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1: =(16x^2-8x+1)-y^2
=(4x-1)^2-y^2
=(4x-1-y)(4x-1+y)
2: =(x^2-2xy+y^2)-z^2
=(x-y)^2-z^2
=(x-y-z)(x-y+z)
3: =(x^2+4xy+4y^2)-16
=(x+2y)^2-4^2
=(x+2y-4)(x+2y+4)
4: =(x^2-4xy+4y^2)-16
=(x-2y)^2-4^2
=(x-2y-4)(x-2y+4)
3x^2+3y^2+4xy-2x+2y+2=0
=>2x^2+4xy+2y^2+x^2-2x+1+y^2+2y+1=0
=>x=1 và y=-1
M=(1-1)^2017+(1-2)^2018+(-1+1)^2015=1
\(P=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\left(\frac{1}{4xy}+4xy\right)\)
\(\Rightarrow P\ge\frac{4}{x^2+y^2+2xy}+2\sqrt{\frac{4xy}{4xz}}=\frac{4}{1^2}+4=8\)
Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)
\(\dfrac{1}{x^2+2xy+y^2}-\dfrac{1}{x^2-y^2}:\dfrac{4xy}{y^2-x^2}\) \(\left(x,y\ne0;x\ne\pm y\right)\)
\(=\dfrac{1}{\left(x+y\right)^2}+\dfrac{1}{y^2-x^2}.\dfrac{y^2-x^2}{4xy}\)
\(=\dfrac{1}{x^2+2xy+y^2}+\dfrac{1}{4xy}\)
\(=\dfrac{6xy+x^2+y^2}{4xy\left(x+y\right)^2}\)
Ta có: \(\dfrac{1}{x^2+2xy+y^2}-\dfrac{1}{x^2-y^2}:\dfrac{4xy}{y^2-x^2}\)
\(=\dfrac{1}{\left(x+y\right)^2}+\dfrac{1}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x+y\right)\left(x-y\right)}{4xy}\)
\(=\dfrac{1}{\left(x+y\right)^2}+\dfrac{1}{4xy}\)
\(=\dfrac{4xy}{4xy\left(x+y\right)^2}+\dfrac{x^2+2xy+y^2}{4xy\left(x+y\right)^2}\)
\(=\dfrac{x^2+6xy+y^2}{4xy\left(x+y\right)^2}\)
\(1,=\left(x-2\right)\left(5-y\right)\\ 2,=2\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(2x-2y-z\right)\\ 3,=5xy\left(x-2y\right)\\ 4,=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-4z^2\right]\\ =3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\\ 6,=-\left(6x^2-3x-4x+2\right)=-\left(2x-1\right)\left(3x-2\right)\\ 7,=\left(2x+y\right)\left(2x+y+x\right)=\left(2x+y\right)\left(3x+y\right)\\ 8,=\left(x-y\right)\left(x+5\right)\\ 9,=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 10,=\left(x^2-9\right)x=x\left(x-3\right)\left(x+3\right)\\ 11,=\left(x-2\right)\left(y+1\right)\\ 12,=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\\ 13,=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)
Câu 2:
\(B=x^2+2x+y^2-2x-2xy+37\)
\(=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=7^2+2\cdot7+37=49+37+14=100\)
Câu 3:
\(C=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2\cdot5+10=25\)
Ta có (x - y)2 - (x + y)2
= (x - y + x + y)(x - y - x - y)
= 2x.(-2y)
= -4xy (đpcm)
VT=(x-y)2-(x+y)2
=x2-2xy+y2-x2-2xy-y2
=-4xy=VP (đccm)
#H