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

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

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 \(x\in\left\{-5;2\right\}\)

\(a,\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\left(x-1\right)\left(5x+3-3x+8\right)=0\)

\(\left(x-1\right)\left(2x+11\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)

\(b,3x\left(25x+15\right)-35\left(5x+3\right)=0\)

\(15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\left(5x+3\right).5\left(3x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)

\(c,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\left(3x-2\right)\left(2-5x\right)+\left(3x-2\right)\left(x+11\right)=0\)

\(\left(3x-2\right)\left(2-5x+x+11\right)=0\)

\(\left(3x-2\right)\left(13-4x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\13-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=2\\4x=13\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{13}{4}\end{cases}}}\)

còn đâu tự lm lười :_# 

a) x(x-1) - (x+1)(x+2) = 0

    x\(^2\)- x -x\(^{^2}\)-2x +x+2=0

     -2x+2=0

      -2x=0+2

       -2x=2

         x=-1

Vậy x bằng -1

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

a) \(x^3=x^5\)

=> \(x^3-x^5=0\)

=> \(x^3\left(1-x^2\right)=0\)

=> \(\orbr{\begin{cases}x^3=0\\1-x^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

b) \(4x\left(x+1\right)=x+1\)

=> \(4x^2+4x-x-1=0\)

=> \(4x\left(x+1\right)-1\left(x+1\right)=0\)

=> \(\left(x+1\right)\left(4x-1\right)=0\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{4}\end{cases}}\)

c) \(x\left(x-1\right)-2\left(1-x\right)=0\)

=> \(x\left(x-1\right)-\left[-2\left(x+1\right)\right]=0\)

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

=> \(\left(x-1\right)\left(x+2\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

d) Kết quả ?

e) \(\left(x-3\right)^2+3-x=0\)

=> \(x^2-6x+9+3-x=0\)

=> \(x^2-7x+12=0\)

=> \(x^2-3x-4x+12=0\)

=> \(x\left(x-3\right)-4\left(x-3\right)=0\)

=> (x - 4)(x - 3) = 0

=> \(\orbr{\begin{cases}x=4\\x=3\end{cases}}\)

f) Tương tự

a) Ta có: \(x^2-9x+20=0\)

\(\Leftrightarrow x^2-5x-4x+20=0\)

\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)

Vậy: x∈{4;5}

b) Ta có: \(x^3-4x^2+5x=0\)

\(\Leftrightarrow x\left(x^2-4x+5\right)=0\)(1)

Ta có: \(x^2-4x+5\)

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

Ta có: \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-2\right)^2+1\ge1>0\forall x\)

hay \(x^2-4x+5>0\forall x\)(2)

Từ (1) và (2) suy ra x=0

Vậy: x=0

c) Sửa đề: \(x^2-2x-15=0\)

Ta có: \(x^2-2x-15=0\)

\(\Leftrightarrow x^2+3x-5x-15=0\)

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

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

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

Vậy: x∈{-3;5}

d) Ta có: \(\left(x^2-1\right)^2=4x+1\)

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

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

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

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

\(\Leftrightarrow x\cdot\left[x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\right]=0\)

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

Ta có: \(x^2+2x+2\)

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

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^2+1\ge1>0\forall x\)

hay \(x^2+2x+2>0\forall x\)(4)

Từ (3) và (4) suy ra

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

Vậy: x∈{0;2}

\(A.\left(2,3x-6,5\right)\left(0,1x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2,3x-6,5=0\\0,1x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2,3x=6,5\\0,1x=-2\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{6,5}{2,3}\\x=-20\end{cases}}\)