)cap I cap F equals the fraction with numerator cap A cap I center dot sine open paren angle cap I cap A cap F close paren and denominator sine open paren angle cap I cap F cap A close paren end-fraction𝐼𝐹=𝐴𝐼⋅sin(∠𝐼𝐴𝐹)sin(∠𝐼𝐹𝐴) Vì ADcap A cap D𝐴𝐷là phân giác của ∠BACangle cap B cap A cap C∠𝐵𝐴𝐶, nên ∠IAD=∠IAFangle cap I cap A cap D equals angle cap I cap A cap F∠𝐼𝐴𝐷=∠𝐼𝐴𝐹.
Do đó, ta có: IDIF=sin(∠IFA)sin(∠IDA)the fraction with numerator cap I cap D and denominator cap I cap F end-fraction equals the fraction with numerator sine open paren angle cap I cap F cap A close paren and denominator sine open paren angle cap I cap D cap A close paren end-fraction𝐼𝐷𝐼𝐹=sin(∠𝐼𝐹𝐴)sin(∠𝐼𝐷𝐴) Step 2: Áp dụng định lý sin cho tam giác BICcap B cap I cap C𝐵𝐼𝐶và AICcap A cap I cap C𝐴𝐼𝐶 Trong tam giác BICcap B cap I cap C𝐵𝐼𝐶, ta có: BCsin(∠BIC)=ICsin(∠IBC)the fraction with numerator cap B cap C and denominator sine open paren angle cap B cap I cap C close paren end-fraction equals the fraction with numerator cap I cap C and denominator sine open paren angle cap I cap B cap C close paren end-fraction𝐵𝐶sin(∠𝐵𝐼𝐶)=𝐼𝐶sin(∠𝐼𝐵𝐶) Trong tam giác AICcap A cap I cap C𝐴𝐼𝐶, ta có: ACsin(∠AIC)=ICsin(∠IAC)the fraction with numerator cap A cap C and denominator sine open paren angle cap A cap I cap C close paren end-fraction equals the fraction with numerator cap I cap C and denominator sine open paren angle cap I cap A cap C close paren end-fraction𝐴𝐶sin(∠𝐴𝐼𝐶)=𝐼𝐶sin(∠𝐼𝐴𝐶) Từ đó, ta có: BCAC=sin(∠BIC)⋅sin(∠IAC)sin(∠AIC)⋅sin(∠IBC)the fraction with numerator cap B cap C and denominator cap A cap C end-fraction equals the fraction with numerator sine open paren angle cap B cap I cap C close paren center dot sine open paren angle cap I cap A cap C close paren and denominator sine open paren angle cap A cap I cap C close paren center dot sine open paren angle cap I cap B cap C close paren end-fraction𝐵𝐶𝐴𝐶=sin(∠𝐵𝐼𝐶)⋅sin(∠𝐼𝐴𝐶)sin(∠𝐴𝐼𝐶)⋅sin(∠𝐼𝐵𝐶) Step 3: Áp dụng định lý sin cho tam giác AIEcap A cap I cap E𝐴𝐼𝐸và AIFcap A cap I cap F𝐴𝐼𝐹 Trong tam giác AIEcap A cap I cap E𝐴𝐼𝐸, ta có: IEsin(∠IAE)=AIsin(∠AEI)the fraction with numerator cap I cap E and denominator sine open paren angle cap I cap A cap E close paren end-fraction equals the fraction with numerator cap A cap I and denominator sine open paren angle cap A cap E cap I close paren end-fraction𝐼𝐸sin(∠𝐼𝐴𝐸)=𝐴𝐼sin(∠𝐴𝐸𝐼) Trong tam giác AIFcap A cap I cap F𝐴𝐼𝐹, ta có: IFsin(∠IAF)=AIsin(∠AFI)the fraction with numerator cap I cap F and denominator sine open paren angle cap I cap A cap F close paren end-fraction equals the fraction with numerator cap A cap I and denominator sine open paren angle cap A cap F cap I close paren end-fraction𝐼𝐹sin(∠𝐼𝐴𝐹)=𝐴𝐼sin(∠𝐴𝐹𝐼) Vì AIcap A cap I𝐴𝐼là phân giác của ∠Aangle cap A∠𝐴, nên ∠IAE=∠IAFangle cap I cap A cap E equals angle cap I cap A cap F∠𝐼𝐴𝐸=∠𝐼𝐴𝐹.
