\(A=\frac{x+y+z}{\left(x+y\right)^2-\left(x+y\right)z}:\frac{x^2+y^2-z^2+2xy}{2x+2y}\)

\(=\frac{x+y+z}{\left(x+y\right)\left(x+y-z\right)}:\frac{\left(x+y\right)^2-z^2}{2\left(x+y\right)}\)

\(=\frac{x+y+z}{\left(x+y\right)\left(x+y-z\right)}:\frac{\left(x+y-z\right)\left(x+y+z\right)}{2\left(x+y\right)}\)

\(=\frac{x+y+z}{\left(x+y\right)\left(x+y-z\right)}.\frac{2\left(x+y\right)}{\left(x+y-z\right)\left(x+y+z\right)}\)

\(=\frac{2}{\left(x+y-z\right)^2}\)

Bổ sung nha

ĐKXĐ : \(x\ne-y;z\ne\pm\left(x+y\right)\)

#Học tốt

g) (x - 3)(x² + 3x + 9) = x³ - 3³ = x³ - 27

h) x² + 2x + 1 = (x  + 1)² 

i) x² - 1= (x - 1)(x + 1)

k) x² + 6x + 9 = (x + 3)²

l) 4x² - 9 = (2x)² - 3² = (2x - 3)(2x + 3) 

m) 16x² - 8x + 1 = (4x - 1)² 

n) 9x² + 6x + 1 = (3x + 1)²

o) 36x² + 36x + 9 = (6x + 3)²

p) x³ + 27 = (x + 3)(x² - 3x + 9) 

Trả lời:

g, ( x - 3 ) ( x² + 3x + 9 ) = x³ - 27

h, x² + 2x + 1 = ( x + 1 )²

i, x² - 1 = ( x - 1 ) ( x + 1 )

k, x² + 6x + 9 = ( x + 3 )²

l, 4x² - 9 = ( 2x - 3 ) ( 2x + 3 )

m, 16x² - 8x + 1 = ( 4x - 1 )²

n, 9x² + 6x + 1 = ( 3x + 1 )²

o, 36x² + 36x + 9 = ( 6x + 3 )²

p, x³ + 27 = ( x + 3 ) ( x² - 3x + 9 )

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2n-1}+\frac{1}{2n}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2n}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2n-1}+\frac{1}{2n}-\left(1+\frac{1}{2}+...+\frac{1}{n}\right)\)

\(=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\)(đpcm) 

ta có : \(A+B+C+D=360^0\text{ mà }\hept{\begin{cases}A=C\\B=D\end{cases}\Rightarrow2A+2B=360^0}\)

hay \(A+B=180^0\) vậy A và B là hai góc bù nhau

do đó AD//BC, chứng minh tương tự ta có AB//CD

vậy tứ giác ABCD là hình bình hành

nên AB=CD và AD=BC