a: \(\Leftrightarrow x^2-2x-8-x^2=36\)

=>-2x=44

hay x=-22

b: \(\Leftrightarrow4x^2+x-8x-2-4x^2-27x=1\)

=>-34x=3

hay x=-3/34

c: =>(x-10)(x-1)=0

=>x=10 hoặc x=1

a, Vô lý biểu thức phụ thuộc vào biến x.

b, Ta có : \(\left(2x+6\right)\left(4x^2-12x+36\right)-8x^3+10\)

\(=8x^3-24x^2+72x+24x^2-72x+216-8x^3+10\)

\(=216+10=226\)

\(1,=\left(x^2-2x+6x-12+2\right):\left(x-2\right)\\ =\left[x\left(x-2\right)+6\left(x-2\right)+2\right]:\left(x-2\right)\\ =x+6\left(\text{dư }2\right)\\ 2,=\left(x^3-x^2+5x^2-5x+6x-6\right):\left(x-1\right)\\ =\left[x^2\left(x-1\right)+5x\left(x-1\right)+6\left(x-1\right)\right]:\left(x-1\right)\\ =x^2+5x+6\)

Bạn lm kiểu đặt tính đc k chứ như này mk vẫn chx hiểu lắm 

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

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

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

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

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

a: \(B=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(=\dfrac{10}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)

b: \(\dfrac{x^2-36}{2x+10}\cdot\dfrac{3}{6-x}\)

\(=\dfrac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\dfrac{-3}{x-6}\)

\(=\dfrac{-3\left(x+6\right)}{2\left(x+5\right)}\)

c: \(\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)

\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-10}{4}=\dfrac{-5}{2}\)

d: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)

\(=\dfrac{1-4x^2}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)

\(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x+4}\cdot\dfrac{3}{2\left(1-2x\right)}=\dfrac{3\left(2x+1\right)}{x+4}\)

a.(x+10) /(4*x)-8* 4 -(2*x)/x+2

-(127*x-10)/(4*x)

(5/2-127*x/4)/x

Câu a