`(3x^2+10x-8)=(5x^2-2x+10)^2`

`<=>(3x^2+10x-8+5x^2-2x+10)(3x^2+10x-8-5x^2+2x-10)=0`

`<=> (8x^2+8x+2)(-2x^2+12x-18)=0`

\(\Leftrightarrow\left[{}\begin{matrix}8x^2+8x+2=0\\-2x^2+12x-18=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

Vậy `S={-1/2 ; 3}`.

\(\left(5x^2-2x+19\right)^2=\left(3x^2+10x-8\right)^2\)

\(\Leftrightarrow\left(5x^2-2x+19\right)^2-\left(3x^2+10x-8\right)^2=0\)

\(\Leftrightarrow\left(5x^2-2x+19-3x^2-10x+8\right)\left(5x^2-2x+19+3x^2+10x-8\right)=0\)

\(\Leftrightarrow\left(2x^2-12x+27\right)\left(8x^2+8x+11\right)=0\)

....

\(\left(5x^2-2x+10\right)^2=\left(3x^2+10x-8\right)^2\)

\(\Leftrightarrow\left(5x^2-2x+10\right)^2-\left(3x^2+10x-8\right)^2=0\)

\(\Leftrightarrow\left(5x^2-2x+10-3x^2-10x+8\right).\left(5x^2-2x+10+3x^2+10x-8\right)=0\)

\(\Leftrightarrow\left(2x^2-12x+18\right).\left(8x^2+8x+2\right)=0\)

\(\Leftrightarrow2.\left(x^2-6x+9\right).2.\left(4x^2+4x+1\right)=0\)

\(\Leftrightarrow2.\left(x-3\right)^2.2.\left(2x+1\right)^2=0\)

\(\Leftrightarrow4.\left(x-3\right)^2.\left(2x+1\right)^2=0\)

Vì \(4\ne0.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left(2x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\frac{1}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{3;-\frac{1}{2}\right\}.\)

Chúc bạn học tốt!

\(\Leftrightarrow\left(5x^2-2x+10-3x^2-10x+8\right)\left(5x^2-2x+10+3x^2+10x-8\right)=0\)

\(\Leftrightarrow\left(2x^2-12x+18\right)\left(8x^2+8x+2\right)=0\)

\(\Leftrightarrow\left(2x+1\right)^2=0\)

hay x=-1/2

\(PT\Leftrightarrow\left(3x^2+10x-8\right)^2-\left(5x^2-2x+10\right)^2=0\)

\(\Leftrightarrow\left(3x^2+10x-8-5x^2+2x-10\right)\left(3x^2+10x-8+5x^2-2x+10\right)=0\)

\(\Leftrightarrow\left(-2x^2+12x-18\right)\left(8x^2+8x+2\right)=0\)

\(\Leftrightarrow-4\left(x-3\right)^2\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left(2x+1\right)^2=0\end{matrix}\right.\)

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

Vậy: \(S=\left\{3;-\dfrac{1}{2}\right\}\)

1)6x-8=3x+1

6x-3x=1+8

3x=9

x=3

Vậy x=3

2: 12-10x=25-30x

=>20x=13

=>x=13/20

3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)

=>6x+9-8x+10=10x+21

=>10x+21=-2x+19

=>12x=-2

=>x=-1/6

4: \(\Leftrightarrow25x-15-6x+12=11-5x\)

=>19x-3=11-5x

=>24x=14

=>x=7/12

5: \(\Leftrightarrow8-12x-5+10x=4-6x\)

=>4-6x=-2x+3

=>-4x=-1

=>x=1/4

6: \(\Leftrightarrow32x-24-6+9x=13-40x\)

=>41x-30=13-40x

=>81x=43

=>x=43/81

7: \(\Leftrightarrow10x-5+20x=5x-11\)

=>30x-5=5x-11

=>25x=-6

=>x=-6/25

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

TH1: 5x² - 2x + 10 = 3x² + 10x - 8

      => 2x² - 12x + 18 = 0

      => x² - 6x + 9 = 0

      => (x - 3)² = 0

      => x = 3

TH2: 5x² - 2x + 10 = - 3x² - 10x + 8

      => 8x² + 8x + 2 = 0

      => 4x² + 4x + 1 = 0

      => (2x + 1)² = 0 

      => x = -1/2

Vậy x = 3 , x = -1/2

\(\left(5x^2-2x+10\right)^2=\left(3x^2+10x-8\right)^2\)

\(\Leftrightarrow\left(5x^2-2x+10\right)^2-\left(3x^2+10x-8\right)^2=0\)

\(\Leftrightarrow\left(5x^2-2x+10+3x^2+10x-8\right)\left(5x^2-2x+10-3x^2-10x+8\right)=0\)

\(\Leftrightarrow\left(8x^2+8x+2\right)\left(2x^2-12x+18\right)=0\)

\(\Leftrightarrow2\left(4x^2+4x+1\right).2\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow4\left(2x+1\right)^2\left(x-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\)

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

Vậy ..........

a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)

c: Ta có: \(\left(x-3\right)^2-4=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

b. 

PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$

$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$

$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$

$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$

$\Leftrightarrow (x-3)^2(2x+1)^2=0$

$\Leftrightarrow (x-3)(2x+1)=0$

$\Leftrightarrow x-3=0$ hoặc $2x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$

d.

$x^2-2x=24$

$\Leftrightarrow x^2-2x-24=0$

$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$

$\Leftrightarrow x=-4$ hoặc $x=6$

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

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

\(\Leftrightarrow\left(x-7\right)\left(3x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

Bài 2:

\(3x^2+10x-8=5x^2-2x+10\)

\(\Leftrightarrow 2x^2-12x+18=0\)

\(\Leftrightarrow x^2-6x+9=0\)

\(\Leftrightarrow (x-3)^2=0\Rightarrow x=3\)

Bài 3:

\((x^2-2x+1)-4=0\)

\(\Leftrightarrow (x-1)^2-2^2=0\)

\(\Leftrightarrow (x-1-2)(x-1+2)=0\Leftrightarrow (x-3)(x+1)=0\)

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