1,  = [(x-1)^2-y^2] : (x-y-1)

= (x+y-1).(x-y-1) : (x-y-1) = x+y-1

2, x^2-x-y^2+y = (x^2-y^2)-(x-y) = (x-y).(x+y) - (x-y) = (x-y).(x+y+1)

k mk nha

bài 2

1,35-x=-16-(-7)

35-x=-9

x=35-(-9)=44

2,/x-2/=9 suy ra x-2=9 hoặc =-9 rồi tự tính

3, suy ra 4-2x=0 hoặc x-3 =0 tụ tính x

4, (x-2)^2=64 = 8^2 suy ra x-2 =8 hoặc -8

Bài 2: Thực hiện phép tính (tính bằng cách hợp lí nếu có thể)

-16 - (35 - x) = -7

-16 - 35 + x = -7

-51 + x = -7

x = -7 - (-51)

x = 44

lx - 2l = 9

=> \(\orbr{\begin{cases}x-2=9\\x-2=-9\end{cases}}\)

=> \(\orbr{\begin{cases}x=9+2\\x=-9+2\end{cases}}\)

=> \(\orbr{\begin{cases}x=11\\x=-7\end{cases}}\)

(4 - 2x) . (x - 3) = 0

=> \(\orbr{\begin{cases}4-2x=0\\x-3=0\end{cases}}\)

=> \(\orbr{\begin{cases}2x=4-0\\x=0+3\end{cases}}\)

=> \(\orbr{\begin{cases}2x=4\\x=3\end{cases}}\)

=> \(\orbr{\begin{cases}x=4\div2\\x=3\end{cases}}\)

=> \(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

(x - 2)² - 64 = 0

=> (x - 2)² = 0 + 64

=> (x - 2)² = 64

=> (x - 2)² = 8²

=> x - 2 = 8

=> x = 8 + 2

=> x = 10

Bài 2:

1: \(A=\left(x+2\right)\left(x^2-2x+4\right)+2\left(x+1\right)\left(1-x\right)\)

\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)

\(=x^3+2^3-2\left(x^2-1\right)\)

\(=x^3+8-2x^2+2=x^3-2x^2+10\)

\(B=\left(2x-y\right)^2-2\left(4x^2-y^2\right)+\left(2x+y\right)^2+4\left(y+2\right)\)

\(=\left(2x-y\right)^2-2\cdot\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2+4\left(y+2\right)\)

\(=\left(2x-y-2x-y\right)^2+4\left(y+2\right)\)

\(=\left(-2y\right)^2+4\left(y+2\right)\)

\(=4y^2+4y+8\)

2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)

3: \(B=4y^2+4y+8\)

\(=4y^2+4y+1+7\)

\(=\left(2y+1\right)^2+7>=7>0\forall y\)

=>B luôn dương với mọi y

Bài 1:

5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)

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

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

6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)

\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)

\(=6x^2+23x-55-6x^2-84x-294\)

=-61x-349

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

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

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

\(=x-y+1\)

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

\(=3\left(2x+3y\right)\left(x-y\right)\)

Bài 4:

a: \(=x-\dfrac{3}{2}+2y\)

b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)

b: \(\dfrac{xy}{2x-y}-\dfrac{2x^2}{y-2x}=\dfrac{xy}{2x-y}+\dfrac{2x^2}{2x-y}=\dfrac{xy+2x^2}{2x-y}\)

b: \(\dfrac{3x^2-x}{x-1}+\dfrac{x+2}{1-x}+\dfrac{3-2x^2}{x-1}\)

\(=\dfrac{3x^2-x-x-2+3-2x^2}{x-1}\)

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

a,\(\dfrac{3-x}{x-5}+\dfrac{2x-8}{x-5}=\dfrac{3-x+2x-8}{x-5}=\dfrac{x-5}{x-5}=1\)

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