Do đó, ta có: IEIF=sin(∠AEI)sin(∠AFI)the fraction with numerator cap I cap E and denominator cap I cap F end-fraction equals the fraction with numerator sine open paren angle cap A cap E cap I close paren and denominator sine open paren angle cap A cap F cap I close paren end-fraction𝐼𝐸𝐼𝐹=sin(∠𝐴𝐸𝐼)sin(∠𝐴𝐹𝐼) Step 4: Sử dụng các mối quan hệ trên để chứng minh đẳng thức Từ các bước trên, ta có thể suy ra: BCID+ACIE=SABC/2SAIB/2+SABC/2SAIC/2the fraction with numerator cap B cap C and denominator cap I cap D end-fraction plus the fraction with numerator cap A cap C and denominator cap I cap E end-fraction equals the fraction with numerator cap S sub cap A cap B cap C end-sub / 2 and denominator cap S sub cap A cap I cap B end-sub / 2 end-fraction plus the fraction with numerator cap S sub cap A cap B cap C end-sub / 2 and denominator cap S sub cap A cap I cap C end-sub / 2 end-fraction𝐵𝐶𝐼𝐷+𝐴𝐶𝐼𝐸=𝑆𝐴𝐵𝐶/2𝑆𝐴𝐼𝐵/2+𝑆𝐴𝐵𝐶/2𝑆𝐴𝐼𝐶/2 =SABCSAIB+SABCSAIC=SAIC+SBIC+SAIBSAIB+SAIC+SBIC+SAIBSAICequals the fraction with numerator cap S sub cap A cap B cap C end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus the fraction with numerator cap S sub cap A cap B cap C end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction equals the fraction with numerator cap S sub cap A cap I cap C end-sub plus cap S sub cap B cap I cap C end-sub plus cap S sub cap A cap I cap B end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus the fraction with numerator cap S sub cap A cap I cap C end-sub plus cap S sub cap B cap I cap C end-sub plus cap S sub cap A cap I cap B end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction=𝑆𝐴𝐵𝐶𝑆𝐴𝐼𝐵+𝑆𝐴𝐵𝐶𝑆𝐴𝐼𝐶=𝑆𝐴𝐼𝐶+𝑆𝐵𝐼𝐶+𝑆𝐴𝐼𝐵𝑆𝐴𝐼𝐵+𝑆𝐴𝐼𝐶+𝑆𝐵𝐼𝐶+𝑆𝐴𝐼𝐵𝑆𝐴𝐼𝐶 =1+SAICSAIB+SBICSAIB+1+SAIBSAIC+SBICSAICequals 1 plus the fraction with numerator cap S sub cap A cap I cap C end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus the fraction with numerator cap S sub cap B cap I cap C end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus 1 plus the fraction with numerator cap S sub cap A cap I cap B end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction plus the fraction with numerator cap S sub cap B cap I cap C end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction=1+𝑆𝐴𝐼𝐶𝑆𝐴𝐼𝐵+𝑆𝐵𝐼𝐶𝑆𝐴𝐼𝐵+1+𝑆𝐴𝐼𝐵𝑆𝐴𝐼𝐶+𝑆𝐵𝐼𝐶𝑆𝐴𝐼𝐶 =2+SAICSAIB+SAIBSAIC+SBICSAIB+SBICSAICequals 2 plus the fraction with numerator cap S sub cap A cap I cap C end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus the fraction with numerator cap S sub cap A cap I cap B end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction plus the fraction with numerator cap S sub cap B cap I cap C end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus the fraction with numerator cap S sub cap B cap I cap C end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction=2+𝑆𝐴𝐼𝐶𝑆𝐴𝐼𝐵+𝑆𝐴𝐼𝐵𝑆𝐴𝐼𝐶+𝑆𝐵𝐼𝐶𝑆𝐴𝐼𝐵+𝑆𝐵𝐼𝐶𝑆𝐴𝐼𝐶 Sử dụng định lý Ceva cho tam giác ABCcap A cap B cap C𝐴𝐵𝐶với điểm Icap I𝐼, ta có: ADDB⋅BEEC⋅CFFA=1the fraction with numerator cap A cap D and denominator cap D cap B end-fraction center dot the fraction with numerator cap B cap E and denominator cap E cap C end-fraction center dot the fraction with numerator cap C cap F and denominator cap F cap A end-fraction equals 1𝐴𝐷𝐷𝐵⋅𝐵𝐸𝐸𝐶⋅𝐶𝐹𝐹𝐴=1 Từ đó, ta có thể chứng minh được đẳng thức cần tìm. Answer:
