a: Ta có: \(\sqrt[3]{x+1}=-5\)

\(\Leftrightarrow x+1=-125\)

hay x=-126

b: Ta có: \(\sqrt[3]{x+1}-1=x\)

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

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

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

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

Ta có (\(^{x^{2^{ }}^{ }+3x}\)) (\(^{x^{2^{ }}+3x+4}\))

Đặt \(x^{2^{ }^{ }}+3x\) là a ta có

a.(a+4)=-4

4a+\(a^2\) -4=0

\(^{ }\left(a-2\right)^2\)=0

Suy ra a=2

hay \(x^{2^{ }^{ }^{ }}+3x=2\)

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

𝑥=−3±17√/2

a) (x+2)³−x+1=(x−1)(x+1)

=>(x+2)(x²-2x+4)-x+1=(x²-1²)

=>x³-2x²+4x+2x²-4x+8-x+1=(x²-1)

=>x³-2x²+4x+2x²-4x+8+1-x²+1=0

=>x³-x²+10=0

=>x³-x²=-10

=>x²(x-1)=-10

=>....

1:

a: =>(|x|+4)(|x|-1)=0

=>|x|-1=0

=>x=1; x=-1

b: =>x^2-4>=0

=>x>=2 hoặc x<=-2

d: =>|2x+5|=2x-5

=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0

=>x=0(loại)

a.

\(\dfrac{x+1}{x-1}>0\Rightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)

b.

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+2\right)}{x-9}< 0\Rightarrow\left[{}\begin{matrix}x< -2\\1< x< 9\end{matrix}\right.\)

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

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

\(ĐK:x\ne1;-3\)

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

\(\Leftrightarrow2x\left(x+3\right)+4=\left(2x-5\right)\left(x-1\right)\)

\(\Leftrightarrow2x^2+6x+4=2x^2-2x-5x+5\)

\(\Leftrightarrow13x=1\)

\(\Leftrightarrow x=\dfrac{1}{13}\left(tm\right)\)

a, \(-2x\ge-5\Leftrightarrow x\le\dfrac{5}{2}\)

b, TH1 : \(\left\{{}\begin{matrix}x-1>0\\x+3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>1\\x>-3\end{matrix}\right.\Leftrightarrow x>1\)

TH2 : \(\left\{{}\begin{matrix}x-1< 0\\x+3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 1\\x< -3\end{matrix}\right.\Leftrightarrow x< -3\)

bạn tự kl nhé 

a, \(\left|x+5\right|=3x+1\)

TH1 : \(x+5=3x+1\Leftrightarrow-2x=-4\Leftrightarrow x=2\)

TH2 : \(x+5=-3x-1\Leftrightarrow4x=-6\Leftrightarrow x=-\dfrac{3}{2}\)( ktm )

b, \(\left|-5x\right|=2x+21\)

TH1 : \(5x=2x+21\Leftrightarrow3x=21\Leftrightarrow x=7\)

TH2 : \(5x=-2x-21\Leftrightarrow7x=-21\Leftrightarrow x=-3\)

khó nhìn quá ko

a) Ta có: |x+5|=3x+1

\(\Leftrightarrow\left[{}\begin{matrix}x+5=3x+1\left(x\ge-5\right)\\x+5=-3x-1\left(x< -5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3x=1-5\\x+3x=-1-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=-4\\4x=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{3}{2}\left(loại\right)\end{matrix}\right.\)

Vậy: S={2}