Rút gọn biểu thức \(\frac{\left|x\right|+\left|y\right|}{x+y}\)
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quy đồng mẫu thức ta được
\(\frac{yz\left(z-y\right)+xz\left(x-z\right)+xy\left(y-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)\(=\frac{yz\left(z-y\right)+xz\left(x-z\right)-xy\left[\left(z-y\right)+\left(x-z\right)\right]}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{y\left(z-y\right)\left(z-x\right)+x\left(x-z\right)\left(z-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(z-y\right)\left(z-x\right)\left(y-x\right)}{xyz\left(z-y\right)\left(z-x\right)\left(y-x\right)}=\frac{1}{xyz}\)
a: \(A=\dfrac{x^5}{x^3}\cdot\dfrac{y^{-2}}{y}=x^2\cdot y^{-1}=\dfrac{x^2}{y}\)
b: \(B=\dfrac{x^2\cdot y^{-3}}{x^3\cdot y^{-12}}=\dfrac{x^2}{x^3}\cdot\dfrac{y^{-3}}{y^{-12}}=\dfrac{1}{x}\cdot y^{-3+12}=\dfrac{y^9}{x}\)
a) \(A=\dfrac{x^5y^{-2}}{x^3y}=\dfrac{x^5}{x^3}.\dfrac{1}{y^{2-1}}=x^{5-3}y^{-1}=x^2y^{-1}\).
b) \(B=\dfrac{x^2y^{-3}}{\left(x^{-1}y^4\right)^{-3}}=\dfrac{x^2y^{-3}}{x^3y^{-12}}=x^{2-3}y^{-3-\left(-12\right)}=\dfrac{1}{xy^9}\)
\(\left(x+y\right)^2+\left(x-y\right)^2+\left(x-y\right)\left(x+y\right)-3x^2\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)+\left(x^2-y^2\right)-3x^2\)
\(=x^2+2xy+y^2+x^2-2xy+y^2+x^2-y^2-3x^2\)
\(=3x^2+y^2-3x^2\)
\(=y^2\)
Dat (x-y)2+(y-z)2+(x-z)2=A
=(x+y)3+z3-3x2y-3xy2-3xyz / A
=(x+y+z).(x2+2xy+y2-xy-yz+z2)-3xy(x+y+z) / A
=(x+y+z).(x2+y2+z2-xy-yz-xz) /A
=2(x+y+z).(x2+y2+z2-xy-yz-xz) /2A
=(x+y+z)[ (x2-2xy+y2)+(y2-2yz+z2)+(x2-2xz+z2) / 2A
=(x+y+z).[ (x-y}2+(y-z)2+(x-z)2 ] /2A
=(x+y+z). A /2A
=x+y+z /2
\(\frac{x\left(x+5\right)+y\left(y+5\right)+2\left(xy-3\right)}{x\left(x+6\right)+y\left(y+6\right)2xy}\)
=\(\frac{x^2+5x+y^2+5y+2xy-3}{x^2+6x+y^2+6y+2xy}\)
triệt tiêu x2;y2;2xy ta được:
\(\frac{5x+5y-3}{6x+6y}=\frac{5\left(x+y\right)-3}{6\left(x+y\right)}\)
=\(\frac{5.2010-3}{6.2010}=\frac{3349}{4020}\)
\(D=\frac{x^6-y^6}{\left(x-y\right)\left(x^4+x^2y^2+y^4\right)}=\frac{\left(x^2\right)^3-\left(y^2\right)^3}{\left(x-y\right)\left(x^4+x^2y^2+y^4\right)}=\frac{\left(x^2-y^2\right)\left[\left(x^2\right)^2+x^2y^2+\left(y^2\right)^2\right]}{\left(x-y\right)\left(x^4+x^2y^2+y^4\right)}=\frac{\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)}{\left(x-y\right)\left(x^4+x^2y^2+y^4\right)}=\frac{x^2-y^2}{x-y}\)
Gọn thế này được chưa???