\(ĐK:x\ge-\frac{3}{2}\)

Ta có:

\(x^2+5x+8=3\sqrt{2x^3+5x^2+7x+6}\)

\(\Leftrightarrow\left(x^2+x+2\right)+2\left(2x+3\right)=3\sqrt{2x^3+5x^2+7x+6}\)

\(\Leftrightarrow\left(x^2+x+2\right)+2\left(2x+3\right)=3\sqrt{\left(x^2+x+2\right)\left(2x+3\right)}\)

Đặt \(\sqrt{x^2+x+2}=a;\sqrt{2x+3}=b\)

Khi đó: \(a^2+2b^2=3ab\Leftrightarrow\left(a-b\right)\left(a-2b\right)=0\)

\(\Leftrightarrow\sqrt{x^2+x+2}=\sqrt{2x+3}\left(hoac\right)\sqrt{x^2+x+2}=2\sqrt{2x+3}\)

Với \(\sqrt{x^2+x+2}=\sqrt{2x+3}\Rightarrow x^2+x+2=2x+3\Leftrightarrow x^2-x-1=0\Leftrightarrow x=\frac{1+\sqrt{5}}{2};x=\frac{1-\sqrt{5}}{2}\)Tự đối chiếu điều kiện xác định -,-

\(\sqrt{x^2+x+2}=2\sqrt{2x+3}\Rightarrow x^2+x+2=4\left(2x+3\right)\Leftrightarrow x^2-7x-10=0\)

Tới đây bí rồi huhu

bình phương hai vế rồi rút gọn, phân tích thành nhân tử

\(\left(x+1\right)\left(x^3-9x^2+7x+10\right)=0\)0

a)

`2x-3=2-x`

`<=>2x+x=2+3`

`<=>3x=5`

`<=>x=5/3`

b)

`3x+3=7+5x`

`<=>3x-5x=7-3`

`<=>-2x=4`

`<=>x=-2`

c)

`7x-3=3x+13`

`<=>7x-3x=13+3`

`<=>4x=16`

`<=>x=4`

d)

`(5x-2)/3=(5-3x)/2`

`<=>10x-4=15-9x`

`<=>10x+9x=15+4`

`<=>19x=19`

`<=>x=1`

3:

a: u+v=14 và uv=40

=>u,v là nghiệm của pt là x^2-14x+40=0

=>x=4 hoặc x=10

=>(u,v)=(4;10) hoặc (u,v)=(10;4)

b: u+v=-7 và uv=12

=>u,v là các nghiệm của pt:

x^2+7x+12=0

=>x=-3 hoặc x=-4

=>(u,v)=(-3;-4) hoặc (u,v)=(-4;-3)

c; u+v=-5 và uv=-24

=>u,v  là các nghiệm của phương trình:

x^2+5x-24=0

=>x=-8 hoặc x=3

=>(u,v)=(-8;3) hoặc (u,v)=(3;-8)

2x³+5x²=7x

-7x+5x²+2x³=7x+(-7x)

ta có 7x+(-7x)

-7x + 5x2 + 2x3 = 0

x(-7 + 5x + 2x2) = 0

x[(-7 + -2x)(1 + -1x)] = 0

2x^3 + 5x^2 = 7x

<=> 2x^3 + 5x^2 - 7x = 0

<=> x(2x^2 + 5x - 7) = 0

<=> x(2x^2 + 7x - 2x - 7) = 0

<=> x[x(2x + 7) - (2x + 7)] = 0

<=> x(x - 1)(2x + 7) = 0

<=> x = 0 hoặc x - 1 = 0 hoặc 2x + 7 = 0

<=> x = 0 hoặc x = 1 hoặc x = -7/2

\(a,\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\\ \Rightarrow7.\left(2x-1\right)-3.\left(5x+2\right)=21.\left(x+13\right)\\ \Rightarrow14x-7-15x-6=21x+273\\\Rightarrow -x-21x=273+13\\ \Rightarrow-22x=286\\ \Rightarrow x=-13\\ b,\dfrac{3\left(x+3\right)}{4}+\dfrac{1}{2}=\dfrac{5x+9}{3}-\dfrac{7x-9}{4}=0\\ \Rightarrow9.\left(x+3\right)+6=4.\left(5x+9\right)-3.\left(7x-9\right)=0\\\Rightarrow 9x+27+6=20x+36-21x+27\\ \Rightarrow9x+33=-x+63\\ \Rightarrow10x=30\\ \Rightarrow x=3\)

