a) ( x - 3 )² - 4 = 0

<=> ( x - 3 )² = 4

<=> \(\orbr{\begin{cases}\left(x-3\right)^2=2^2\\\left(x-3\right)^2=\left(-2\right)\end{cases}}\)

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

<=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

Vậy S = { 5 ; 1 }

b) x² - 9 = 0

<=> x² = 9

<=> \(\orbr{\begin{cases}x^2=3^2\\x^2=\left(-3\right)^2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

Vậy S = { 3 ; -3 }

c) x( x - 2x ) - x² - 8 = 0

<=> x² - 2x² - x² - 8 = 0

<=> -2x² - 8 = 0

<=> -2x² = 8

<=> x² = -4 ( vô lí )

<=> x = \(\varnothing\)

Vậy S = { \(\varnothing\)}

d) 2x( x - 1 ) - 2x² + x - 5 = 0

<=> 2x² - 2x - 2x² + x - 5 = 0

<=> -x - 5 = 0

<=> -x = 5

<=> x = -5

Vậy S = { -5 }

e) x( x - 3 ) - ( x + 1 )( x - 2 ) = 0 

<=> x² - 3x - ( x² - x - 2 ) = 0

<=> x² - 3x - x² + x + 2 = 0

<=> - 2x + 2 = 0

<=> -2x = -2

<=> x = 1

Vậy S = { 1 }

f) x( 3x - 1 ) - 3x² - 7x = 0

<=> 3x² - x - 3x² - 7x = 0

<=> -8x = 0

<=> x = 0

Vậy S = { 0 } 

Câu a:

(x + 2)(x -3) - (x -2)(x + 5) = 0

x^2 + 2x - 3x - 6 - x^2 - 5x + 2x + 10 = 0

(x^2 - x^2) + (2x - 3x - 5x + 2x) + (10 - 6) = 0

0 + (-x - 5x + 2x) + 4 = 0

-6x + 2x + 4 = 0

-4x + 4 = 0

4x = 4

x = 4 : 4

x = 1

Vậy x = 1

Câu b:

(2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)

2x^2 - 8x + 3x - 12+ x^2 - 2x - 5x + 10 = 3x^2 - 12x - 5x + 20

(2x^2 + x^2) - (8x+5x-3x +2x) - (12 - 10) = 3x^2 - (12x + 5x) + 20

3x^2 - (13x - 3x + 2x) - 2 = 3x^2 - 17x + 20

3x^2 - (10x + 2x) - 2 = 3x^2 - 17x + 20

3x^2 - 12x - 2 = 3x^2 - 17x + 20

3x^2 - 12x - 2 - 3x^2 + 17x - 20 = 0

(3x^2 - 3x^2) + (-12x + 17x) - (2 + 20) = 0

0 + 5x - 22 = 0

5x = 22

x = 22/5

Vậy x = 22/5

a) \(x\left(5+3x\right)-\left(x+1\right)\left(3x-2\right)=12\)

\(5x+3x^2-3x^2+2x-3x+2=12\)

\(4x=10\)

\(x=\frac{5}{2}\)

vậy \(x=\frac{5}{2}\)

\(13x\left(x-8\right)-x+8=0\)

\(13x\left(x-8\right)-\left(x-8\right)=0\)

\(\left(13x-1\right)\left(x-8\right)=0\)

\(\Rightarrow\orbr{\begin{cases}13x-1=0\\x-8=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{13}\\x=8\end{cases}}\)

 vậy \(\orbr{\begin{cases}x=\frac{1}{13}\\x=8\end{cases}}\)

a. 3(2x - 1)(3x - 1) - (2x - 3)(9x - 1) = 0

<=> 3(6x²-5x+1)-(18x²-29x+3)=0

<=> 14x=0

<=> x=0

b. (x - 3)(x - 5) + 3 (x - 1) = (x - 1)(x - 3)

<=> (x-3)(x-5-x+1)+3(x-1)=0

<=> -4(x-3)+3(x-1)=0

<=> -x+9=0

<=> x=9

c. (x - 1)(x - 2) - (x + 2)(x + 1) = 8

<=> x²-3x+2-(x²+3x+2)=8

<=> -6x=8

<=> \(x=\frac{-4}{3}\)

a/ Ta có : \(49.x^2-4=0\)

\(\Rightarrow49x^2=4\)

\(\Rightarrow x^2=\frac{4}{49}\Rightarrow\orbr{\begin{cases}x=\frac{-2}{7}\\x=\frac{2}{7}\end{cases}}\)

b/ \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=11\)

\(\left(x+3\right)\left(x+3\right)-\left(x+2\right)\left(x-2\right)=11\)

\(\Rightarrow\left(x^2+2.3.x+3^2\right)-\left(x^2-2^2\right)=11\)

\(\Rightarrow x^2+6x+9-x^2+4=11\)

\(\Rightarrow6x+13=11\)

\(\Rightarrow6x=11-13\)

\(\Rightarrow x=\frac{-2}{6}=\frac{-1}{3}\)

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

\(\Rightarrow\left(2x+1\right)\left(2x+1\right)-\left(x-3\right)\left(x-3\right)-3\left[\left(x+5\right)\left(x-5\right)\right]=5\)

\(\Rightarrow\left(4x^2+2.2x+1\right)-\left(x^2-2.3x+9\right)-3\left(x^2-25\right)\)\(=5\)

\(\Rightarrow\left(4x^2+4x+1\right)-\left(x^2-6x+9\right)-\left(3x^2-75\right)=5\)

\(\Rightarrow4x^2+4x+1-x^2+6x-9-3x^2+75=5\)

\(\Rightarrow\left(4x^2-x^2-3x^2\right)+\left(4x+6x\right)+\left(1-9+75\right)=5\)

\(\Rightarrow10x+67=5\)

\(\Rightarrow10x=5-67=-62\)

\(\Rightarrow x=\frac{-62}{10}=\frac{-31}{5}\)

d/ \(\left(3x+1\right)\left(3x-1\right)=8\)

\(\Rightarrow9x^2-1=8\)

\(\Rightarrow9x^2=8+1=9\)

\(\Rightarrow x^2=\frac{9}{9}=1\Leftrightarrow\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)

Ai đó bấm hộ mình cái nút đúng đi!

Ta có : 49x² - 4 = 0

=> 49x² = 4

=> x² = 196

=> x² = 14² ; (-14)²

=> x = 14 ; -14

1)

\(\left(x-3\right)^2+\left(x+5\right)\left(2-x\right)=-\left(9x-19\right)\)

\(-\left(9x-19\right)=0\)

\(9x=-19\)

\(\Rightarrow x=\frac{19}{9}=2\frac{1}{9}\)

\(x_1+x_2=-\frac{b}{a}=-2\)

\(x_1.x_2=\frac{c}{a}=-8\)

\(x\in-4;2\)