\(ĐK:x\in R\)

Đặt \(x^2-2x=a\), PTTT:

\(-a+\sqrt{6a+7}=0\\ \Leftrightarrow\sqrt{6a+7}=a\\ \Leftrightarrow a^2-6a-7=0\\ \Leftrightarrow\left[{}\begin{matrix}a=7\\a=-1\left(loại.do.a=\sqrt{6a+7}\ge0\right)\end{matrix}\right.\\ \Leftrightarrow a=7\\ \Leftrightarrow x^2-2x-7=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1+2\sqrt{2}\\x=1-2\sqrt{2}\end{matrix}\right.\)

Ta có

\(\left\{{}\begin{matrix}\sqrt{2x^2-4x+3}=\sqrt{2\left(x-1\right)^2+1}\ge\sqrt{1}=1\\\sqrt{3x^2-6x+7}=\sqrt{3\left(x-1\right)^2+4}\ge\sqrt{4}=2\end{matrix}\right.\) \(\forall x\)

\(\Rightarrow VT=\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}\ge3\) \(\forall x\)

Lại có \(VP=2-x^2+2x=3-\left(x-1\right)^2\le3\) \(\forall x\)

\(\Rightarrow\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}=2-x^2+2x\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2\left(x-1\right)^2+1}=1\\\sqrt{3\left(x-1\right)^2+4}=2\\3-\left(x-1\right)^2=3\end{matrix}\right.\)

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

Vậy pt có nghiệm duy nhất \(x=1\)

\(3x^3+6x^2-12x+8=0\)

\(\Leftrightarrow4x^3=x^3-6x^2+12x-8\)

\(\Leftrightarrow4x^3=\left(x-2\right)^3\)

\(\Rightarrow\sqrt[3]{4}.x=x-2\)

\(\Rightarrow x=\dfrac{2}{1-\sqrt[3]{4}}\)

=>|x-1|+|x-3|=1

TH1: x<1

Pt sẽ la 1-x+3-x=1

=>4-2x=1

=>x=3/2(loại)

TH2: 1<=x<3

Pt sẽ là x-1+3-x=1

=>2=1(loại)

TH3: x>=3

Pt sẽ là x-1+x-3=1

=>2x-4=1

=>2x=5

=>x=5/2(loại)

ĐK: \(x^2-1\ge0\)

pt <=> \(\left(x^2+2x+1\right)-2\left(x+1\right)\sqrt{x^2-1}+\left(x^2-1\right)-4x^2+4x-1=0\)

<=> \(\left[\left(x+1\right)^2-2\left(x+1\right)\sqrt{x^2-1}+\left(x^2-1\right)\right]-\left(2x-1\right)^2=0\)

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

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

Phương trình tích. Dễ rồi đúng ko? Tự làm tiếp nhé!

a) 2x⁴ - x³ -2x² -x +2=0

=> (2x⁴- 2x³) +(x³-x²) -(x² -x) -(2x-2)=0

=>(x-1)(2x³+x²-x-2)=0

=>(x-1)²( 2x²+3x+2)=0 ( vì 2x²+3x+2>0)

=> x-1=0 => x =1

chia cho x² , rồi đặt ẩn