Ta có : x² + 8x - 7 = 0

=> x² + 8x + 16 - 9 = 0

=> (x + 4)² - 9 = 0

=>  (x + 4)² = 9

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

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-7\end{cases}}\)

a) (3-x)²-(5+x)²=0 => x=-1

b) (x-3)²-x(x+2)=7 => x=0,25

c) x²+8x-7=0   (ko biết làm)

d) x²+4x-5=0  => x=1

thank you 2 ban nhieu !!!

Ta có 

x*(x-3)+5*(x-3)=0

=>(x-3)(x+5)=0

=> x-3=0 hoặc x+5=0

=> x=3 hoặc x=-5

Ta có

(x+3)*(x55) là tương tự trên

Ta có

7*(x-3)-4(x-3)=0

=>(7-4)(x-3)=0

=>x=3

KL

Vì (3.x-5)-(2.x-7)=0 nên 3.x-5=2.x-7( vì 2 số bằng nhau trừ cho nhau bằng 0)

Có 3.x-5=2.x-7 ( bên dưới mình áp dụng quy tắc chuyển vế nhé)

3.x-2.x=5-7

x=-2

Tick cho mình nhé. Chắc chắn đúng

a) Ta có: \(\left(x-3\right)\left(x+4\right)>0\)

Nếu: \(\hept{\begin{cases}x-3>0\\x+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>3\\x>-4\end{cases}}\Rightarrow x>3\)

Nếu: \(\hept{\begin{cases}x-3< 0\\x+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 3\\x< -4\end{cases}}\Rightarrow x< -4\)

Vậy \(\orbr{\begin{cases}x>3\\x< -4\end{cases}}\)

b) Ta có: \(\left|\frac{5}{7}x-4\right|< \frac{2}{7}\)

\(\Leftrightarrow-\frac{2}{7}< \frac{5}{7}x-4< \frac{2}{7}\)

\(\Leftrightarrow\frac{26}{7}< \frac{5}{4}x< \frac{30}{7}\)

\(\Leftrightarrow\frac{104}{35}< x< \frac{24}{7}\)

a) x(x-8)-x(x+1)=2

    x² -8x -x²-x=2

    -9x=2

    \(x=-\frac{2}{9}\)

b) (x+3)5 - 7(x+9)=0

 5x + 15 -7x -63=0

-2x - 48 =0

-2x=48

x=-24

c)4(x-7)+7(x-2)=11

 4x -28 + 7x -14=11

11x -42=11

11x=11+42

11x=53

x=\(\frac{53}{11}\)

a) \(x\left(x-5\right)-4x+20=0\)

\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-4=0\\x-5=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=5\end{array}\right.\)

b) \(x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+6=0\\x-7=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-6\\x=7\end{array}\right.\)

d) \(x^2-9x+8=0\)

\(\Leftrightarrow x^2-x-8x+8=0\)

\(\Leftrightarrow x\left(x-1\right)-8\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-8=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=8\end{array}\right.\)

g) \(3x^2-5x+2=0\)

\(\Leftrightarrow3x^2-3x-2x+2=0\)

\(\Leftrightarrow3x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=\frac{2}{3}\end{array}\right.\)

x + y = x.y

=> xy - x - y = 0

=> (xy - x) - y + 1 = 1

=> x(y - 1) - (y - 1) = 1

=> (x - 1)(y - 1) = 1

=> x - 1 = y - 1 = 1 hoặc x - 1 = y - 1 = -1

=> x = y = 2 hoặc x = y = 0