xàm quá

A B C D H K

Cho hình thang cân ABCD như hình vẽ với AH và BK là đường cao. Áp dụng pitago ta có:

\(\hept{\begin{cases}AC^2=AH^2+HC^2\\AD^2=AH^2+HD^2\end{cases}}\)

\(\Rightarrow AC^2-AD^2=HC^2-HD^2=\left(HC+HD\right)\left(HC-HD\right)=DC.AB\)

\(\Rightarrow AC^2=AD^2+AB.DC\)

PS: Bài có mấy dòng tự làm đi chứ nhok

bình phương của bn là tổng 2 bình phương đúng ko ?

nếu vậy thì đề bài là 2 lần tích 2 đáy chứ ????

a, (x-2)^2 - (x-3)(x+3)=6

x^2-4x+4-(x^2-9)=6

x^2-4x+4-x^2+9=6

(x^2-x^2)-4x+13=6

-4x=-7

x=1,75

b, 4(x-3)^2 - (2x-1)(2x+1)=10

4(x^2-6x+9)-(4x^2-1)=10

4x^2-24x+36-4x^2+1=10

-24x+37=10

x=9/8

c,(x-4)^2 - (x+2)(x-2)=6

x^2-8x+16-(x^2-4)=6

x^2-8x+16-x^2+4=6

-8x+20=6

x=7/4

d, 9(x+1)^2 - (3x-2)(3x+2)=10

9(x^2+2x+1)-(9x^2-4)=10

9x^2+18x+9-9x^2+4=10

18x+13=10

x=-1/6

\(a,\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(-4x+13=6\)

\(-4x=6-13\)

\(-4x=-7\)

\(x=\frac{-7}{-4}\)

\(x=\frac{7}{4}\)

Vậy \(x=\frac{7}{4}\)

\(b,4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)

\(4\left(x^2-6x+9\right)-\left(4x^2-1\right)=10\)

\(4x^2-24x+36-4x^2+1=10\)

\(-24x+37=10\)

\(x=\frac{9}{8}\)

Vậy \(x=\frac{9}{8}\)

\(c,\left(x-4\right)^2-\left(x+2\right)\left(x-2\right)=6\)

\(x^2-8x+16-\left(x^2-4\right)=6\)

\(x^2-8x+16-x^2+4=6\)

\(-8x+20=6\)

\(x=\frac{7}{4}\)

Vậy \(x=\frac{7}{4}\)

\(d,9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)

\(9\left(x^2+2x+1\right)-\left(9x^2-4\right)=10\)

\(9x^2+18x+9-9x^2+4=10\)

\(18x+13=10\)

\(x=\frac{-1}{6}\)

Vậy \(x=\frac{-1}{6}\)

dưới mẫu nè: (2+1)(2^2+1)(2*4+1)(2*8+1)(2*16+1)=(2*4-1)(2*4+1)(2*8+1)(2*16+1)(*vì 2+1=2*2-1)

cứ như thế thì được: 2*32-1

Ta có : \(\frac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\frac{\left(2^4\right)^8-1}{\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\frac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\frac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\frac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\frac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)

\(=\frac{2^{32}-1}{2^{32}-1}=1\)

dễ mak

Bài 1 :

1) a² - 4 + y ( a - 2 )

= ( a + 2 ) ( a - 2 ) + y ( a - 2 )

= ( a - 2 ) ( a + 2 + y )

2) ( x - 2 )² - 9y²

= ( x - 2 - 3y ) ( x - 2 + 3y )

Bài 2 :

1) 3 ( x + 4 ) - 2x = 5

=> 3x + 12 - 2x = 5

=> x + 12 = 5

=> x = 5 - 12 = - 7

Vậy x = - 7

2) x ( x - 2 ) - x² - 6 = 0

=> x² - 2x - x² - 6 = 0

=> - 2x - 6 = 0

=> 2x = - 6

=> x = \(-\frac{6}{2}=3\)

Vậy x = 3

3 ) x² - 3x = 0

=> x ( x - 3 ) = 0

=> \(\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Vậy \(x\in\left\{0;3\right\}\)

4) 5 - 3 ( x - 6 ) = 4

=> 5 - 3x + 18 = 4

=> 3x = 5 + 18 - 4

=> 3x = 19

=> x = \(\frac{19}{3}\)

Vậy \(x=\frac{19}{3}\)

a) \(\dfrac{2}{x+3}+\dfrac{1}{x}\) MTC: \(x\left(x+3\right)\)

\(=\dfrac{2x}{x\left(x+3\right)}+\dfrac{x+3}{x\left(x+3\right)}\)

\(=\dfrac{2x+x+3}{x\left(x+3\right)}\)

\(=\dfrac{3x+3}{x\left(x+3\right)}\)

b) \(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}\)

\(=\dfrac{x+1}{2\left(x-1\right)}+\dfrac{-2x}{\left(x-1\right)\left(x+1\right)}\) MTC: \(2\left(x-1\right)\left(x+1\right)\)

\(=\dfrac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}+\dfrac{-2x.2}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x+1\right)^2-4x}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x+1\right)-4x}{2\left(x-1\right)}\)

\(=\dfrac{x+1-4x}{2\left(x-1\right)}\)

\(=\dfrac{1-3x}{2\left(x-1\right)}\)

c) \(\dfrac{y-12}{6y-36}+\dfrac{6}{y^2-6y}\)

\(=\dfrac{y-12}{6\left(y-6\right)}+\dfrac{6}{y\left(y-6\right)}\) MTC: \(6y\left(y-6\right)\)

\(=\dfrac{y\left(y-12\right)}{6y\left(y-6\right)}+\dfrac{6.6}{6y\left(y-6\right)}\)

\(=\dfrac{y\left(y-12\right)+6^2}{6y\left(y-6\right)}\)

\(=\dfrac{y^2-12y+6^2}{6y\left(y-6\right)}\)

\(=\dfrac{\left(y-6\right)^2}{6y\left(y-6\right)}\)

\(=\dfrac{y-6}{6y}\)

Bạn Nguyễn Nam làm sai câu b rồi , làm lại cho tất nè

a) \(\dfrac{2}{x+3}+\dfrac{1}{x}=\dfrac{2x+x+3}{x\left(x+3\right)}=\dfrac{3x+3}{x\left(x+3\right)}\)

b) \(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x+1\right)^2-4x}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+2x+1-4x}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{2\left(x+1\right)}\)

c) \(\dfrac{y-12}{6y-36}+\dfrac{6}{y^2-6y}=\dfrac{y-12}{6\left(y-6\right)}+\dfrac{6}{y\left(y-6\right)}\)

\(=\dfrac{y^2-12y+36}{6y\left(y-6\right)}=\dfrac{\left(y-6\right)^2}{6y\left(y-6\right)}=\dfrac{y-6}{6y}\)

d) \(\dfrac{6x}{x+3}+\dfrac{3}{2x+6}=\dfrac{6x}{x+3}+\dfrac{3}{2\left(x+3\right)}=\dfrac{12x}{2\left(x+3\right)}\)( sửa đề )

mình đội toán thcs vũ quý

tui đội tuyển thcs thanh hà nhưng đề vĩnh tường tui làm nòi giúp đc gì ko

nó không cho mik gửi hình ảnh thế nên các bạn xem hình này nha

ở đây học chứ có phải nơi bán đồ