7(2x - 5) - 5(7x - 2) + 2(5x - 7) = (x + 2) - (x + 4)

=> 14x - 35 - 35x + 10 + 10x -14 = x + 2 - x - 4

=> (14x - 35x + 10x) + (-35 + 10 - 14) = -2

=> -11x + (-39) = -2

=> -11x = -2 - (-39)

=> -11x = 37

=> x = \(\frac{-37}{11}\)

7(2x - 5) - 5(7x - 2) + 2(5x - 7)             = (x + 2) - (x + 4)

=> 14x - 35 - 35x + 10 + 10x -14         = x + 2 - x - 4

=> (14x - 35x + 10x) + (-35 + 10 - 14) = -6
=> -11x - 39                                         = -6
=> -11x                                                = -6+39
=> -11x                                                = 33
=>      x                                                = 33:(-11)
=>      x                                                = -3

X = 0 

nha

kb nha

(x+1)²-(x+1)=0

<=> (x+1)² hoặc x+1=0

(x+1)²=0 => x=-1

x+1=0 => x=-1

Vậy x=-1

b) 5x²-13x=0

x(5x-13)=0

<=> x=0 hoặc 5x-13=0

5x-13=0 => 5x=13 => x=13/5

Vậy x=13/5

c) x²-7x³=0

<=> x(x-7x²)=0

=> x=0 hoặc

1) x² - 9x = 0

=> x.(x - 9) = 0

=> \(\orbr{\begin{cases}x=0\\x-9=0\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\x=9\end{cases}}\)

2) x⁴ - 4x² = 0

=> x².(x² - 4) = 0

=> \(\orbr{\begin{cases}x^2=0\\x^2-4=0\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\x^2=4\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\x\in\left\{2;-2\right\}\end{cases}}\)

3) x² - 4x + 3 = 0

=> x² - x - 3x + 3 = 0

=> x.(x - 1) - 3.(x - 1) = 0

=> (x - 1).(x - 3) = 0

=> \(\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)=> \(\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

x = -1

tick cko mk nha mk lảm mãi ms ra đó bao jo có bài khó mk jaai jup

a) Ta có: \(x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Vậy: S={-5;2}

b) Ta có: \(3x^2-7x+1=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)

c) Ta có: \(3x^2-7x+8=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)

Vậy: \(x\in\varnothing\)

ko bt

\(a.x^3+3x^2+4x+2\)

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

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

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

\(b.6x^4-x^3-7x^2+x+1\)

\(=6x^4-6x^3+5x^3-5x^2-2x^2+2x-x+1\)

\(=6x^3\left(x-1\right)+5x^2\left(x-1\right)-2x\left(x-1\right)-\left(x-1\right)\)

\(=\left(x-1\right)\left(6x^3+5x^2-2x-1\right)\)

\(=\left(x-1\right)\left(6x^3+6x^2-x^2-x-x-1\right)\)

\(=\left(x-1\right)\left[6x^2\left(x+1\right)-x\left(x+1\right)-\left(x+1\right)\right]\)

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

\(=\left(x-1\right)\left(x+1\right)\left(6x^2-3x+2x-1\right)\)

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

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

k giùm cái cho đỡ buồn!

bài 1 

a) 

x : 13 = 41

x = 41 x 13

x = 533

b) 1428 : x = 14

x = 1428 : 14

x = 102

c) 4x : 17 = 0

4x = 0 x 17

4x = 0

x = 0 : 4

x = 0

d) 7x - 8 = 713

7x = 713+ 8

7x = 721

x = 721 : 7

x = 103

e) 8 [ x - 3 ] = 0

x - 3 = 0 : 8

x - 3 = 0

x = 0 + 3

x = 3

x =