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a: \(=\dfrac{2x^2+2xy-xy-y^2}{2x^2-2xy-xy+y^2}=\dfrac{\left(x+y\right)\left(2x-y\right)}{\left(x-y\right)\left(2x-y\right)}=\dfrac{x+ỹ}{x-y}\)
b: Sửa đề:\(\dfrac{\left(x+y\right)^2}{2y^2+xy-x^2}\)
\(=\dfrac{\left(x+y\right)^2}{2y^2+2xy-xy-x^2}\)
\(=\dfrac{\left(x+y\right)^2}{\left(x+y\right)\left(2y-x\right)}=\dfrac{x+y}{2y-x}\)
Bạn nhân 2 cả 3 câu rồi phân tích ra hằng đẳng thức là được
a,\(x^2+2xy+7x+7y+y^2+10=\left(x^2+2xy+y^2\right)+7\left(x+y\right)+10\)
\(=\left(x+y\right)^2+2\left(x+y\right)+5\left(x+y\right)+10\)
\(=\left(x+y\right)\left(x+y+2\right)+5\left(x+y+2\right)\)
\(=\left(x+y+2\right)\left(x+y+5\right)\)
b,\(x^2y+xy^2+x+y=2010\Rightarrow xy\left(x+y\right)+x+y=2010\)
\(\Rightarrow12\left(x+y\right)=2010\Rightarrow x+y=167,5\)
Ta có:\(x^2+y^2=x^2+2xy+y^2-2xy=\left(x+y\right)^2-2xy=\left(167,5\right)^2-2.11=28034,25\)
a)\(\left(x^2+4-4x\right)\left(x^2+4+4x\right)\)
b)\(x\left(y+1\right)+\left(y+1\right)=\left(y+1\right)\left(x+1\right)\)
c)\(\left(x+y\right)^2-2\left(x+y\right)=\left(x+y\right)\left(x+y-2\right)\)
Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)
\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)
\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)
Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)
\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)
\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)
\(M=x^2\left(x+y-2\right)-y\left(x+y-2\right)+y+x-2+1\)
\(=1\)
\(N=x^2\left(x-2\right)-xy^2+2xy+2\left(x+y-2\right)+2\)
Ta có : \(x+y-2=0\Rightarrow x+2=-y\)
\(\Rightarrow N=-x^2y-xy^2+2xy+2\)
\(N=-xy\left(x+y-2\right)+2=2\)
\(P=x^3\left(x+y-2\right)+x^2y\left(x+y-2\right)-x\left(x+y-2\right)+3=3\)
Bài làm
a) x2 - 2xy + y2 - zx + yz
= ( x2 - 2xy + y2 ) - ( zx - yz )
= ( x - y )2 - z( x - y )
= ( x - y )( x - y - z )
b) x3 - x2 - 5x + 125
= ( x3 + 125 ) - ( x2 + 5x )
= ( x + 5 )( x2 -.5x + 25 ) - x( x + 5 )
= ( x + 5 )( x2 - 5x + 25 - x )
= ( x + 5 )( x2 - 6x + 25 )
# Học tốt #
câu a nhầm đề à bạn,mk nghĩ -xz chứ ko phải -xy.
\(1\hept{\begin{cases}6x^2-8x+3x-4\\2x\left(3x-4\right)+\left(3x-4\right)\\\left(3x-4\right)\left(2x+1\right)\end{cases}}\)
\(2\hept{\begin{cases}7x^2-7xy-5x+5y+6xy\\7x\left(x-y\right)-5\left(x-y\right)+\frac{6xy\left(x-y\right)}{\left(x-y\right)}\\\left(x-y\right)\left(7x-5+\frac{6xy}{\left(x-y\right)}\right)\end{cases}}\)
\(3\hept{\begin{cases}5x\left(x-y\right)-15\left(x-y\right)\\\left(x-y\right)\left(5x-15\right)\end{cases}}\)
\(4,,2x^2+x=x\left(2x+1\right)\)
\(5\hept{\begin{cases}x^3-4x-3x^2+12\\x\left(x^2-4\right)-3\left(x^2-4\right)\\\left(x+2\right)\left(x-2\right)\left(x-3\right)\end{cases}}\)
\(6\hept{\begin{cases}2x+2y+x^2-y^2\\2\left(x+y\right)+\left(x+y\right)\left(x-y\right)\\\left(x+y\right)\left(2+x-y\right)\end{cases}}\)
\(7\hept{\begin{cases}\left(x^2y-2xy\right)-\left(xy-2y\right)+\left(xy-y\right)\\xy\left(x-2\right)-y\left(x-2\right)+y\left(x-1\right)\\y\left(X-2\right)\left(x-1\right)+y\left(x-1\right)\end{cases}}\Leftrightarrow y\left(x-1\right)\left(x-2+1\right)\)
\(8\hept{\begin{cases}x\left(2-y\right)+z\left(2-y\right)\\\left(2-y\right)\left(x+1\right)\end{cases}}\)
\(P+R=-xy\cdot(x-y)\\\Leftrightarrow R=-xy(x-y)-P\\\Leftrightarrow R=-x^2y+xy^2-(5x^2y-2xy^2+xy-x+y-2)\\\Leftrightarrow R=-x^2y+xy^2-5x^2y+2xy^2-xy+x-y+2\\\Leftrightarrow R=(-x^2y-5x^2y)+(xy^2+2xy^2)-xy+x-y+2\\\Leftrightarrow R=-6x^2y+3xy^2-xy+x-y+2\)
Ta có:
\(P+R=-xy\cdot\left(x-y\right)\)
\(\Leftrightarrow\left(5x^2y-2xy^2+xy-x+y-2\right)+R=-x^2y+xy^2\)
\(\Leftrightarrow R=-x^2y+xy^2-5x^2y+2xy^2+xy+x-y+2\)
\(\Leftrightarrow R=\left(-x^2y-5x^2y\right)+\left(xy^2+2xy^2\right)+xy+x-y+2\)
\(\Leftrightarrow R=-6x^2y+3xy^2+xy+x-y+2\)