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Ta có :
\(0,0\left(8\right)=\dfrac{1}{10}.0,\left(8\right)=\dfrac{1}{10}.0,\left(1\right).8=\dfrac{1}{10}.\dfrac{1}{9}.8=\dfrac{4}{45}\)
\(0,1\left(2\right)=0,1+0,0\left(2\right)\)
\(=\dfrac{1}{10}+\dfrac{1}{10}.0,\left(2\right)=\dfrac{1}{10}+\dfrac{1}{10}.0,\left(1\right).2\)
\(=\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{9}.2=\dfrac{9}{90}+\dfrac{2}{90}=\dfrac{11}{90}\)
\(0,1\left(23\right)=0,1+0,0\left(23\right)=\dfrac{1}{10}+\dfrac{1}{10}.0,23\)
\(=\dfrac{1}{10}+\dfrac{1}{10}.0,\left(01\right).23\)
\(\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{99}.23=\dfrac{99}{990}+\dfrac{23}{990}=\dfrac{122}{990}=\dfrac{61}{495}\)
a) \(9\cdot3^3\cdot\frac{1}{81}\cdot3^2\)
\(=\frac{3^2\cdot3^3\cdot3^2}{3^4}\)
\(=3^3=27\)
b) \(4\cdot2^5:\left(2^3\cdot\frac{1}{16}\right)\)
\(=\frac{2^2\cdot2^2\cdot2^4}{2^3}\)
\(=2^5=32\)
c) \(3^2\cdot2^5\cdot\left(\frac{2}{3}\right)^2\)
\(=\frac{3^2\cdot2^5\cdot2^4}{3^2}\)
\(=2^9=512\)
d) \(\left(\frac{1}{3}\right)^2\cdot\frac{1}{3}\cdot9^2\)
\(=\frac{1^2\cdot1\cdot3^4}{3^2}\)
\(=3^2=9\)
3,
a) (−23+37):45+(−13+47):45
= \(-\frac{5}{21}:\frac{4}{5}+\frac{5}{21}:\frac{4}{5}\)
= \(\left(-\frac{5}{21}+\frac{5}{21}\right):\frac{4}{5}\)
= \(0:\frac{4}{5}=0\)
2,
a) \(\frac{-3}{4}\).\(\frac{12}{-5}\).(\(\frac{-25}{6}\))
= \(\frac{-3.4.3.\left(-5\right).5}{4.\left(-5\right).3.3}\)
= \(-5\)
b) (−2).\(\frac{-38}{21}\).\(\frac{-7}{4}\).(\(\frac{-3}{8}\))
= \(\frac{-2.\left(-38\right)\left(-7\right)\left(-3\right)}{\left(-7\right)\left(-3\right)\left(-2\right)\left(-2\right).8}\)
= \(\frac{19}{8}\)
c) (\(\frac{11}{12}:\frac{33}{16}\)).\(\frac{3}{5}\)
= \(\left(\frac{11}{12}.\frac{16}{33}\right).\frac{3}{5}\)
= \(\frac{4}{9}.\frac{3}{5}\)
= \(\frac{4}{15}\)
d) \(\frac{7}{23}\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left(\frac{-41}{10}\right)\)
= \(\frac{-287}{203}\)
3. Tính:
a) (\(\frac{-2}{3}+\frac{3}{7}\)):\(\frac{4}{5}\)+(\(\frac{-1}{3}+\frac{4}{7}\)):\(\frac{4}{5}\)
= (\(\frac{-2}{3}+\frac{3}{7}\)\(+\)\(\frac{-1}{3}+\frac{4}{7}\)) : \(\frac{4}{5}\)
= 0 : \(\frac{4}{5}\)
= 0
b) \(\frac{5}{9}\):(\(\frac{1}{11}-\frac{5}{22}\))+\(\frac{5}{9}\):(\(\frac{1}{15}-\frac{2}{3}\))
= \(\frac{5}{9}\): \(\frac{-3}{22}\)+ \(\frac{5}{9}\): \(\frac{-3}{5}\)
= \(\frac{5}{9}\): \(\frac{-81}{110}\)
= \(\frac{-550}{729}\)
a) \(25^3:5^2=5^6:5^2=5^4=625\)
b) \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
c) \(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3-1+\frac{1}{8}=\frac{17}{8}\)
a) \(3^2.2^5.\left(\frac{2}{3}\right)^2=2^5.\frac{2^2.3^2}{3^2}=2^5.2^2=2^7\)
b) \(\left(\frac{1}{3}\right)^2.\frac{1}{3}.9^2=\frac{1^2.1}{3^2.3}.\left(3^2\right)^2=\frac{1.3^{2.2}}{3^3}=\frac{3^4}{3^3}=3^1\)
AI HỎI ??
