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`@` `\text {Ans}`
`\downarrow`
\((x+y)(x-y)+(xy^4-x^3y^2) \div (xy^2) \)
`= x(x-y) + y(x-y) + xy^4 \div xy^2 - x^3y^2 \div xy^2`
`= x^2 - xy + xy - y^2 + y^2 - x^2`
`= (x^2 - x^2) + (-xy + xy) + (-y^2 + y^2)`
`= 0`
1) \(A=\left(x+y\right)^2+4xy=x^2+2xy+y^2+4xy=x^2+6xy+y^2\)
2) \(B=\left(6x-2\right)^2+4\left(3x-1\right)\left(2+y\right)+\left(y+2\right)^2\)
\(=\left(6x-2\right)^2+2\left(6x-2\right)\left(y+2\right)+\left(y+2\right)^2\)
\(=\left(6x-2+y+2\right)^2=\left(6x+y\right)^2=36x^2+12xy+y^2\)
3) \(C=\left(x-y\right)^2+2\left(x^2-y^2\right)+\left(x+y\right)^2\)
\(=\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x-y+x+y\right)^2=\left(2x\right)^2=4x^2\)
Ta có: \(x^2+y^2-z^2\)
\(=\left(x+y\right)^2-z^2-2xy\)
\(=\left(x+y+z\right)\left(x+y-z\right)-2xy\)
\(=-2xy\)
Ta có: \(x^2+z^2-y^2\)
\(=\left(x+z\right)^2-y^2-2xz\)
\(=\left(x+y+z\right)\left(x+z-y\right)-2xz\)
\(=-2xz\)
Ta có: \(y^2+z^2-x^2\)
\(=\left(y+z\right)^2-x^2-2yz\)
\(=\left(x+y+z\right)\left(y+z-x\right)-2yz\)
\(=-2yz\)
Ta có: \(\dfrac{xy}{x^2+y^2-z^2}+\dfrac{xz}{x^2+z^2-y^2}+\dfrac{yz}{y^2+z^2-x^2}\)
\(=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}\)
\(=\dfrac{1}{-2}+\dfrac{1}{-2}+\dfrac{1}{-2}\)
\(=\dfrac{-3}{2}\)
\(\left(x+2\right)^2+4\left(x+2\right)\left(x-2\right)+\left(x-4\right)^2\\ =x^2+4x+4+4x^2-16+x^2-8x+16\\ =6x^2-4x+4\)
(x + 2)2 + 4(x + 2)(x - 2) + (x - 4)2
<=> x2 + 4x + 4 + 4(x2 - 4) + x2 - 8x + 16
<=> x2 + 4x + 4 + 4x2 - 16 + x2 - 8x + 16
<=> x2 + 4x2 + x2 + 4x - 8x + 4 - 16 + 16
<=> 6x2 - 4x + 4
a) (2x^2 +2xy - xy -y^2 ) / (2x^2 - 2xy - xy +y^2)
= 2x(x+y) - y(x+y) / 2x(x-y) - y(x-y)
= (2x-y)(x+y) / (2x-y)(x-y)
= x+y/x-y
Rút gọn cái sau:
\(\frac{32x+4x^2+2x^3}{x^3+64}\)
\(=\frac{2x\left(x^2+2x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}\)
Đề có vẻ sai sai ?
\(A=\left(x-y\right)^2-2\left(x^2-xy-y^2\right)=x^2-2xy+y^2-2x^2+2xy+2y^2\)
\(=-x^2+3y^2\)
\(\frac{x^3+y^3}{x^2+xy+y^2}\)
\(=\frac{\left(x+y\right)\left(x^2+xy+y^2\right)}{x^2+xy+y^2}\)
\(=x+y\)