cmr
(x-y)2 + 4xy = (x+y)2
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a) \(\left(x-y\right)^2+2xy\)
\(=x^2-2xy+y^2+2xy\)
\(=x^2+y^2\left(đpcm\right)\)
b) \(\left(x-y\right)^2+4xy\)
\(=x^2-2xy+y^2+4xy\)
\(=x^2+2xy+y^2\)
\(=\left(x+y\right)^2\left(đpcm\right)\)
a, Ta có:\(\left(x-y\right)^2=x^2-2xy+y^2\)
\(\Leftrightarrow x^2+y^2=\left(x-y\right)^2+2xy\left(ĐCCM\right)\)
b,Ta có:\(\left(x+y\right)^2=x^2+2xy+y^2\)
\(\Leftrightarrow\left(x+y\right)^2=x^2-2xy+4xy+y^2\)
\(\Leftrightarrow\left(x+y\right)^2=\left(x-y\right)^2+4xy\left(ĐCCM\right)\)
Vt = (x - y)^2 + 4xy = x^2 -2xy + y^2 + 4xy = x^2 +2xy+ y^2 = ( x+y)^2 = VP
=> ĐPCM
b, (x + y)^2 = ( x - y)^2 + 4xy = 5^2 + 4.3 = 25 + 12 = 37
Ah đã có mặt :)
\(\left(1+x^2\right)\left(1+y^2\right)+4xy+2\left(x+y\right)\left(1+xy\right)\)
\(=1+x^2+y^2+x^2y^2+4xy+2\left(x+y\right)\left(1+xy\right)\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2y^2+2xy+1\right)+2\left(x+y\right)\left(1+xy\right)\)
\(=\left(x+y\right)^2+\left(1+xy\right)^2+2\left(x+y\right)\left(1+xy\right)\)
\(=\left(x+y+xy+1\right)^2\)là số CP (đpcm)
a)
VT=(x-y)2+4xy=x2-2xy+y2+4xy=x2+2xy+y2=(x+y)2=VP
=> (x-y)2+4xy=(x+y)2
b) (x+y)2=x2+2xy+y2
=x2-2xy+y2+4xy
=(x-y)2+4xy
=52+4.3
=25+12
=37
a ) Ta có :
\(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]\)
\(=\left(2x\right)\left(2y\right)\)
\(=4xy\)
\(\Rightarrow DPCM\)
\(f\left(x,y\right)=\left(x^2+4y^2-4xy\right)+\left(2x-4y\right)+1+\left(y^2-2y+1\right)+1\)
\(f\left(x,y\right)=\left(x-2y\right)^2+2\left(x-2y\right)+1+\left(y-1\right)^2+1\)
\(f\left(x,y\right)=\left(x-2y+1\right)^2+\left(y-1\right)^2+1\)
\(\left\{{}\begin{matrix}\left(x-2y+1\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\)=> f(x;y) >=1 >0 => dpcm
\(VT=\left(x-y\right)^2+4xy=x^2-2xy+y^2+4xy\)
\(=x^2+2xy+y^2=\left(x+y\right)^2=VP\)
p/s: chúc bạn học tốt
(x-y)2+4xy =x2-2xy+y2 +4xy=x2+2xy+y2=(x+y)2