AI HỎI ??
bọn 2007
a)5+0,(9)
b)-2+0,(35)
c)0+0,(21)
d)-6+0,32(49)
Ta cùng làm từng phần nhé 👇
a) \(5 , \left(\right. 9 \left.\right)\)
Ta đặt \(x = 5 , \left(\right. 9 \left.\right)\)
Khi đó:
\(x = 5 , 9999...\)
Ta biết \(0 , 9999... = 1\)
\(\Rightarrow x = 5 + 1 = 6\)
✅ Kết quả:
\(5 , \left(\right. 9 \left.\right) = 6 = \frac{6}{1}\)
b) \(- 2 , \left(\right. 35 \left.\right)\)
Đặt \(x = - 2 , \left(\right. 35 \left.\right) = - 2 , 353535...\)
Xét phần dương \(2 , \left(\right. 35 \left.\right)\):
Gọi \(y = 2 , \left(\right. 35 \left.\right)\)
Hai chữ số lặp ⇒ nhân cả hai vế với 100:
\(100 y = 235 , \left(\right. 35 \left.\right)\)
Trừ vế theo vế:
\(100 y - y = 235 , \left(\right. 35 \left.\right) - 2 , \left(\right. 35 \left.\right)\)\(99 y = 233\)\(y = \frac{233}{99}\)
Vì có dấu “–” ở đầu nên:
\(x = - \frac{233}{99}\)
✅ Kết quả:
\(- 2 , \left(\right. 35 \left.\right) = - \frac{233}{99}\)
c) \(0 , 1 \left(\right. 21 \left.\right)\)
Gọi \(x = 0 , 1 \left(\right. 21 \left.\right) = 0 , 1212121...\)
Phần thập phân gồm 1 chữ không lặp (1) và 2 chữ lặp (21).
Công thức tổng quát:
\(x = \frac{\text{Ph} \overset{ˋ}{\hat{\text{a}}} \text{n}\&\text{nbsp};\text{kh} \hat{\text{o}} \text{ng}\&\text{nbsp};\text{l}ặ\text{p}\&\text{nbsp};\text{v} \overset{ˋ}{\text{a}} \&\text{nbsp};\text{l}ặ\text{p} - \text{Ph} \overset{ˋ}{\hat{\text{a}}} \text{n}\&\text{nbsp};\text{kh} \hat{\text{o}} \text{ng}\&\text{nbsp};\text{l}ặ\text{p}}{9 \ldots 90 \ldots 0}\)
(với số 9 bằng số chữ lặp, số 0 bằng số chữ không lặp)
Ta có:
\(x = \frac{121 - 1}{990} = \frac{120}{990} = \frac{4}{33}\)
✅ Kết quả:
\(0 , 1 \left(\right. 21 \left.\right) = \frac{4}{33}\)
d) \(- 6 , 32 \left(\right. 49 \left.\right)\)
Xét phần dương \(6 , 32 \left(\right. 49 \left.\right)\):
Gọi \(y = 6 , 32 \left(\right. 49 \left.\right)\)
Phần thập phân: 2 chữ không lặp (32) và 2 chữ lặp (49)
Công thức tương tự:
\(y = \frac{\text{S} \overset{ˊ}{\hat{\text{o}}} \&\text{nbsp};\text{g} \overset{ˋ}{\hat{\text{o}}} \text{m}\&\text{nbsp};\text{ph} \overset{ˋ}{\hat{\text{a}}} \text{n}\&\text{nbsp};\text{kh} \hat{\text{o}} \text{ng}\&\text{nbsp};\text{l}ặ\text{p}\&\text{nbsp};\text{v} \overset{ˋ}{\text{a}} \&\text{nbsp};\text{l}ặ\text{p} - \text{ph} \overset{ˋ}{\hat{\text{a}}} \text{n}\&\text{nbsp};\text{kh} \hat{\text{o}} \text{ng}\&\text{nbsp};\text{l}ặ\text{p}}{99 \ldots 90 \ldots 0}\)
Cụ thể:
\(y = \frac{63249 - 632}{9900} = \frac{62617}{9900}\)
Vậy:
\(x = - \frac{62617}{9900}\)
✅ Kết quả:
\(- 6 , 32 \left(\right. 49 \left.\right) = - \frac{62617}{9900}\)
🔹Tóm tắt kết quả:
Số thập phân tuần hoàn
Dạng phân số
a) 5,(9)
\(\frac{6}{1}\)6116
b) -2,(35)
\(- \frac{233}{99}\)−23399−99233
c) 0,1(21)
\(\frac{4}{33}\)433334
d) -6,32(49)
\(- \frac{62617}{9900}\)−626179900−990062617
a: \(5,\left(9\right)=5+\frac99=5+1=6\)
b: \(-2,\left(35\right)=-2-\frac{35}{99}=\frac{-198-35}{99}=\frac{-233}{99}\)
c: \(0,1\left(21\right)=0,1+0,0\left(21\right)=\frac{1}{10}+\frac{21}{990}=\frac{99}{990}+\frac{21}{990}=\frac{120}{990}=\frac{4}{33}\)
d: \(-6,32\left(49\right)=-6,32-0,00\left(49\right)\)
\(=-\frac{632}{100}-\frac{49}{9900}=\frac{-62568-49}{9900}=\frac{-62617}{9900}\)
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