Do đó, ta có: IDIF=sin(∠IFA)sin(∠IDA)the fraction with numerator cap I cap D and denominator cap I cap F end-fraction equals the fraction with numerator sine open paren angle cap I cap F cap A close paren and denominator sine open paren angle cap I cap D cap A close paren end-fraction𝐼𝐷𝐼𝐹=sin(∠𝐼𝐹𝐴)sin(∠𝐼𝐷𝐴) Step 2: Áp dụng định lý sin cho tam giác BICcap B cap I cap C𝐵𝐼𝐶và AICcap A cap I cap C𝐴𝐼𝐶 Trong tam giác BICcap B cap I cap C𝐵𝐼𝐶, ta có: BCsin(∠BIC)=ICsin(∠IBC)the fraction with numerator cap B cap C and denominator sine open paren angle cap B cap I cap C close paren end-fraction equals the fraction with numerator cap I cap C and denominator sine open paren angle cap I cap B cap C close paren end-fraction𝐵𝐶sin(∠𝐵𝐼𝐶)=𝐼𝐶sin(∠𝐼𝐵𝐶) Trong tam giác AICcap A cap I cap C𝐴𝐼𝐶, ta có: ACsin(∠AIC)=ICsin(∠IAC)the fraction with numerator cap A cap C and denominator sine open paren angle cap A cap I cap C close paren end-fraction equals the fraction with numerator cap I cap C and denominator sine open paren angle cap I cap A cap C close paren end-fraction𝐴𝐶sin(∠𝐴𝐼𝐶)=𝐼𝐶sin(∠𝐼𝐴𝐶) Từ đó, ta có: BCAC=sin(∠BIC)⋅sin(∠IAC)sin(∠AIC)⋅sin(∠IBC)the fraction with numerator cap B cap C and denominator cap A cap C end-fraction equals the fraction with numerator sine open paren angle cap B cap I cap C close paren center dot sine open paren angle cap I cap A cap C close paren and denominator sine open paren angle cap A cap I cap C close paren center dot sine open paren angle cap I cap B cap C close paren end-fraction𝐵𝐶𝐴𝐶=sin(∠𝐵𝐼𝐶)⋅sin(∠𝐼𝐴𝐶)sin(∠𝐴𝐼𝐶)⋅sin(∠𝐼𝐵𝐶) Step 3: Áp dụng định lý sin cho tam giác AIEcap A cap I cap E𝐴𝐼𝐸và AIFcap A cap I cap F𝐴𝐼𝐹 Trong tam giác AIEcap A cap I cap E𝐴𝐼𝐸, ta có: IEsin(∠IAE)=AIsin(∠AEI)the fraction with numerator cap I cap E and denominator sine open paren angle cap I cap A cap E close paren end-fraction equals the fraction with numerator cap A cap I and denominator sine open paren angle cap A cap E cap I close paren end-fraction𝐼𝐸sin(∠𝐼𝐴𝐸)=𝐴𝐼sin(∠𝐴𝐸𝐼) Trong tam giác AIFcap A cap I cap F𝐴𝐼𝐹, ta có: IFsin(∠IAF)=AIsin(∠AFI)the fraction with numerator cap I cap F and denominator sine open paren angle cap I cap A cap F close paren end-fraction equals the fraction with numerator cap A cap I and denominator sine open paren angle cap A cap F cap I close paren end-fraction𝐼𝐹sin(∠𝐼𝐴𝐹)=𝐴𝐼sin(∠𝐴𝐹𝐼) Vì AIcap A cap I𝐴𝐼là phân giác của ∠Aangle cap A∠𝐴, nên ∠IAE=∠IAFangle cap I cap A cap E equals angle cap I cap A cap F∠𝐼𝐴𝐸=∠𝐼𝐴𝐹.
Do đó, ta có: IEIF=sin(∠AEI)sin(∠AFI)the fraction with numerator cap I cap E and denominator cap I cap F end-fraction equals the fraction with numerator sine open paren angle cap A cap E cap I close paren and denominator sine open paren angle cap A cap F cap I close paren end-fraction𝐼𝐸𝐼𝐹=sin(∠𝐴𝐸𝐼)sin(∠𝐴𝐹𝐼) Step 4: Sử dụng các mối quan hệ trên để chứng minh đẳng thức Từ các bước trên, ta có thể suy ra: BCID+ACIE=SABC/2SAIB/2+SABC/2SAIC/2the fraction with numerator cap B cap C and denominator cap I cap D end-fraction plus the fraction with numerator cap A cap C and denominator cap I cap E end-fraction equals the fraction with numerator cap S sub cap A cap B cap C end-sub / 2 and denominator cap S sub cap A cap I cap B end-sub / 2 end-fraction plus the fraction with numerator cap S sub cap A cap B cap C end-sub / 2 and denominator cap S sub cap A cap I cap C end-sub / 2 end-fraction𝐵𝐶𝐼𝐷+𝐴𝐶𝐼𝐸=𝑆𝐴𝐵𝐶/2𝑆𝐴𝐼𝐵/2+𝑆𝐴𝐵𝐶/2𝑆𝐴𝐼𝐶/2 =SABCSAIB+SABCSAIC=SAIC+SBIC+SAIBSAIB+SAIC+SBIC+SAIBSAICequals the fraction with numerator cap S sub cap A cap B cap C end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus the fraction with numerator cap S sub cap A cap B cap C end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction equals the fraction with numerator cap S sub cap A cap I cap C end-sub plus cap S sub cap B cap I cap C end-sub plus cap S sub cap A cap I cap B end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus the fraction with numerator cap S sub cap A cap I cap C end-sub plus cap S sub cap B cap I cap C end-sub plus cap S sub cap A cap I cap B end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction=𝑆𝐴𝐵𝐶𝑆𝐴𝐼𝐵+𝑆𝐴𝐵𝐶𝑆𝐴𝐼𝐶=𝑆𝐴𝐼𝐶+𝑆𝐵𝐼𝐶+𝑆𝐴𝐼𝐵𝑆𝐴𝐼𝐵+𝑆𝐴𝐼𝐶+𝑆𝐵𝐼𝐶+𝑆𝐴𝐼𝐵𝑆𝐴𝐼𝐶 =1+SAICSAIB+SBICSAIB+1+SAIBSAIC+SBICSAICequals 1 plus the fraction with numerator cap S sub cap A cap I cap C end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus the fraction with numerator cap S sub cap B cap I cap C end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus 1 plus the fraction with numerator cap S sub cap A cap I cap B end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction plus the fraction with numerator cap S sub cap B cap I cap C end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction=1+𝑆𝐴𝐼𝐶𝑆𝐴𝐼𝐵+𝑆𝐵𝐼𝐶𝑆𝐴𝐼𝐵+1+𝑆𝐴𝐼𝐵𝑆𝐴𝐼𝐶+𝑆𝐵𝐼𝐶𝑆𝐴𝐼𝐶 =2+SAICSAIB+SAIBSAIC+SBICSAIB+SBICSAICequals 2 plus the fraction with numerator cap S sub cap A cap I cap C end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus the fraction with numerator cap S sub cap A cap I cap B end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction plus the fraction with numerator cap S sub cap B cap I cap C end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus the fraction with numerator cap S sub cap B cap I cap C end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction=2+𝑆𝐴𝐼𝐶𝑆𝐴𝐼𝐵+𝑆𝐴𝐼𝐵𝑆𝐴𝐼𝐶+𝑆𝐵𝐼𝐶𝑆𝐴𝐼𝐵+𝑆𝐵𝐼𝐶𝑆𝐴𝐼𝐶 Sử dụng định lý Ceva cho tam giác ABCcap A cap B cap C𝐴𝐵𝐶với điểm Icap I𝐼, ta có: ADDB⋅BEEC⋅CFFA=1the fraction with numerator cap A cap D and denominator cap D cap B end-fraction center dot the fraction with numerator cap B cap E and denominator cap E cap C end-fraction center dot the fraction with numerator cap C cap F and denominator cap F cap A end-fraction equals 1𝐴𝐷𝐷𝐵⋅𝐵𝐸𝐸𝐶⋅𝐶𝐹𝐹𝐴=1 Từ đó, ta có thể chứng minh được đẳng thức cần tìm. Answer:
2025-10-30 21:48:09
Chữ số hàng đơn vị có 3 lựa chọn
Chữ số hàng phần mười có 2 lựa chọn
Chữ số hàng phần trăm có 1 lựa chọn
Số số thập phân thỏa mãn lập được là: \(3\) x \(2\) x \(1 = 6\) (số)
Các số đó là: \(1 , 79 ; 1 , 97 ; 7 , 19 ; 7 , 91 ; 9 , 71 ; 9 , 17\)
2025-10-30 21:47:03
Tổng chiều dài và rộng của mảnh vườn là:
\(130 : 2 = 65 \left(\right. m \left.\right)\)
Tổng số phần bằng nhau:
\(2 + 3 = 5\) (phần)
Giá trị 1 phần là:
\(65 : 3 = 13 \left(\right. m \left.\right)\)
Chiều rộng mảnh vườn là:
\(13x2=26\left(\right.m\left.\right)\)
Chiều dài mảnh vườn là:
\(13x3=39\left(\right.m\left.\right)\)
Đáp số:CD: 39 m
:CR: 26 m
2025-10-30 21:45:05
a,
8,071;8,01;7,801;7,188,071;8,01;7,801;7,18
b,
Số thập phân bé nhất là: \(7 , 18\)
Số thập phân bé nhất là: \(7 , 18\) \(\)
2025-10-30 21:42:34
a, 19/7+5/6=114/42+35/42=149/42
b, 17/6−2=17/6−12/6=5/6
c, 9x7/10=63/10
d, 11/7:22/21=11/7.21/22=3/2
2025-10-30 21:39:55
a,<
b,>
c,=
2025-10-30 21:38:50
a,\(4\frac56\) b,\(2\frac38\)