1) ĐKXĐ: \(x\notin\left\{-2;2\right\}\)

Ta có: \(\dfrac{x-1}{x+2}-\dfrac{9}{x^2-4}=\dfrac{-3}{x-2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{9}{\left(x-2\right)\left(x+2\right)}=\dfrac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2-3x+2-9=-3x-6\)

\(\Leftrightarrow x^2-3x-7+3x+6=0\)

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

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

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

Vậy: S={1;-1}

2)

Sửa đề: \(\dfrac{3x-3}{x^2-9}-\dfrac{1}{x-3}=\dfrac{x+1}{x+3}\)

ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

Ta có: \(\dfrac{3x-3}{x^2-9}-\dfrac{1}{x-3}=\dfrac{x+1}{x+3}\)

\(\Leftrightarrow\dfrac{3x-3}{\left(x-3\right)\left(x+3\right)}-\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

Suy ra: \(3x-3-x-3=x^2-3x+x-3\)

\(\Leftrightarrow x^2-2x-3=2x-6\)

\(\Leftrightarrow x^2-2x-3-2x+6=0\)

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

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

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

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

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

Vậy: S={1}

`1)(x-1)/(x+2)-9/(x^2-4)=-3/(x-2)(x ne 2)`

`<=>x^2-3x+2-9=-3x-6`

`<=>x^2-1=0`

`<=>x=+-1`

a) (x-2)³+6(x+1)²-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6x²+12x+6-x³+12=0

\(\Rightarrow\)24x+10=0

\(\Rightarrow\)24x=-10

\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)

b)(x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x-4)(x+4)+3x²

\(\Rightarrow\)x²-25-(x²+6x+9)+3(x²-4x+4)=x²+2x+1-(x²-16)+3x²

\(\Rightarrow\)x²​-25-x²-6x-9+3x²-12x+12=x²+2x+1-x²+16+3x²

\(\Rightarrow\)3x²-18x-22=3x²+2x+17

\(\Rightarrow\)3x²-18x-22-3x²-2x-17=0

\(\Rightarrow\)-20x-39=0

\(\Rightarrow\)-20x=39

\(\Rightarrow\)x=\(-\dfrac{39}{20}\)

c) (2x+3)² +(x-1)(x+1)=5(x+2)²-(x-5)(x+1)+(x+4)

⇒4x²+12x+9+x²-1=5(x²+4x+4)-(x²+x-5x-5)+x+4

⇒5x²+12x+8=5x²+20x+20-x²-x+5x+5+x+4

⇒5x²+12x+8-5x²-20x-20+x²+x-5x-5-x-4=0

⇒x²-13x-21=0

\(b,P=\left[\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-1\right]:\dfrac{9-x^2+\left(x-3\right)\left(x+3\right)-\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}\left(x\ne\pm3;x\ne2\right)\\ P=\left(\dfrac{x}{x+3}-1\right)\cdot\dfrac{\left(x-2\right)\left(x+3\right)}{9-x^2+x^2-9-\left(x-2\right)^2}\\ P=\dfrac{x-x-3}{x+3}\cdot\dfrac{\left(x-2\right)\left(x+3\right)}{-\left(x-2\right)^2}\\ P=\dfrac{-3}{-\left(x-2\right)}=\dfrac{3}{x-2}\)

Với \(x^3-4x=0\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\left(ktm\right)\\x=-2\end{matrix}\right.\)

Với \(x=0\Leftrightarrow P=\dfrac{3}{0-2}=-\dfrac{3}{2}\)

Với \(x=-2\Leftrightarrow P=\dfrac{3}{-2-2}=-\dfrac{3}{4}\)

a) x = 1; x = - 1 3                 b) x = 2.

c) x = 3; x = -2.                 d) x = -3; x = 0; x = 2.

a) ${}^{}-25=0$

${}^{}=0+25=25$

${}^{}={}^{}={}^{}$

$\Rightarrow [\begin{array}{c}2x-1=5\\ 2x-1=-5\end{array}\Rightarrow [\begin{array}{c}2x=6\\ 2x=-4\end{array}\Rightarrow [\begin{array}{c}x=3\\ x=-2\end{array}$

b) $8{}^{}-50x=0$

$2x(4{}^{}$

câu b thiếu bn ơi

ok

a) ${}^{}-25=0$

${}^{}=0+25=25$

${}^{}={}^{}={}^{}$

$\Rightarrow [\begin{array}{c}2x-1=5\\ 2x-1=-5\end{array}\Rightarrow [\begin{array}{c}2x=6\\ 2x=-4\end{array}\Rightarrow [\begin{array}{c}x=3\\ x=-2\end{array}$

b) $8{}^{}-50x=0$

$2x(4{}^{}$

a: Ta có: \(\left(2x-1\right)^2-25=0\)

\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)

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

Bài 1: 

c: x=4

b: x=2

a) (x - 140) : 7 = 3³ - 2³ . 3

(x - 140) : 7 = 27 - 8 . 3 = 27 - 24 = 3

x - 140 = 3 x 7 = 21

x = 21 + 140 = 161

b) x³ . x² = 2⁸ : 2³

x⁵ = 2⁵

=> x = 2

c) (x + 2) . ( x - 4) = 0

x = -2 hoặc 4

d) 3^x-3 - 3² = 2 . 3² =

3^x-3 - 9 = 2 . 9 = 18

3^x-3 = 18 + 9 = 27

3^x-3 = 3³

=> x - 3 = 3

x = 3 + 3 = 6

2.

a) 9 : ( x + 2 )

9 ⋮ 1 ; 9 ⋮ 3 ; 9 ⋮ 9

=> x = -1 ; 1 ; 7

1)    x^2-x-(3x-3)=0

⇔   X^2-x-3x+3=0

⇔  x^2-4x+3     =0

⇔x^2-3x-x+3    =0

⇔ x(x-3)-(x-3)   =0

⇔(x-1)(x-3)       =0

⇔  x-1=0       -> x=1

      x-3=0       ->  x=3

Vậy tập nghiệm S={ 1;3}