\(\frac{1}{x+2}+\frac{5}{x-2}=\frac{3x-12}{x^2-4}\)

ĐKXĐ \(x\ne\pm2\)

\(\frac{1}{x+2}+\frac{5}{x-2}=\frac{3x-12}{x^2-4}\)

\(\Leftrightarrow\frac{x-2+5\left(x+2\right)}{x^2-4}=\frac{3x-12}{x^2-4}\)

\(\Leftrightarrow x-2+5x+10=3x-12\)

\(\Leftrightarrow6x+8=3x-12\)

\(\Leftrightarrow3x=20\Leftrightarrow x=\frac{20}{3}\left(tm\right)\)

Đặt  \(A=x^2+y^2+2x+4y+16\)

\(A=\left(x^2+2x+1\right)+\left(y^2+4y+4\right)+11\)

\(A=\left(x+1\right)^2+\left(y+2\right)^2+11\)

Mà  \(\left(x+1\right)^2\ge0\forall x\)

       \(\left(y+2\right)^2\ge0\forall y\)

\(\Rightarrow A\ge11\)

Dấu "=" xảy ra khi :

\(\hept{\begin{cases}x+1=0\\y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-2\end{cases}}\)

Vậy  \(A_{Min}=11\Leftrightarrow\left(x;y\right)=\left(-1;-2\right)\)

tính HD nha không phải AD hifhi mk nhầm

Làm theo hằng đẳng thức nhé :) Tính thì bạn tự tính :)) 

\(a)\) \(\left(\frac{1}{2}x+y\right)\left(\frac{1}{2}x+y\right)=\left(\frac{1}{2}x+y\right)^2=\frac{1}{4}x^2+xy+y^2\)

\(b)\) \(\left(x-\frac{1}{2}y\right)\left(x-\frac{1}{2}y\right)=\left(x-\frac{1}{2}y\right)^2=x^2-xy+y^2\)

\(c)\) \(\left(a+b\right)^2=a^2+2ab+b^2\)

\(d)\) \(\left(a-b\right)^2=a^2-2ab+b^2\)

\(e)\) \(\left(a+b\right)\left(a-b\right)=a^2-b^2\)

Chúc bạn học tốt ~ 

cảm ơn bạn nha

\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2abc\left(a+b+c\right)\)

thay \(\hept{\begin{cases}a+b+c=0\\a^2+b^2+c^2=0\end{cases}}\)

=>\(a^4+b^4+c^4=14^2-2abc.0=196\)