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a: \(\dfrac{1}{2a}+\dfrac{2}{3b}\)(ĐKXĐ: a<>0 và b<>0)
\(=\dfrac{1\cdot3b+2\cdot2a}{2a\cdot3b}\)
\(=\dfrac{3b+4a}{6ab}\)
b: \(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}\)(ĐKXĐ: \(x\notin\left\{1;-1\right\}\))
\(=\dfrac{\left(x-1\right)^2-\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^2-2x+1-x^2-2x-1}{\left(x+1\right)\left(x-1\right)}=\dfrac{-4x}{x^2-1}\)
c: \(\dfrac{x+y}{xy-y}+\dfrac{z}{yz}\)(ĐKXĐ: \(\left\{{}\begin{matrix}x< >1\\y< >0\\z< >0\end{matrix}\right.\))
\(=\dfrac{x+y}{y\left(x-1\right)}+\dfrac{1}{y}\)
\(=\dfrac{x+y+x-1}{y\left(x-1\right)}\)
\(=\dfrac{2x+y-1}{y\left(x-1\right)}\)
d: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
\(\dfrac{2}{x-3}-\dfrac{12}{x^2-9}\)
\(=\dfrac{2}{x-3}-\dfrac{12}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x+6-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)
e: ĐKXĐ: x<>2
\(\dfrac{1}{x-2}+\dfrac{2}{x^2-4x+4}\)
\(=\dfrac{1}{x-2}+\dfrac{2}{\left(x-2\right)^2}\)
\(=\dfrac{x-2+2}{\left(x-2\right)^2}=\dfrac{x}{\left(x-2\right)^2}\)
`a, a/(a-3) - 3/(a+3) = (a(a+3) - 3(a-3))/(a^2-9)`
`= (a^2+9)/(a^2-9)`
`b, 1/(2x) + 2/x^2 = x/(2x^2) + 4/(2x^2) = (x+4)/(2x^2)`
`c, 4/(x^2-1) - 2/(x^2+x) = (4x)/(x(x-1)(x+1)) - (2(x-1))/(x(x+1)(x-1))`
`= (2x+2)/(x(x-1)(x+1)`
`= 2/(x(x-1))`
a: \(=\dfrac{3b+4a}{6ab}\)
b: \(=\dfrac{x^2-2x+1-x^2-2x-1}{x^2-1}=\dfrac{-4x}{x^2-1}\)
c: \(=\dfrac{xz+yz-xy-xz}{xyz}=\dfrac{yz-xy}{xyz}=\dfrac{z-x}{xz}\)
d: \(=\dfrac{2x+6-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)
e: \(=\dfrac{x-2+2}{\left(x-2\right)^2}=\dfrac{x}{\left(x-2\right)^2}\)
Câu 1 : x2-y2+2yz-z2=-(y2-2yz+z2-x2) Câu 2: x2-2xy+y2-xz+yz=(x2-2xy+y2)-xz+yz
=-(y-z)2 -x2 =(x-y)2-z(x-y)
=-(y-z-x)(y-z+x) =(x-y)(x-y-z)
a, \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}=\frac{x+1}{2\left(x+3\right)}+\frac{3x+2}{x\left(x+3\right)}\)
\(=\frac{x^2+x}{2x\left(x+3\right)}+\frac{6x+4}{2x\left(x+3\right)}=\frac{x^2+7x+4}{2x\left(x+3\right)}\)
b, Sua de : \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)
\(=\frac{3x}{2x\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{2x+6}{2x\left(x+3\right)}=\frac{1}{x}\)
\(\dfrac{1}{x^2-x+1}-\dfrac{x^2+2}{x^3+1}=\dfrac{1}{x^2-x+1}-\dfrac{x^2+2}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x^2+2}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x+1-x^2-2}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
cảm ơn :))