Nhận thấy \(x=0\) không phải nghiệm, chia 2 vế cho \(x^2\)

\(6x^2+7x-36+\frac{7}{x}+\frac{6}{x^2}=0\)

\(\Leftrightarrow6\left(x^2+\frac{1}{x^2}\right)+7\left(x+\frac{1}{x}\right)-36=0\)

Đặt \(x+\frac{1}{x}=a\) (\(\left|a\right|\ge2\)) \(\Rightarrow x^2+\frac{1}{x^2}=a^2-2\)

\(6\left(a^2-2\right)+7a-36=0\)

\(\Leftrightarrow6a^2+7a-48=0\)

Nghiệm xấu

ĐKXĐ:    \(0\le x\le\frac{3}{2}\)

ĐẶT:    \(\hept{\begin{cases}\sqrt{x}=a\\\sqrt{3-2x}=b\end{cases}\Rightarrow}a;b\ge0\)

=>   \(\hept{\begin{cases}x=a^2\\3-2x=b^2\end{cases}}\)

=>    \(2a^2+b^2=3\)

KHI ĐÓ PT BAN ĐẦU SẼ ĐƯỢC:     \(9+3ab=7a+5b\)

<=>     \(6+3+3ab=7a+5b\)     (*)

THAY    \(2a^2+b^2=3\)vào PT (*) TA SẼ ĐƯỢC:   

=>    \(2a^2+b^2+3ab+6=2\left(2a+b\right)+3\left(a+b\right)\)

<=>   \(\left(a+b\right)\left(2a+b\right)+6=2\left(2a+b\right)+3\left(a+b\right)\)

<=>    \(\left(a+b-2\right)\left(2a+b-3\right)=0\)

<=>    \(\orbr{\begin{cases}a+b=2\\2a+b=3\end{cases}}\)

TH1:     \(a+b=2\Rightarrow\sqrt{x}+\sqrt{3-2x}=2\)

=>    \(x+3-2x+2\sqrt{x\left(3-2x\right)}=4\)

<=>  \(2\sqrt{3x-2x^2}=x+1\)

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

<=>  \(12x-8x^2=x^2+2x+1\)

<=>  \(9x^2-10x+1=0\)

<=>  \(\left(x-1\right)\left(9x-1\right)=0\)

<=>   \(\orbr{\begin{cases}x=1\\x=\frac{1}{9}\end{cases}}\)

=> TA THẤY CÁC GIÁ TRỊ x đều TMĐK.

BẠN TỰ XÉT NỐT TRƯỜNG HỢP 2:     \(2a+b=3\Rightarrow2\sqrt{x}+\sqrt{3-2x}=3\)      nha

đạt 

\(\hept{\begin{cases}\sqrt{a}=f\\\sqrt{3-2a}=h\end{cases}}\Rightarrow3ab+9=7f+5h\)

\(a,x^2< 1=1^2=>x< 1\) thỏa mãn bất phương trình

\(b,2x+5\ge7=>2x\ge7-5=2=>x\ge1\) thỏa mãn bất phương trình

\(Dk:x,y\ge\frac{-5}{4}\)

\(\left\{{}\begin{matrix}\left(2x-3\right)^2=4y+5\\\left(2y-3\right)^2=4x+5\end{matrix}\right.\Rightarrow\left(2y-3\right)^2-\left(2x-3\right)^2=4x-4y\Leftrightarrow\left(2y-2x\right)\left(2x+2y-6\right)=4\left(x-y\right)\Leftrightarrow4\left(y-x\right)\left(x+y-3\right)=4\left(x-y\right)\Leftrightarrow-4\left(x-y\right)\left(x+y-3\right)=4\left(x-y\right)\)

\(+,x=y\Rightarrow\left(2x-3\right)^2=4x+5\Leftrightarrow4x^2-12x+9=4x+5\Leftrightarrow4x^2-16x+4=0\Leftrightarrow x^2-4x+1=0\)

\(\Delta=16-4=12>0\Rightarrow\left[{}\begin{matrix}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y=2+\sqrt{3}\left(tm\right)\\x=y=2-\sqrt{3}\left(tm\right)\end{matrix}\right.\)

\(+,x\ne y\Rightarrow-4\left(x+y-3\right)=4\Leftrightarrow x+y-3=-1\Leftrightarrow x+y=2\)

\(\Leftrightarrow x=2-y\Rightarrow\left(1-2y\right)^2=4y+5\Leftrightarrow1-4y+4y^2=4y+5\Leftrightarrow4y^2-8y-4=0\Leftrightarrow y^2-2y-1=0;\Delta=\left(-2\right)^2-\left(-1\right).1.4=4-\left(-4\right)=8>0\Rightarrow\left[{}\begin{matrix}x=1+\sqrt{2}\\x=1-\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=1-\sqrt{2};x=1+\sqrt{2}\left(tm\right)\\x=1-\sqrt{2};y=1+\sqrt{2}\left(tm\right)\end{matrix}\right.\)