a) \(5x\left(x+4\right)-x\left(5x+1\right)=0\)

\(\Leftrightarrow x\left[5\left(x+4\right)-5x-1\right]=0\)

\(\Leftrightarrow x\left(5x+20-5x-1\right)=0\Leftrightarrow x=0\)

b) \(3x\left(5-x\right)+4\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(4-3x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{4}{3}\end{cases}}\)

c) \(x\left(x-3\right)+4x-12=0\)

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

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

d) \(x^2-36=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

e) \(x^2+3x+1=2\)

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

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

\(\Leftrightarrow x^2+3x+\frac{3}{2}-\frac{5}{2}=0\)

\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2-\frac{5}{2}=0\)

\(\Leftrightarrow\left(x+\frac{3}{2}+\frac{\sqrt{5}}{\sqrt{2}}\right)\left(x+\frac{3}{2}-\frac{\sqrt{5}}{\sqrt{2}}\right)=0\)

Còn lại ........... Tự lm nất nha 

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

\(\Leftrightarrow3\left(x^2-2.x.1+1^2\right)-3x^2+15x=1\)

\(\Leftrightarrow3x^2-2x+1-3x^2+15x=1\)

\(\Leftrightarrow13x+1=1\)

\(\Leftrightarrow13x=0\)

\(\Leftrightarrow x=0\)

Mấy bạn còn lại cũng như vậy 

a) \(x^2-36=0\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow x=\pm\sqrt{36}=\pm6\)

b) \(\left(3x-5\right)^2-\left(x+6\right)^2=0\)

\(\Leftrightarrow\left(3x-5-x-6\right)\left(3x-5+x+6\right)=0\)

\(\Leftrightarrow\left(2x-11\right)\left(4x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=\frac{-1}{4}\end{cases}}\)

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

Answer:

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

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

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

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

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

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

\(x^2-5x+6=0\)

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

\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)

\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

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

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

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)

\(x^2-2x+5=0\)

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

\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)

Vậy không có giá trị \(x\) thoả mãn

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

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

\(\Rightarrow x.\left(x+3\right)-2\left(x+3\right)=0\)

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

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

a. \(x\left(x^2-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)=17\)

\(x^3-25x-\left(x^3+8\right)=17\)

\(x^3-25x-x^3-8=17\)

\(-25x=25\)

\(x=-1\)

c. \(6x^2-\left(6x^2-4x+15x-10\right)=7\)

\(6x^2-6x^2-11x+10=7\)

\(-11x=-3\)

\(x=\frac{3}{11}\)

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