a/\(\Leftrightarrow\left(12x^2+12x+11\right)\left(y^2-2y+2\right)=\left(4x^2+4x+3\right)\left(5y^2-10y+9\right)\)

\(\Leftrightarrow12x^2y^2-24x^2y+24x^2+12xy^2-24xy+24x+11y^2-22y+22=20x^2y^2-40x^2y+36x^2+20xy^2-40xy+36x+15y^2-30y+36\)

Có sai đề ko cậu

a/ ĐKXĐ: ...

Đặt \(x^2-x=t\)

\(\frac{t}{t+1}-\frac{t+2}{t-2}=1\Leftrightarrow t\left(t-2\right)-\left(t+1\right)\left(t+2\right)=\left(t+1\right)\left(t-2\right)\)

\(\Leftrightarrow t^2+4t=0\Rightarrow\left[{}\begin{matrix}t=0\\t=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-x=0\\x^2-x=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0;1\\x^2-x+4=0\left(vn\right)\end{matrix}\right.\)

b.

\(\Leftrightarrow\frac{3\left(2x+1\right)^2+8}{\left(2x+1\right)^2+2}=\frac{5\left(y-1\right)^2+4}{\left(y-1\right)^2+1}\)

Đặt \(\left\{{}\begin{matrix}2x+1=a\\y-1=b\end{matrix}\right.\)

\(\Rightarrow\frac{3a^2+8}{a^2+2}=\frac{5b^2+4}{b^2+1}\Leftrightarrow\left(3a^2+8\right)\left(b^2+1\right)=\left(a^2+2\right)\left(5b^2+4\right)\)

\(\Leftrightarrow3a^2b^2+3a^2+8b^2=5a^2b^2+4a^2+10b^2\)

\(\Leftrightarrow2a^2b^2+a^2+2b^2=0\Leftrightarrow\left\{{}\begin{matrix}a=0\\b=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\y-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=1\end{matrix}\right.\)

\(\Leftrightarrow x^4-4x^3+12x^2-32x+32=\left(y-5\right)^2\)

\(\Leftrightarrow\left(x-2\right)^2\left(x^2+8\right)=\left(y-5\right)^2\)

- Với \(x=2\Rightarrow y=5\)

- Với \(x\ne2\Rightarrow x-2\) là ước của \(y-5\) 

Đặt \(y-5=n\left(x-2\right)\)

\(\Rightarrow\left(x-2\right)^2\left(x^2+8\right)=n^2\left(x-2\right)^2\)

\(\Rightarrow x^2+8=n^2\)

\(\Rightarrow\left(n-x\right)\left(n+x\right)=8\)

\(\Rightarrow\left[{}\begin{matrix}x=1;n=-3\Rightarrow y=8\\x=-1;n=-3\Rightarrow y=14\\x=1;n=3\Rightarrow y=2\\x=-1;n=3\Rightarrow y=-4\end{matrix}\right.\) 

Lời giải:

PT $\Leftrightarrow (4x^2+y^2-4xy)+9y^2+12x+6y+13=0$

$\Leftrightarrow (2x-y)^2+6(2x-y)+9y^2+12y+13=0$

$\Leftrightarrow (2x-y)^2+6(2x-y)+9+(9y^2+12y+4)=0$

$\Leftrightarrow (2x-y+3)^2+(3y+2)^2=0$

$\Rightarrow (2x-y+3)^2=(3y+2)^2=0$

$\Rightarrow y=-\frac{2}{3}; x=\frac{-11}{6}$

`a, <=> 5/3 . 3sqrt(x^2+2) + 3/2.2sqrt(x^2+2)-7sqrt6=sqrt(x^2+2)`

`= (5+3-1)sqrt(x^2+2)=7sqrt6`

`<=> 7sqrt(x^2+2)=7sqrt6`.

`<=> x^2+2=36`.

`<=> x^2=34`.

`<=> x=+-sqrt(34)`.

Vậy...

`b, sqrt(4x^2-12x+9)-6=0`

`<=> |2x-3|=6`.

`@ x >=3/2 <=> 2x-3=6.`

`<=> x=9/2 (tm)`.

`@x <3/2 <=> 3-2x=6`

`<=> 2x=-3`

`<=> x=-3/2.`

Vậy...

b) \(1+4x-3|x+2|+4=0\)

\(\Leftrightarrow4x-3|x+2|=-5\left(1\right)\)

TH1: Với \(|x+2|=x+2\)thay vào (1) ta được:

\(4x-3\left(x+2\right)=-5\)

\(\Leftrightarrow4x-3x-6=-5\)

\(\Leftrightarrow x=1\)(chọn tự thử lại nhé nó =0 )

TH2: Với \(|x+2|=-x-2\)thay vào (1) ta được: 

\(4x-3\left(-x-2\right)=-5\)

\(\Leftrightarrow4x+3x+6=-5\)

\(\Leftrightarrow7x=-11\)

\(\Leftrightarrow x=\frac{-11}{7}\)( loại tự thử lại nhé nó ko =0 )

Vậy x=1