Lời giải:

$\frac{x^3+8}{x^2-2x+1}.\frac{x^2+3x+2}{1-x^2}=\frac{(x^3+8)(x^2+3x+2)}{(x^2-2x+1)(1-x^2)}$

$=\frac{(x+2)(x^2-2x+4)(x+1)(x+2)}{(x-1)^2(1-x)(x+1)}$

$=\frac{(x+2)^2(x^2-2x+4)}{-(x-1)^3}$
 

\(2x-24=8^5:8^3\)

\(2x-24=8^2\)

\(2x-24=64\)

\(2x=88\)

\(x=44\)

\(\left(2x-24\right).8^3=8^5\)

\(\left(2x-24\right)=8^5:8^3\)

\(\left(2x-24\right)=8^2\)

\(\left(2x-24\right)=64\)

\(2x=64+24\)

\(2x=88\)

\(\Rightarrow x=44\)

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

\(\Rightarrow x+2-2=0\)

\(\Rightarrow x=0\)

\(\left(x+3\right)+1=7\)

\(\Rightarrow x+3+1=7\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=3\)

\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)

\(\Rightarrow3x=12\)

\(\Rightarrow x=4\)

\(\left(5x+4\right)-1=13\)

\(\Rightarrow5x+4-1=13\)

\(\Rightarrow5x+3=13\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

\(\left(4x-8\right)-3=5\)

\(\Rightarrow4x-8-3=5\)

\(\Rightarrow4x-11=5\)

\(\Rightarrow4x=16\)

\(\Rightarrow x=4\)

\(8-\left(2x+4\right)=2\)

\(\Rightarrow8-2x-4=2\)

\(\Rightarrow4-2x=2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

\(7+\left(5x+2\right)=14\)

\(\Rightarrow7+5x+2=14\)

\(\Rightarrow9+5x=14\)

\(\Rightarrow5x=5\)

\(\Rightarrow x=1\)

\(5-\left(3x-11\right)=1\)

\(\Rightarrow5-3x+11=1\)

\(\Rightarrow16-3x=1\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

1) (3x-1)(-1/2x+5)=0

TH1: 3x-1=0

        3x   = 1

          x    = 1/3

TH2:  -1/2x+5=0

         -1/2x     =-5

               x     = 10

2)  (3/4-x)^3=-8

     (3/4-x)^3=(-2)^3

=> 3/4-x=-2

           x=3/4+2

           x= 11/4

3)  |2x-1|=-4^2

     |2x-1|=16

=> 2x-1=-16 hoặc 2x-1=16

TH1: 2x-1=-16

        2x   =-15

          x    = -15/2

TH2:  2x-1=16

          2x   =17

            x    = 17/2

\(2x^2y+4xy^2+2y^3-8\) 

\(=2y\left(x^2+2xy+y^2\right)-8\)

\(=2y\left(x+y\right)^2-8\) 

\(=2\left[y\left(x+y\right)^2-4\right]\)

thank you huỳnh thanh phong

\(\dfrac{2x}{3y}=-\dfrac{1}{3}\\ \Rightarrow3y=2x:-\dfrac{1}{3}=\dfrac{2x.3}{-1}=-6x\\ \Rightarrow y=-\dfrac{6x}{3}=-2x\)

Thế \(y=-2x\) vào \(2x+3y^2=\dfrac{161}{4}\) được:

\(2x+3.\left(-2x\right)^2=\dfrac{161}{4}\\ \Leftrightarrow2x+12x^2-\dfrac{161}{4}=0\\ \Leftrightarrow48x^2+8x-161=0\\ \Leftrightarrow\left(48x^2+92x\right)+\left(-84x-161\right)=0\\ \Leftrightarrow4x\left(12x+23\right)-7\left(12x+23\right)=0\\ \Leftrightarrow\left(4x-7\right)\left(12x+23\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{4}\Rightarrow y=-\dfrac{2.7}{4}=-\dfrac{7}{2}\\x=-\dfrac{23}{12}\Rightarrow y=-2.-\dfrac{23}{12}=\dfrac{23}{6}\end{matrix}\right.\)

Vậy phương trình có nghiệm \(\left\{x;y\right\}=\left\{\dfrac{7}{4};-\dfrac{7}{2}\right\}\) hoặc \(\left\{x;y\right\}=\left\{-\dfrac{23}{12};\dfrac{23}{6}\right\}\)

\(\left(2x+1\right)^2\ge4x^2-3\)

\(\Leftrightarrow4x^2+2x+2x+1\ge4x^2-3\)

\(\Leftrightarrow4x^2-4x^2+2x+2x\ge-3-1\)

\(\Leftrightarrow4x\ge-4\\ \Leftrightarrow x\ge-1\)

Vậy phương trình có tập nghiệm là:\(S=\left\{x|x\ge-1\right\}\)

\(\left(2x+1\right)^2\ge4x^2-3\)

\(\Leftrightarrow4x^2+4x+1\ge4x^2-3\)

\(\Leftrightarrow4x^2-4x^2+4x\ge-3-1\)

\(\Leftrightarrow4x\ge-4\)

\(\Leftrightarrow x\ge-1\)

\(\left(2x+1\right)^2\ge4x^2-3\\ \Leftrightarrow4x^2+4x+1-4x^2+3\ge0\\ \Leftrightarrow4x\ge-4\\ \Leftrightarrow x\ge-1\)

Vậy: S= [-1; +∞)

5.(3x+8)-7.(2x+3)=16

=> 15x+40-14x-21=16

=>(15x-14x)+(40-21)

=> x+19=16

=>x=-3. Vậy x=-3

k cho mik nha

Chúc bạn hok tốt

\(5\left(3x+8\right)-7\left(2x+3\right)=16\)

\(15x+40-14x-21=16\)

\(x+19=16\)

\(x+19-16=0\)

\(x+3=0\)

\(x=-3\)