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 \(x\in\left\{-5;2\right\}\)

\(3x^3-48x=0\)

\(3x\cdot\left(x^2-16\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\left\{\pm4\right\}\end{cases}}\)

Vậy,............

Ta có

x 3   –   12 x 2   +   48 x   –   64   =   0     ⇔   x 3   –   3 . x 2 . 4   +   3 . x . 4 2   –   4 3   =   0     ⇔   ( x   –   4 ) 3   =   0

ó x – 4 = 0 ó x = 4

Vậy x = 4

Đáp án cần chọn là: B

\(\Delta=b^2-4ac=\left(-48\right)^2-4.1.\left(-25\right)=2400>0\)

do đó pt có 2 nghiệm phân biệt là:

\(•x_1=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{48-\sqrt{2400}}{2}=24-10\sqrt{6}\\ •x_2=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{48+\sqrt{2400}}{2}=24+10\sqrt{6}\)

\(x^2-48x-25=0\)

\(\Leftrightarrow x^2-2.x.24+24^2-601=0\)

\(\Leftrightarrow\left(x-24\right)^2-601=0\)

\(\Leftrightarrow\left(x-24\right)^2=601\)

\(\Leftrightarrow x-24=\sqrt{601}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-24=\sqrt{601}\\x-24=-\sqrt{601}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=24+\sqrt{601}\\x=24-\sqrt{601}\end{matrix}\right.\)

\(x=0\)

\(3x^3-75x=0\Leftrightarrow3x\left(x^2-25\right)=0\Leftrightarrow3x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow x=0;x=-5;x=5\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)

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

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

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

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

Vậy \(x\in\left\{-3;1\right\}\)