\(\left(x^2-5x+8\right)^2-\left(5x-17\right)^2=0\)

\(\Leftrightarrow\left(x^2-5x+8-5x+17\right)\left(x^2-5x+8+5x-17\right)=0\)

\(\Leftrightarrow\left(x^2-10x+25\right)\left(x^2-9\right)=0\)

\(\Leftrightarrow\left(x^2-5x-5x+25\right)\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[x\left(x-5\right)-5\left(x-5\right)\right]\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x-5\right)^2.\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^2=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\\x=-3\end{matrix}\right.\)

C2: (2x - 3)³ + (6x - 17)³

= (2x - 3 + 6x - 17)\(\left[\left(2x-3\right)^2-\left(2x-3\right)\left(6x-17\right)+\left(6x-17\right)^2\right]\)

= (8x - 20)(4x² - 12x + 9 - 12x² + 34x + 18x - 51 + 36x² - 204x + 289)

= (8x - 20)(4x² - 12x² + 36x² - 12x + 34x + 18x - 204x + 9 - 51 + 289)

= (8x - 20)(28x² - 164x + 247)

Câu 1: 

Ta có: \(3x^3-5x-2\)

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

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

5x - 7 = (-14)+(-8)

5x - 7 = -22

5x = (-22)+7

5x = - 15

x = (-15) : 5

x = -3

vây...............

5x-7=(-14)+(-8)

5x-7=-22

5x=-22+7

5x=-15

x=-15/5=-3

Cảm  ơn

1) Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

2) Ta có: \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

3) Ta có: \(\left(2x-1\right)^2-\left(2x+5\right)^2=11\)

\(\Leftrightarrow4x^2-4x-1-4x^2-20x-25=11\)

\(\Leftrightarrow-24x=11+1+25=37\)

hay \(x=-\dfrac{37}{24}\)

5) Ta có: \(3x^2-5x-8=0\)

\(\Leftrightarrow3x^2+3x-8x-8=0\)

\(\Leftrightarrow3x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)

8) Ta có: \(\left|x-5\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

10) Ta có: \(\left|2x+1\right|=\left|x-1\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x-1\\2x+1=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=-1-1\\2x+x=1-1\end{matrix}\right.\)

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

`15x^2-25x+18`

`=5.(3x^2-5x+6)+12`

`=5.2+12`

`=22`

Ta có: 3x\(^2\) + 5x - 8 = 0
=> 3x\(^2\) + 8x - 3x - 8 = 0
=> 3x(x - 1) + 8(x - 1) = 0
=> (3x + 8)(x - 1) = 0
=> 3x + 8 = 0
hoặc x - 1 = 0
=> 3x = -8
hoặc x = 1
=> x = -8/3
hoặc x = 1
Vậy x = -8/3 và x = 1 là nghiệm của 3x\(^2\)2 + 5x - 8

a =3  b =5  c= -8

a+b+c =0

x1 =1 

x2=\(\dfrac{-8}{3}\)

a) \(3x^2-5x+2=0\)

Vì \(a+b+c=3-5+2=0\)

\(\Rightarrow\) pt co 2 ngiệm pb : \(x_1=1\) ; \(x_2=\frac{2}{3}\)

Vậy \(S=\left\{1;\frac{2}{3}\right\}\)

b) \(-3x^2+14x-8=0\)

\(\Delta'=7^2-\left(-3\right)\times\left(-8\right)=49-24=25\)

\(\Rightarrow\) pt có 2 nghiệm pb : \(x_1=4\) ; \(x_2=\frac{2}{3}\)

Vậy \(S=\left\{4;\frac{2}{3}\right\}\)

=> 4x = 64 : 16 = 4

=> x = 4 : 4 = 1

k mk nha

 5x -x = 8^2 : 2^4

5x-x=64-16

5x-x=48

4x=48

x=12

\(5x-x=8^2:2^4\)

\(4x=2^6:2^4\)

\(4x=2^2\)

\(x=4:4\)

\(x=1\)

vậy \(x=1\)