\(\left(x^2+8x\right)+8\left(x^2+8x\right)=48\)

Đặt: \(u=x^2+8x\)

\(\Rightarrow u^2+8u=48\)

\(\Leftrightarrow u^2+8u-48=0\)

\(\Leftrightarrow u^2-4u+12u-48=0\)

\(\Leftrightarrow u\left(u-4\right)+12\left(u-4\right)=0\)

\(\Leftrightarrow\left(u+12\right)\left(u-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}u+12=0\Leftrightarrow u=-12\\u-4=0\Leftrightarrow u=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+8x=-12\\x^2+8x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+8x+12=0\\x^2+8x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4+2\sqrt{5}\\x=-4-2\sqrt{5}\\x=-2\\x=-6\end{matrix}\right.\)

\(\Leftrightarrow x^4+16x^3+64x^2+8x^2+64x=48\\ \Leftrightarrow x^4+16x^3+72x^2+64x-48=0\\ \Leftrightarrow\left(x+2\right)\left(x+6\right)\left(x^2+8x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\x+6=0\\x^2+8x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-6\\x=-4\pm2\sqrt{5}\end{matrix}\right.\)

Vậy...

\(2x^3+3x^2-32x=48\)

\(2x^3+3x^2-32x-48=0\)

\(\left(2x^3+3x^2\right)-\left(32x+48\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2-16=0\\2x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}\left(x+4\right)\left(x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=-4\\x=4\end{cases}}\\x=-\frac{3}{2}\end{cases}}}\)\(\left(x+4\right)\left(x-4\right)\left(2x+3\right)=0\)

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

\(2x^3+3x^2-32x=48\)

\(\Leftrightarrow2x^3+3x^2-32x-48=0\)

\(\Leftrightarrow\left(2x^3-32x\right)+\left(3x^2-48\right)=0\)

\(\Leftrightarrow2x\left(x^2-16\right)+3\left(x^2-16\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-4=0;x+4=0\\2x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\pm4\\x=\frac{-3}{2}\end{cases}}\)

Vậy tập nghiệm của pt là S={4;-4;-3/2}

_Học tốt_

Ta có \(2x^3+3x^2-32x=48\)

\(\Leftrightarrow2x^3+3x^2-32x-48=0\)

\(\Leftrightarrow2x^2\left(x+\frac{3}{2}\right)-32\left(x+\frac{3}{2}\right)=0\)

\(\Leftrightarrow\left(x+\frac{3}{2}\right)\left(2x^2-32\right)=0\)

\(\Leftrightarrow\left(x+\frac{3}{2}\right)2\left(x^2-16\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{2}=0\\x^2-16=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=\pm4\end{cases}}}\)

Vậy đccm

\(\dfrac{1}{x^2+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)

\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)

\(\Leftrightarrow\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)+\left(\dfrac{1}{x+2}-\dfrac{1}{x+4}\right)+\left(\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)+\left(\dfrac{1}{x+6}-\dfrac{1}{x+8}\right)=\dfrac{8}{105}\)

\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)

\(\Leftrightarrow\dfrac{8}{x\left(x+8\right)}=\dfrac{8}{105}\)

\(\Leftrightarrow x\left(x+8\right)=105\)

\(\Leftrightarrow x^2+8x-105=0\)

\(\Leftrightarrow x^2-7x+15x-105=0\)

\(\Leftrightarrow x\left(x-7\right)+15\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)

Thử lại ta có nghiệm của phương trình trên là \(x=7\text{v}à\text{x}=15\)

ĐKXĐ: \(x\neq 0\).

Đặt \(\dfrac{x}{3}-\dfrac{4}{x}=t\).

PT đã cho tương đương:

\(3t^2+8-10t=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=\dfrac{4}{3}\end{matrix}\right.\).

Với t = 2 ta có \(\dfrac{x}{3}-\dfrac{4}{x}=2\Leftrightarrow\dfrac{x^2-12}{3x}=2\Leftrightarrow x^2-6x-12=0\Leftrightarrow x=\pm\sqrt{21}+3\).

Với t = \(\frac{4}{3}\) ta có \(\dfrac{x}{3}-\dfrac{4}{x}=\dfrac{4}{3}\Leftrightarrow\dfrac{x^2-12}{3x}=\dfrac{4}{3}\Leftrightarrow x^2-12=4x\Leftrightarrow x^2-4x-12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\).

Vậy...