a/ \(x^2+xy+y^2+1=\left(x^2+xy+\frac{y^2}{4}\right)+\frac{3y^2}{4}+1=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1>0\)

b/ \(x^2+5y^2+2x-4xy-10y+14\)

\(=\left(x^2-4xy+4y^2\right)+2\left(x-2y\right)+1+\left(y^2-6y+9\right)+4\)

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

\(=\left(x-2y+1\right)^2+\left(y-3\right)^2+4>0\)

\(a)\) \(\frac{x^2y-xy}{x-1}=xy\)

\(\Leftrightarrow\)\(\frac{xy\left(x-1\right)}{x-1}=xy\)

\(\Leftrightarrow\)\(xy=xy\) ( đpcm ) 

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

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

\(\Leftrightarrow\)\(\frac{x+y}{x^2+xy^2}=\frac{1}{x}\)

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

\(\Leftrightarrow\)\(x^2+xy=x^2+xy^2\)

\(\Leftrightarrow\)\(xy=xy^2\)

\(\Leftrightarrow\)\(y=y^2\) ( đề sai hay mình sai =.= ) 

Chúc bạn học tốt ~ 

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

b,Sửa đề \(\frac{x^2-y^2}{x^2+xy}=\frac{x-y}{x}\)

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

Lời giải:

Ta có:

$x^4+y^4+(x+y)^4=(x^4+y^4+2x^2y^2)-2x^2y^2+[(x+y)^2]^2$

$=(x^2+y^2)^2-2x^2y^2+(x^2+2xy+y^2)^2$
$=(x^2+y^2)^2-2x^2y^2+(x^2+y^2)^2+(2xy)^2+4xy(x^2+y^2)$

$=2(x^2+y^2)^2+2x^2y^2+4xy(x^2+y^2)$

$=2[(x^2+y^2)^2+2xy(x^2+y^2)+(xy)^2]$

$=2(x^2+y^2+xy)^2$

Ta có đpcm.

Đặt \(xy-12x+15y\)là (*)

Từ phương trình (1) ta có \(x^2-3xy+2y^2+x-y=0\Leftrightarrow\left(x-y\right)\left(x-2y\right)+\left(x-y\right)=0\)

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

Với \(x=y\)thay vào (2) ta có \(x^2-2x^2+x^2-5x+7x=0\Leftrightarrow x=0\Rightarrow x=y=0\)

Thay \(x=y=0\)vào (*) ta thấy 0.0-12.0+15.0=0(tm)

Với \(x=2y-1\Rightarrow\left(2y-1\right)^2-2\left(2y-1\right)y+y^2-5\left(2y-1\right)+7y=0\)

\(\Leftrightarrow4y^2-4y+1-4y^2+2y+y^2-10y+5+7y=0\)

\(\Leftrightarrow y^2-5y+6=0\Leftrightarrow\left(y-2\right)\left(y-3\right)=0\Leftrightarrow\orbr{\begin{cases}y=2\\y=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}}\)

Với \(x=3;y=2\)thay vào (*)  ta thấy \(3.2-12.3+15.0=0\left(tm\right)\)

Với \(x=5;y=3\)thay vào (*)  ta thấy \(5.3-12.5+15.3=0\left(tm\right)\)

Vậy .....

2314654564

Hơn 170 điểm đã  là CTV rồi 

_______________Bài làm___________________

a, \(x^2+xy+y^2+1\)

\(=\left(x^2+2x\dfrac{y}{2}+\dfrac{y^2}{4}\right)+\dfrac{3y^2}{4}+1=\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^3}{4}+1\)

Do \(\left(x+\dfrac{y}{2}\right)^2\ge0\forall x,y\)

Và \(\dfrac{3y^2}{4}\ge0\forall y\)

Nên: \(\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1>0\forall x,y=>đpcm\)

b, \(x^2+5y^2+2x-4xy-10y+14\)

\(=\left(x^2-4xy+4y^2\right)+\left(2x-4y\right)+\left(y^2-6y+9\right)+5\)

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

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

Do \(\left(x-2y+1\right)^2\ge0\forall x,y\)

Và \(\left(y-3\right)^2\ge0\forall y\)

Nên \(\left(x-2y+1\right)^2+\left(y-3\right)^2+4>0\)

c, \(5x^2+10y^2-6xy-4x-2y+3\)

\(=\left(x^2-6xy+9y^2\right)+\left(4x^2-2x+1\right)+\left(y^2-2y+1\right)+1\)

\(=\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1\)

Do .........

tự làm ik

Đáp án D

Dễ thấy:

     \(VT\ge\left(x+y\right)^2+1-\dfrac{\left(x+y\right)^2}{4}=\dfrac{3\left(x+y\right)^2}{4}+1\)

Áp dụng Cô-si:

     \(\dfrac{3\left(x+y\right)^2}{4}+1\ge2\sqrt{\dfrac{3\left(x+y\right)^2}{4}.1}=\sqrt{3}\left|x+y\right|\ge\sqrt{3}\left(x+y\right)\)

Do đó:

     \(\left(x+y\right)^2+1-xy\ge\sqrt{3}\left(x+y\right),\forall x,y\in R\)

\(ĐK:x\ne y;x\ne-y;x^2+xy+y^2\ne0;x^2-xy+y^2\ne0\)

\(A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\left[1:\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)}\right]\\ A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+y^2\right)}\\ A=x-y=B\)

\(x=0;y=0\Leftrightarrow B=0\)

Giá trị của A không xác định vì \(x=y\) trái với ĐK:\(x\ne y\)

Vậy \(A\ne B\)