\(a,\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\)

\(\Rightarrow7\left(2x-1\right)-3\left(5x+2\right)-21x-273=0\)

\(\Rightarrow14x-7-15x-6-21x-273=0\)

\(\Rightarrow-22x=286\)

\(\Rightarrow x=-13\)

\(b,\dfrac{3\left(x+3\right)}{4}+\dfrac{1}{2}=\dfrac{5x+9}{3}-\dfrac{7x-9}{4}\)

\(\Rightarrow9\left(x+3\right)+6-4\left(5x+9\right)+3\left(7x-9\right)=0\)

\(\Rightarrow9x+27+6-20x-36+21x-27=0\)

\(\Rightarrow10x=30\Rightarrow x=3\)

`(2x)/(3x^2-x+2)-(7x)/(3x^2+5x+2)=1(x ne -1,-2/3)`

Đặt `a=3x^2+2x+2(a>=5/3)`

`pt<=>(2x)/(a-3x)-(7x)/(a+3x)=1`

`=>2x(a+3x)-7x(a-3x)=a^2-9x^2`

`<=>2ax+6x^2-7ax+21x^2=a^2-9x^2`

`<=>-5ax+27x^2=a^2-9x^2`

`<=>a^2-36x^2+5ax=0`

`<=>a^2-4ax+9ax-36x^2=0`

`<=>a(a-4x)+9x(a-4x)=0`

`<=>(a-4x)(a+9x)=0`

`+)a=4x`

`=>3x^2+2x+2=4x`

`=>3x^2-2x+2=0`

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

`=>(x-1/3)^2+5/9=0` vô lý

`+)a+9x=0`

`=>3x^2+2x+2+9x=0`

`=>3x^2+11x+2=0`

`=>x^2+11/3x+2/3=0`

`=>x=(-11+-\sqrt{97})/6`

ĐKXĐ: \(x\ne-1;x\ne-\dfrac{2}{3}\)

Ta có: \(\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{3x^2+5x+2}=1\)(1)

\(\Leftrightarrow\dfrac{2}{3x-1+\dfrac{2}{x}}-\dfrac{7}{3x+5+\dfrac{2}{x}}=1\)

Đặt: \(3x+\dfrac{2}{x}=a\)  (x khác 0) thì pt(1) trở thành:

\(\dfrac{2}{a-1}-\dfrac{7}{a+5}=1\)

\(\Leftrightarrow\dfrac{2\left(a+5\right)-7\left(a-1\right)}{\left(a-1\right)\left(a+5\right)}=1\)

\(\Leftrightarrow2\left(a+5\right)-7\left(a-1\right)=\left(a-1\right)\left(a+5\right)\)

\(\Leftrightarrow-5a+17=a^2+4a-5\)

\(\Leftrightarrow a^2+4a+5-5-17=0\)

\(\Leftrightarrow a^2+9a-22=0\)

\(\Leftrightarrow\left(a-2\right)\left(a+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2\\a=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{2}{x}=2\\3x+\dfrac{2}{x}=-11\end{matrix}\right.\)

Vì \(\left\{{}\begin{matrix}3x^2+2-2x\ne0\\3x^2+11x+2\ne0\end{matrix}\right.\)

=> PT vô nghiệm 

Ủa hình như sai:vvv

Câu 1: 

a) Ta có: 7x+21=0

\(\Leftrightarrow7x=-21\)

hay x=-3

Vậy: S={-3}

b) Ta có: 3x-2=2x-3

\(\Leftrightarrow3x-2-2x+3=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

Vậy: S={-1}

c) Ta có: 5x-2x-24=0

\(\Leftrightarrow3x=24\)

hay x=8

Vậy: S={8}

Câu 2: 

a) Ta có: \(\left(2x+1\right)\left(x-1\right)=0\)

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

Vậy: \(S=\left\{-\dfrac{1}{2};1\right\}\)

b) Ta có: \(\left(2x-3\right)\left(-x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\-x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\-x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=7\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};7\right\}\)

c) Ta có: \(\left(x+3\right)^3-9\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)^2-9\right]=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-6\end{matrix}\right.\)

Vậy: S={0;-3;-6}