\(9x^2-6x+1-3\left(x^2+2x-3\right)\)

\(=6x^2-12x+10=6\left(x^2-2x+1-1\right)+10\)

\(\Leftrightarrow6\left(x-1\right)^2+4>0\)(luôn đúng) 

\(\left(2x-1\right)\left(2x-3\right)\left(x+1\right)^2=18\)

\(\Leftrightarrow\left(2x-1\right)\left(2x-3\right)\left(2x+2\right)^2=72\) (*)

Đặt \(a=2x+2\)

(*) \(\Leftrightarrow\left(a-3\right)\left(a-5\right).a^2=0\)

\(\Leftrightarrow\left(a^3-5a^2\right)\left(a-3\right)=0\)

\(\Leftrightarrow a^4-8a^3+15a^2=0\)

\(\Leftrightarrow a^4-5a^3-3a^3+15a^2=0\)

\(\Leftrightarrow a^3.\left(a-5\right)-3a^2.\left(a-5\right)=0\)

\(\Leftrightarrow\left(a-5\right)\left(a-3\right).a^2=0\)

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

\(\left(2-3x\right)^2=\dfrac{1}{36}=\left(\dfrac{1}{6}\right)^2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{18}\\x=\dfrac{13}{18}\end{matrix}\right.\)

\(\left[{}\begin{matrix}2-3x=\dfrac{1}{6}\\2-3x=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{11}{6}\\3x=\dfrac{13}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{6}:3=\dfrac{11}{18}\\x=\dfrac{13}{6}:3=\dfrac{13}{18}\end{matrix}\right.\)

\(\Leftrightarrow\left(5x-3\right)^2-2\left(5x-3\right)=0\)

\(\Leftrightarrow\left(5x-3\right)\left(5x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=1\end{matrix}\right.\)

mấy bn dưới lớp 8 ko bt thì ko đc lm bài này

alo

a) \(\left(2a+1\right)\left(4a-1\right)< 2a\left(4a+1\right)\)

\(\Leftrightarrow8a^2+4a-2a-1< 8a^2+2a\)

\(\Leftrightarrow8a^2+4a-2a-1-8a^2-2a< 0\)

\(\Leftrightarrow-1< 0\) (Luôn đúng với mọi \(a\))

b) \(\left(3-b\right)^2>b\left(b-6\right)\)

\(\Leftrightarrow9-6b+b^2>b^2-6b\)

\(\Leftrightarrow9-6b+b^2-b^2+6b>0\)

\(\Leftrightarrow9>0\) (Luôn đúng với mọi \(b\))

\(\left|2x+1\right|=3x-2\)đk : x >= 2/3 

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

\(\left|2x+1\right|=3x-2\)

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

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

\(\dfrac{x\left(x-8\right)+3\left(x+6\right)}{\left(x+6\right)\left(x-8\right)}=\dfrac{-12x+33}{\left(x+6\right)\left(x-8\right)}\left(đk:x\ne-6;8\right)\)

\(x^2-8x+3x+18=-12x+33\)

\(x^2-5x+18+12x-33=0\)

\(x^2+7x+15=0\)

\(\text{∆}=7^2-4.15=-11< 0\)

⇒ pt vô nghiệm

đk : x khác -6 ; 8 

\(x^2-8x+3x+18=-12x+33\Leftrightarrow x^2+7x-25=0\)

\(\Leftrightarrow x=\dfrac{-7\pm\sqrt{149}}{2}\)