Ta có :

\(x^2-3\left|x\right|=18\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-3x=18\\x^2-3\left(-x\right)=18\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-3x-18=0\\x^2+3x-18=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-6x+3x-18=0\\x^2+6x-3x-18=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)\left(x-6\right)=0\\\left(x-3\right)\left(x+6\right)=0\end{cases}}\)

Từ đây có thể suy ra được tập nghiệm của phương trình là \(S=\left\{-3;6;3;-6\right\}\)

Chúc bạn học tốt !

       (2x+3)²(x+1)(x+2)=18

<=> (4x²+12x+9)(x²+3x+2)=18

<=> (4x⁴+12x³+8x²+12x³+36x²+24x+9x²+27x+18=18

<=> 4x⁴+24x³+53x²+51x=0

<=> x(4x³+24x²+53x+51)=0

<=> x=0 hoặc 4x³+24x²+53x+51=0 (1)

Giải (1): 

        4x³+24x²+53x+51=0

<=> (x+3)(4x²+12x+7)=0

<=> x+3=0 (2) hoặc 4x²+12x+7=0 (3)

Giải (2):

         x+3=0

<=> x= -3

Giải (3): 

Ta có: Δ=b²-4ac=12²-4.4.7=-128<0

nên (4x²+12x+7)>0 (Δ<0 và a=4 =>a>0)

do đó: phương trình (3) vô nghiệm

Vậy: phương trình đã cho có nghiệm : x= 0 hoặc x= -3

\(\left(4x^2+12x+9\right)\left(x^2+3x+2\right)=18\)

Đặt \(x^2+3x+2=t\)

t(4t+1) =18

 tự giải tiếp nhé.

\(x^2\left(x+4,5\right)=13,5\)

<=>\(x^3+4,5x^2-13,5=0\)

<=> \(x^3+3x^2+1,5x^2+4,5x-4,5x-13,5=0\)

<=>\(x^2\left(x+3\right)+1,5x\left(x+3\right)-4,5\left(x+3\right)=0\)

<=>\(\left(x+3\right)\left(x^2+1,5x-4,5\right)=0\)

<=>\(\left(x+3\right)\left[x^2+3x-1,5-4,5\right]=0\)

<=>\(\left(x+3\right)\left[x\left(x+3\right)-1,5\left(x+3\right)\right]=0\)

<=>\(\left(x+3\right)^2\left(x-1,5\right)=0\)

<=> \(\left[{}\begin{matrix}\left(x+3\right)^2=0\\x-1,5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1,5\end{matrix}\right.\)

Vậy...

Ta có: \(x^2\left(x+4.5\right)=13.5\)

\(\Leftrightarrow x^3+\dfrac{9}{2}x^2-\dfrac{27}{2}=0\)

\(\Leftrightarrow2x^3+9x^2-27=0\)

\(\Leftrightarrow2x^3-3x^2+12x^2-18x+18x-27=0\)

\(\Leftrightarrow x^2\left(2x-3\right)+12x\left(2x-3\right)+9\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x^2+12x+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x^2+12x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\\left(x+6\right)^2=27\end{matrix}\right.\)

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

Vậy: \(S=\left\{\dfrac{3}{2};3\sqrt{3}-6;-3\sqrt{3}-6\right\}\)

a: =>3x+3=4x-4

=>-x=-7

hay x=7(nhận)

b: (x-1)(x-3)=0

=>x-1=0 hoặc x-3=0

=>x=1 hoặc x=3

c: 2(x-1)+x=0

=>2x-2+x=0

=>3x-2=0

hay x=2/3

a, ĐKXĐ : x ≠ 1 ; x ≠ -1

\(\Rightarrow3\left(x+1\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x+3=4x-4\)

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\left(N\right)\)

b,

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

c,

\(\Leftrightarrow2x-2+x=0\)

\(\Leftrightarrow3x=2\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

Đặt \(a=\sqrt{2x^2+16x+18};b=\sqrt{x^2-1}\left(a,b\ge0\right);\)

Ta có: \(a+b=\sqrt{a^2+2b^2}\Rightarrow a^2+2ab+b^2=a^2+2b^2\)

\(\Leftrightarrow b\left(2a-b\right)=0\)

TH1: \(\sqrt{x^2-1}=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=1\end{cases}\left(TM\right)}\)

TH2: \(2\sqrt{2x^2+16x+18}=\sqrt{x^2-1}\Leftrightarrow7x^2+64x+72=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-32+3\sqrt{57}}{7}\left(TM\right)\\x=\frac{-32-3\sqrt{57}}{7}\left(KTM\right)\end{cases}}\)

a: 2/(x-2)=3/(x+2)

=>3x-6=2x+4

=>x=10

b: (x-2)(x+5)=0

=>x-2=0 hoặc x+5=0

=>x=2 hoặc x=-5

c: 2(x+2)-x=4

=>2x+4-x=4

=>x=0

\(a,\dfrac{2}{x-2}=\dfrac{3}{x+2}\)

\(\Leftrightarrow\dfrac{2}{x-2}-\dfrac{3}{x+2}=0\)

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

\(\Leftrightarrow2x+4-3x+6=0\)

\(\Leftrightarrow-x+10=0\)

\(\Leftrightarrow-x=-10\)

\(\Leftrightarrow x=10\)

\(b,\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)

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

\(c,2\left(x+2\right)-x=4\)

\(\Leftrightarrow2x+4-x-4=0\)

\(\Leftrightarrow x=0\)