Question

Giải pt

a)  x\(^{^2}\)+ 3 - \(\sqrt{2x^2-3x+2}\)= \(\frac{3}{2}\)(x+1)

b)  2x\(^{^2}\)- 9x + \(\sqrt{9x-2x^2}\)+ 6 = 0

Answer 1

\(\Leftrightarrow x^2-1+1-\sqrt{2x^2-3x+2}-\frac{3}{2}\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)+\frac{\left(2x-1\right)\left(x-1\right)}{1+\sqrt{2x^2-3x+2}}-\frac{3}{2}\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-\frac{1}{2}+\frac{2\left(x-\frac{1}{2}\right)}{1+\sqrt{2x^2-3x+2}}\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-\frac{1}{2}\right)\left(1+\frac{2}{1+\sqrt{2x^2-3x+2}}\right)=0\)

Do \(\left(1+\frac{2}{1+\sqrt{2x^2-3x+2}}\right)>0\left(\forall x\right)\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow2x^2-9x+9-3+\sqrt{9x-2x^2}=0\)

\(\Leftrightarrow2x\left(x-3\right)-3\left(x-3\right)+\frac{\left(x-3\right)\left(-2x+3\right)}{\sqrt{9x-2x^2}+3}=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x-3-\frac{2x-3}{\sqrt{9x-2x^2}+3}\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x-3\right)\left(1-\frac{1}{\sqrt{9x-2x^2}+3}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{3}{2}\end{cases}}\)

TH còn lại loại bạn tự giải nha

Answer 2

a) đK:\(2x^2-3x+2\ge0\)

 \(x^2+3-\sqrt{2x^2-3x+2}=\frac{3}{2}\left(x+1\right)\)

<=> \(2x^2+6-2\sqrt{2x^2-3x+2}=3\left(x+1\right)\)

<=> \(2x^2-3x+3-2\sqrt{2x^2-3x+2}=0\)

Đặt: \(t=\sqrt{2x^2-3x+2}\left(t\ge0\right)\)

Ta có phương trình:

\(t^2-2+3-2t=0\Leftrightarrow t^2-2t+1=0\Leftrightarrow t=1\)

Với t=1 ta có phương trình:

 \(\sqrt{2x^2-3x+2}=1\Leftrightarrow2x^2-3x+1=0\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=\frac{1}{2}\left(tm\right)\end{cases}}\)

Vậy:...

Câu b tương tự.