(3x² + 10x - 8)² = (5x² - 2x + 10)²

<=> (3x² + 10x - 8)² - (5x² - 2x + 10)² = 0

<=> (3x² + 10x - 8 - 5x² + 2x - 10)(3x² + 10x - 8 + 5x² - 2x + 10) = 0

<=> (-2x² + 12x - 18)(8x² + 8x + 2) = 0

<=> -4(x² - 6x + 9)(4x²  + 4x + 1) = 0

<=> (x - 3)²(2x + 1)² = 0

<=> \(\orbr{\begin{cases}x-3=0\\2x+1=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=3\\x=-\frac{1}{2}\end{cases}}\)

Vậy S = {3; -1/2}

\(\frac{x^2-5x+4}{x^2-2}=5\left(x-1\right)\)

\(\Rightarrow\frac{x^2-x-4x+4}{x^2-2}=5\left(x-1\right)\)

\(\Rightarrow\frac{x\left(x-1\right)-4\left(x-1\right)}{x^2-2}=5\left(x-1\right)\)

\(\Rightarrow\frac{\left(x-1\right)\left(x-4\right)}{x^2-2}=5\left(x-1\right)\)

Với x = 1

=> x - 1 = 0

=> \(\frac{0.\left(x-4\right)}{x^2-2}=5.0\)

=> 0 = 0 ( luôn đúng )

Với x khác 1

=> x - 1 khác 0

=> \(\frac{x-4}{x^2-2}=5\)( chia cả hai vế cho x - 1 )

=> \(x-4=5x^2-10\)

=> \(5x^2-x-6=0\)

=> \(5x^2+5x-6x-6=0\)

=> \(5x\left(x+1\right)-6\left(x+1\right)=0\)

=> \(\left(x+1\right)\left(5x-6\right)=0\)

=> \(\orbr{\begin{cases}x+1=0\\5x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{6}{5}\end{cases}}}\)

Vậy  \(x\in\left\{1;-1;\frac{6}{5}\right\}\)

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

\(\Leftrightarrow9x^2-1+3x^2+6x-x-2=0\)

\(\Leftrightarrow9x^2+3x^2+6x-x=0+1+2\)

\(\Leftrightarrow12x^2+5x=3\)

\(\Leftrightarrow12x^2+5x-3=0\)

\(\Leftrightarrow12x^2-4x+9x-3=0\)

\(\Leftrightarrow4x\left(3x-1\right)+3\left(3x-1\right)\)

\(\Leftrightarrow\left(4x+3\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+3=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-3\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{4}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy tập nghiệm phương trình là S = \(\left\{\dfrac{-3}{4};\dfrac{1}{3}\right\}\)

Ta có

4x-8=9x-3-2x+1

<=>-6=-3x(chuyển vế đổi dấu)

<=>x=2

b)

Ta có

Căn cả 2 vế ta đcx-5/ cawn3 =3 

<=>x=10.2

ĐK \(x\ge-3\)

PT <=> \(x^3+5x^2+6x+2=4\sqrt{x+3}+2\sqrt{2x+7}\)

<=> \(2\left(x+3-2\sqrt{x+3}\right)+\left(x+5-2\sqrt{2x+7}\right)+x^3+5x^2+3x-9=0\)

+  Với x=-3 =>thỏa mãn 

+Với \(x>-3\) ta liên hợp

\(2.\frac{x^2+2x-3}{x+3+2\sqrt{x+3}}+\frac{x^2+2x-3}{x+5+2\sqrt{2x+7}}+\left(x+3\right)\left(x^2+2x-3\right)=0\)

<=> \(\left(x^2+2x-3\right)\left(\frac{2}{x+3+2\sqrt{x+3}}+\frac{1}{x+5+2\sqrt{2x+7}}+x+3\right)=0\)

Do \(x>-3\)=> \(\frac{2}{x+3+2\sqrt{x+3}}+\frac{1}{x+5+2\sqrt{2x+7}}+x+3>0\)

=> \(x=1\)(TMĐKXĐ)

Vậy \(x=1;x=-3\)

\(4\left(x+1\right)^2=\sqrt{2\left(x^4+x^2+1\right)}\)

\(\Leftrightarrow16\left(x+1\right)^4=2\left(x^4+x^2+1\right)\)

\(\Leftrightarrow\left(x^2+3x+1\right)\left(7x^2+11x+7\right)=0\)

\(\sqrt{\frac{x+56}{16}+\sqrt{x-8}}=\frac{x}{8}\)

\(\Leftrightarrow2\sqrt{x+56+16\sqrt{x-8}}=x\)

\(\Leftrightarrow2\sqrt{\left(\sqrt{x-8}+8\right)^2}=x\)

\(\Leftrightarrow2\sqrt{x-8}+16=x\)

\(\Leftrightarrow x=24\)

a: Để phương trình có nghiệm duy nhất thì \(\left(m-3\right)\left(m+2\right)< >0\)

hay \(m\notin\left\{3;-2\right\}\)

Để phương trình vô nghiệm thì \(\left\{{}\begin{matrix}\left(m-3\right)\left(m+2\right)=0\\\left(m-3\right)\left(m-1\right)< >0\end{matrix}\right.\Leftrightarrow m=-2\)

Để phương trình có vô số nghiệm thì m=3

\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)

\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)

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

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

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

\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)

\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)

\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)

\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)

\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)

\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)