a: \(x^3+8x=5x^2+4\)

=>\(x^3-5x^2+8x-4=0\)

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

=>\(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x^2-4x+4\right)=0\)

=>\(\left(x-1\right)\left(x-2\right)^2=0\)

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

2: \(x^3+3x^2=x+6\)

=>\(x^3+3x^2-x-6=0\)

=>\(x^3+2x^2+x^2+2x-3x-6=0\)

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

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

=>\(\left[{}\begin{matrix}x+2=0\\x^2+x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1+\sqrt{13}}{2}\\x=\dfrac{-1-\sqrt{13}}{2}\end{matrix}\right.\)

3: ĐKXĐ: x>=0

\(2x+3\sqrt{x}=1\)

=>\(2x+3\sqrt{x}-1=0\)

=>\(x+\dfrac{3}{2}\sqrt{x}-\dfrac{1}{2}=0\)

=>\(\left(\sqrt{x}\right)^2+2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{17}{16}=0\)

=>\(\left(\sqrt{x}+\dfrac{3}{4}\right)^2=\dfrac{17}{16}\)

=>\(\left[{}\begin{matrix}\sqrt{x}+\dfrac{3}{4}=-\dfrac{\sqrt{17}}{4}\\\sqrt{x}+\dfrac{3}{4}=\dfrac{\sqrt{17}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{17}-3}{4}\left(nhận\right)\\\sqrt{x}=\dfrac{-\sqrt{17}-3}{4}\left(loại\right)\end{matrix}\right.\)

=>\(x=\dfrac{13-3\sqrt{17}}{8}\left(nhận\right)\)

4: \(x^4+4x^2+1=3x^3+3x\)

=>\(x^4-3x^3+4x^2-3x+1=0\)

=>\(x^4-x^3-2x^3+2x^2+2x^2-2x-x+1=0\)

=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x^3-2x^2+2x-1\right)=0\)

=>\(\left(x-1\right)\left(x^3-x^2-x^2+x+x-1\right)=0\)

=>\(\left(x-1\right)^2\cdot\left(x^2-x+1\right)=0\)

=>(x-1)^2=0

=>x-1=0

=>x=1

a.

\(x^3+8x=5x^2+4\)

\(\Leftrightarrow x^3-5x^2+8x-4=0\)

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

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

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

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

b.

\(x^3+3x^2-x-6=0\)

\(\Leftrightarrow\left(x^3+x^2-3x\right)+\left(2x^2+2x-6\right)=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1\pm\sqrt{13}}{2}\end{matrix}\right.\)

c.

\(2x+3\sqrt{x}+1=0\)

ĐKXĐ: \(x\ge0\)

Do \(x\ge0\Rightarrow\left\{{}\begin{matrix}2x\ge0\\3\sqrt{x}\ge0\end{matrix}\right.\)

\(\Rightarrow2x+3\sqrt{x}+1>0\)

Pt đã cho vô nghiệm

d.

\(x^4+4x^2+1=3x^3+3x\)

\(\Leftrightarrow x^4-3x^3+4x^2-3x+1=0\)

- Với \(x=0\) ko phải nghiệm

- Với \(x\ne0\) chia cả 2 vế của pt cho \(x^2\)

\(\Rightarrow x^2-3x+4-\dfrac{3}{x}+\dfrac{1}{x^2}=0\)

\(\Leftrightarrow\left(x^2+\dfrac{1}{x^2}+2\right)-3\left(x+\dfrac{1}{x}\right)+2=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-3\left(x+\dfrac{1}{x}\right)+2=0\)

Đặt \(x+\dfrac{1}{x}=t\)

\(\Rightarrow t^2-3t+2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-x+1=0\left(vn\right)\\x^2-2x+1=0\end{matrix}\right.\)

\(\Rightarrow x=1\)

a) 3x-6=0

3x=6 => x=2

b) (3x+2)(4x-5)=0

=> 3x+2=0 => x=-2/3

hoặc 4x-5=0 => x=5/4

câu c ,d thiếu dấu '=" để thành 1 pt rồi bạn

c) \(\dfrac{2x-5}{3}+\dfrac{x-3}{5}=\dfrac{4x+3}{15}\)

=> 10x -25 +3X-9=4X+3

=>9x=37

=>x=37/9

d) \(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-5}{x^2-9}\) ĐK (x khác 3,-3)

=>5x+15+4x-12=x-5

=>8x=-8

=>x=-1

\(a,\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19y=-21\\5x-4y=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{21}{19}\\5x-4\left(-\dfrac{21}{19}\right)=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{19}\\y=-\dfrac{21}{19}\end{matrix}\right.\)

\(c,\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\10x-5y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\13x=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\\ d,\Leftrightarrow\left\{{}\begin{matrix}5x-10y=-30\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\-7y=-35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\ e,\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+3\cdot6=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)

a: =>\(\dfrac{x^2+2x-13-x+1}{x-1}< 0\)

=>\(\dfrac{x^2+x-12}{x-1}< 0\)

=>\(\dfrac{\left(x+4\right)\left(x-3\right)}{x-1}< 0\)

=>1<x<3 hoặc x<-4

b: =>\(\dfrac{3x^2+4x-3x-4}{x-1}< 3\)

=>3x+4<3

=>3x<-1

=>x<-1/3

c: TH1: 2x^2-3x+1>0 và x+2>0

=>(2x-1)(x-1)>0 và x+2>0

=>x>1

TH2: (2x-1)(x-1)<0 và x+2<0

=>x<-2 và 1/2<x<1

=>Loại

a. \(\sqrt[3]{1-2x}+3=0\left(ĐK:x\le\dfrac{1}{2}\right)\)

<=> \(\sqrt[3]{1-2x}=-3\)

<=> \(1-2x=\left(-3\right)^3\)

<=> \(1-2x=-27\)

<=> \(-2x=-28\)

<=> \(x=14\left(TM\right)\)

a: \(x-3\left(2x-6\right)=21-\left(5x+3\right)\)

=>\(x-6x+18=21-5x-3\)

=>18=18(luôn đúng)

=>\(x\in R\)

b: \(\left(x-2\right)\left(x+2\right)-\left(x-1\right)^2=2\left(x+1\right)\)

=>\(x^2-4-x^2+2x-1=2x+2\)

=>2x-5=2x+2

=>-7=0(vô lý)

=>\(x\in\varnothing\)

c: \(\dfrac{9x+4}{6}=1-\dfrac{3x-5}{9}\)

=>\(\dfrac{3\left(9x+4\right)}{18}=\dfrac{18}{18}-\dfrac{2\left(3x-5\right)}{18}\)

=>3(9x+4)=18-2(3x-5)

=>27x+12=18-6x+10

=>27x+12=-6x+28

=>33x=16

=>\(x=\dfrac{16}{33}\left(nhận\right)\)

d: ĐKXĐ: \(x\notin\left\{2;5\right\}\)

\(\dfrac{6x+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)

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

=>\(6x+1+5\left(x-5\right)=3\left(x-2\right)\)

=>6x+1+5x-25=3x-6

=>11x-24=3x-6

=>8x=18

=>\(x=\dfrac{9}{4}\left(nhận\right)\)

a: 

=>

=>18=18(luôn đúng)

=>

b: 

=>

=>2x-5=2x+2

=>-7=0(vô lý)

=>x∈∅

c: 3(9x+4)18=1818−2(3x−5)18

=>3(9x+4)=18-2(3x-5)

=>27x+12=18-6x+10

=>27x+12=-6x+28

=>33x=16

=>

=>