a) \(\Rightarrow x\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)

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

c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

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

d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

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

e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)

vậy giỏi zữ vậy

Bài 2: 

a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

=>-13x=26

hay x=-2

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

hay \(x\in\left\{1;\dfrac{1}{5}\right\}\)

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

hay \(x\in\left\{-5;2\right\}\)

Bài 1.

\(a,x+5x^2=0\\ \Rightarrow a,x\left(1+5x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\\ b,\left(x+3\right)^2+\left(4+x\right)\left(4-x\right)=0\\ \Rightarrow x^2+6x+9+16-x^2=0\\ \Rightarrow6x+25=0\\ \Rightarrow6x=-25\\ \Rightarrow x=-\dfrac{25}{6}\)

\(c,5x\left(x-1\right)=x-1\\ \Rightarrow c,5x\left(x-1\right)-\left(x-1\right)\\ \Rightarrow\left(x-1\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ d,x^2-2x-3=0\\ \Rightarrow\left(x^2-3x\right)+\left(x-3\right)=0\\ \Rightarrow x\left(x-3\right)+\left(x-3\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)

Lời giải:
a. $(x^2-9)(5x+15)=0$

$\Rightarrow x^2-9=0$ hoặc $5x+15=0$
Nếu $x^2-9=0$

$\Rightarrow x^2=9=3^2=(-3)^2$

$\Rightarrow x=3$ hoặc $-3$
Nếu $5x+15=0$

$\Rightarrow x=-3$
b.

$x^2-8x=0$
$\Rightarrow x(x-8)=0$

$\Rightarrow x=0$ hoặc $x-8=0$

$\Rightarrow x=0$ hoặc $x=8$

c. 

$5+12(x-1)^2=53$

$12(x-1)^2=53-5=48$

$(x-1)^2=48:12=4=2^2=(-2)^2$

$\Rightarrow x-1=2$ hoặc $x-2=-2$
$\Rightarrow x=3$ hoặc $x=0$

d.

$(x-5)^2=36=6^2=(-6)^2$
$\Rightarrow x-5=6$ hoặc $x-5=-6$

$\Rightarrow x=11$ hoặc $x=-1$

e.

$(3x-5)^3=64=4^3$

$\Rightarrow 3x-5=4$

$\Rightarrow 3x=9$

$\Rightarrow x=3$

f.

$4^{2x}+2^{4x+3}=144$
$2^{4x}+2^{4x}.8=144$

$2^{4x}(1+8)=144$

$2^{4x}.9=144$

$2^{4x}=144:9=16=2^4$

$\Rightarrow 4x=4\Rightarrow x=1$

a) \(5x\left(x-1\right)=x-1\)

\(\Rightarrow5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right).\left(5x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)

Vậy \(x=1\) hoặc \(x=\frac{1}{5}\)

b) \(2\left(x+5\right)-x^2-5x\)

\(\Rightarrow2\left(x+5\right)-x\left(x+5\right)\)

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

\(\Rightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)

Vậy \(x=-5\)hoặc \(x=2\)

a) 5x(x - 1) = x - 1

    5x(x - 1) - (x - 1) = 0

    (x - 1) (5x - 1) = 0

TH1: x - 1 = 0

        x       = 1

TH2: 5x - 1 = 0

         5x      = 1

          x       = 1/5

Vay x = 1 hoac x = 1/5.

b) 2(x + 5) - x² - 5x = 0

    2(x + 5) - x(x + 5) = 0

    (x + 5) (2 - x) = 0

TH1: x + 5 = 0

        x        = -5

TH2: 2 - x = 0

              x  = 2

Vay x = -5 hoac x = 2

a, x = 1

b , x = -5 hoặc x = 2

a) A = (x - 5)(x² + 5x + 25) - (x - 2)(x + 2) + x(x² + x + 4)

= x³ - 125 - x² + 4 + x³ + x² + 4x

= (x³ + x³) + (-x² + x²) + 4x + (-125 + 4)

= 2x³ + 4x - 121

b) Tại x = -2 ta có:

A = 2.(-2)³ + 4.(-2) - 121

= 2.(-8) - 8 - 121

= -16 - 129

= -145

c) x² - 1 = 0

x² = 1

x = -1; x = 1

*) Tại x = -1 ta có:

A = 2.(-1)³ + 4.(-1) - 121

= 2.(-1) - 4 - 121

= -2 - 125

= -127

*) Tại x = 1 ta có:

A = 2.1³ + 4.1 - 121

= 2.1 + 4 - 121

= 2 - 117

= -115

sửa lại:

a) \(5x\left(x-1\right)=x-1\)

=> \(5x\left(x-1\right)-\left(x-1\right)=0\)

=> \(\left(x-1\right)\left(5x-1\right)=0\)

=> x - 1 = 0   hoặc   5x - 1 = 0

=> x = 1       hoặc    5x = 1  =>  x = 1/5

Vậy x = 1 hoặc x = 1/5

b) \(2\left(5+x\right)-x^2-5x=0\)

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

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

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

=> x + 5 = 0  hoặc   2 - x = 0

=> x = -5   hoặc   x = 2

a, 5x(x-1)=x-1

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

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

<=>5x-1=0 hoặc x-1=0

<=>x=1/5 hoặc x=1

b,2(5+x)-x²-5x=0

<=>2(x+5)-x(x+5)=0

<=>(2-x)(x+5)=0

<=>2-x=0 hoặc x+5=0

<=>x=2 hoặc x=-5

a, 5x(x-1)=x-1

   5x        = (x-1):(x-1)=1

=> x=1/5 

:)