Sửa \(\frac{x}{x+3}-\frac{x-2}{2x-6}=\frac{2x+2}{x^2-9}\)ĐK : \(x\ne\pm3\)

\(\Leftrightarrow\frac{2x\left(x-3\right)}{2\left(x+3\right)\left(x-3\right)}-\frac{\left(x-2\right)\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{4x+4}{2\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow2x^2-6x-x^2-3x+2x+6=4x+4\)

\(\Leftrightarrow x^2-7x+6-4x-4=0\)

\(\Leftrightarrow x^2-11x+2=0\)( bạn kiểm tra lại đề nhé ! )

Ta có: \(x^2+5y^2+2y-4xy-3=0\)

   \(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(y^2+2y+1\right)=4\)

   \(\Leftrightarrow\left(x-4y\right)^2+\left(y+1\right)^2=4\)

Vì \(\hept{\begin{cases}\left(x-4y\right)^2\ge0\forall x,y\\\left(y+1\right)^2\ge0\forall y\end{cases}}\)\(\Rightarrow\)\(\left(x-4y\right)^2+\left(y+1\right)^2\ge0\forall x,y\)

mà \(\left(x-4y\right)^2+\left(y+1\right)^2=4\)\(\Rightarrow\)\(0\le\left(x-4y\right)^2+\left(y+1\right)^2\le4\forall x,y\)

Vì \(x,y\in Z\)\(\Rightarrow\)\(\hept{\begin{cases}\left(x-4y\right)^2\inℤ\\\left(y+1\right)^2\inℤ\end{cases}}\)

+) \(\hept{\begin{cases}\left(x-4y\right)^2=0\\\left(y+1\right)^2=4\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x-4y=0\\y+1=2\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=4\\y=1\end{cases}}\)( TM )

+) \(\hept{\begin{cases}\left(x-4y\right)^2=1\\\left(y+1\right)^2=3\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x-4y=1\\y+1=\sqrt{3}\end{cases}}\)( loại )

+) \(\hept{\begin{cases}\left(x-4y\right)^2=2\\\left(y+1\right)^2=2\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x-4y=\sqrt{2}\\y+1=\sqrt{2}\end{cases}}\)( loại )

+) \(\hept{\begin{cases}\left(x-4y\right)^2=4\\\left(y+1\right)^2=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x-4y=2\\y+1=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=-2\\y=-1\end{cases}}\)( TM )

Vậy \(\left(x,y\right)\in\left\{\left(-2,-1\right);\left(4,1\right)\right\}\)

\(x^2+5y^2+2y-4xy-3=0\)

\(\Leftrightarrow x^2-4xy+4y^2+y^2+2y+1-4=0\)

\(\Leftrightarrow\left(x-2y\right)^2+\left(y+1\right)^2=4\)

Vì \(\hept{\begin{cases}\left(x-2y\right)^2\ge0\\\left(y+1\right)^2\ge0\end{cases}\Leftrightarrow0\le\left(x-2y\right)^2+\left(y+1\right)^2\le4}\)

Ta xét lần lượt là ra nha

ko biết

Đặt A = \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\)

A = \(\left(1+\frac{a+b}{a}\right)\left(1+\frac{a+b}{b}\right)\)(Vì a + b = 1)

A = \(\left(2+\frac{b}{a}\right)\left(2+\frac{a}{b}\right)\)

A = \(4+\frac{2a}{b}+\frac{2b}{a}+1\)

A = \(5+2\left(\frac{a}{b}+\frac{b}{a}\right)\)

Vì a, b dương nên áp dụng BĐT Cô - si cho 2 số dương, ta được :

\(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{ab}{ba}}\)

\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}\ge2.1=2\)

\(\Leftrightarrow2\left(\frac{a}{b}+\frac{b}{a}\right)\ge4\)

\(\Leftrightarrow5+2\left(\frac{a}{b}+\frac{b}{a}\right)\ge4+5\)

\(\Leftrightarrow A\ge9\)

Dấu bằng xảy ra \(\Leftrightarrow\)a = b > 0

Vậy \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\ge9\)với a, b là các số dương và a + b = 1

Tớ quên. Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}a=b>0\\a+b=1\end{cases}}\)

\(\Leftrightarrow a=b=\frac{1}{2}\)

Đặt A =\(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\)

Vì a + b \(\ne\)0 nên A luôn được xác định.

 Giả sử \(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge2\)

\(\Leftrightarrow\frac{\left(a^2+b^2\right)\left(a+b\right)^2}{\left(a+b\right)^2}+\frac{\left(ab+1\right)^2}{\left(a+b\right)^2}-\frac{2\left(a+b\right)^2}{\left(a+b\right)^2}\ge0\)

\(\Leftrightarrow\left(a^2+b^2\right)\left(a+b\right)^2+\left(ab+1\right)^2-2\left(a+b\right)^2\ge0\)(vì a + b \(\ne\)0)

\(\Leftrightarrow[\left(a^2+2ab+b^2\right)-2ab]\left(a+b\right)^2+\left(ab+1\right)^2-2\left(a+b\right)^2\ge0\)

\(\Leftrightarrow[\left(a+b\right)^2-2ab]\left(a+b\right)^2+\left(ab+1\right)^2-2\left(a+b\right)^2\ge0\)

\(\Leftrightarrow\left(a+b\right)^4-2ab\left(a+b\right)^2+\left(ab+1\right)^2-2\left(a+b\right)^2\ge0\)

\(\Leftrightarrow\left(a+b\right)^4-\left[2ab\left(a+b\right)^2+2\left(a+b\right)^2\right]+\left(ab+1\right)^2\ge0\)

\(\Leftrightarrow\left[\left(a+b\right)^2\right]^2-2\left(a+b\right)^2\left(ab+1\right)+\left(ab+1\right)^2\ge0\)

\(\left[\left(a+b\right)^2-\left(ab+1\right)^2\right]^2\ge0\)(luôn đúng)

Dấu bằng xảy ra 

\(\Leftrightarrow\hept{\begin{cases}a+b\ne0\\\Leftrightarrow a=b\end{cases}}\Leftrightarrow a=b\left(a,b\ne0\right)\)

Vậy \(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge\)2 với a, b là các số thỏa mãn a+b \(\ne\)0

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}a=b\\a+b\ne0\end{cases}\Leftrightarrow a=b}\)(a,b \(\ne\)0)

Vậy \(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge2\) với a, b là các số thỏa mãn \(a+b\ne0\)