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\(a,\dfrac{x+2}{x-1}-\dfrac{x-3}{x-1}-\dfrac{x-4}{1-x}\\ =\dfrac{x+2}{x-1}-\dfrac{x-3}{x-1}+\dfrac{x-4}{x-1}\\ =\dfrac{x+2-x+3+x-4}{x-1}\\ =\dfrac{x+1}{x-1}\)
\(b,\dfrac{1}{x+5}-\dfrac{1}{x-5}+\dfrac{2x}{x^2-25}\\ =\dfrac{1}{x+5}-\dfrac{1}{x-5}+\dfrac{2x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{x-5-x-5+2x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2x-10}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2}{x+5}\)
\(c,x+\dfrac{2y^2}{x+y}-y\\ =\dfrac{x\left(x+y\right)+2y^2-y\left(x+y\right)}{x+y}\\ =\dfrac{x^2+xy+2y^2-xy-y^2}{x+y}\\ =\dfrac{x^2+y^2}{x+y}\)
a: \(=2x+x^3-5x^4\)
b: \(=\dfrac{8x^2+4x-7x-3}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{8x^2-3x-3}{\left(2x-1\right)\left(2x+1\right)}\)
`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))`
\(a,\dfrac{8y}{3x^2}.\dfrac{9x^2}{4y^2}=\dfrac{72x^2y}{12x^2y^2}=\dfrac{6}{y}\\b,\dfrac{3x+x^2}{x^2+x+1}.\dfrac{3x^3-3}{x+3}=\dfrac{x\left(x+3\right)3\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x+3\right)}=3x\left(x-1\right)=3x^2-3x \)
\(c,\dfrac{2x^2+4}{x-3}.\dfrac{3x+1}{x-1}.\dfrac{6-2x}{x^2+2}=\dfrac{2\left(x^2+2\right)\left(3x+1\right)2\left(3-x\right)}{\left(x-3\right)\left(x-1\right)\left(x^2+2\right)}=\dfrac{-4\left(3x+1\right)}{x-1}=\dfrac{-12x-4}{x-1}\)
\(d,\dfrac{2x^2}{3y^3}:\left(-\dfrac{4x^3}{21y^2}\right)=\dfrac{-2x^2.21y^2}{3y^3.4x^3}=\dfrac{-42x^2y^2}{12x^3y^3}=\dfrac{-7}{2xy}\)
\(e,\dfrac{2x+10}{x^3-64}:\dfrac{\left(x+5\right)^2}{2x-8}=\dfrac{2\left(x+5\right)}{\left(x-4\right)\left(x^2+4x+16\right)}.\dfrac{2\left(x-4\right)}{\left(x+5\right)^2}=\dfrac{4}{\left(x+5\right)\left(x^2+4x+16\right)}=\dfrac{4}{x^3+9x^2+16x+80}\)
\(f,\dfrac{1}{x+y}\left(\dfrac{x+y}{xy}-x-y\right)-\dfrac{1}{x^2}:\dfrac{y}{x}=\dfrac{1}{x+y}\left(\dfrac{\left(x+y\right)\left(1-xy\right)}{xy}\right)-\dfrac{x}{x^2y}=\dfrac{1-xy}{xy}-\dfrac{x}{x^2y}=\dfrac{-x^2y}{x^2y}=-1\)
\(a,=\dfrac{4y.5x^3}{3x^2.2y^3}=\dfrac{20x^3y}{6x^2y^3}=\dfrac{10x}{3y^2}\\ b,=\dfrac{\left(x-1\right)^2.x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2.x.\left(x+1\right)}{\left(x-1\right)^2.\left(x+1\right)}=x\)
\(c,=\dfrac{x\left(2+x\right).3\left(x^3+1\right)}{\left(x^2-x+1\right).3.\left(x+2\right)}=\dfrac{3x.\left(x+2\right).\left(x+1\right)\left(x^2-x+1\right)}{\left(x^2-x+1\right).3\left(x+2\right)}=x\left(x+1\right)\)
\(\left(\dfrac{x}{x+1}+\dfrac{x-1}{x}\right):\left(\dfrac{x}{x+1}-\dfrac{x-1}{x}\right)\) \(\left(đk:x\ne0;-1\right)\)
\(=\dfrac{x^2+\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}:\left(\dfrac{x^2-\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}\right)\)
\(=\dfrac{x^2+x^2-1}{x\left(x+1\right)}.\dfrac{x\left(x+1\right)}{x^2-x^2+1}\)
\(=\dfrac{\left(2x^2-1\right)x\left(x+1\right)}{x\left(x+1\right)}=2x^2-1\)
a: \(=\dfrac{2x^2-1-x^2-3}{x-2}=\dfrac{x^2-4}{x-2}=x+2\)
b: \(=\dfrac{x\left(x-y\right)+y\left(x+y\right)}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{x^2-xy+xy+y^2}{x^2-y^2}=\dfrac{x^2+y^2}{x^2-y^2}\)
c: \(=\dfrac{x+1-2}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x+1}\)
d: \(=\dfrac{\left(x+2\right)\cdot y-x\left(y-2\right)}{xy\left(x+y\right)}\)
\(=\dfrac{2y+2x}{xy\left(x+y\right)}=\dfrac{2}{xy}\)
e: \(=\dfrac{1}{x\left(2x-3\right)}-\dfrac{1}{\left(2x-3\right)\left(2x+3\right)}\)
\(=\dfrac{2x+3-x}{x\left(2x-3\right)\left(2x+3\right)}=\dfrac{x+3}{x\left(2x-3\right)\left(2x+3\right)}\)
g: \(=\dfrac{-2x+x+3-x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2x+6}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2}{x+3}\)
a: =-1/5x^5y^2
b: =-9/7xy^3
c: =7/12xy^2z
d: =2x^4
e: =3/4x^5y
f: =11x^2y^5+x^6
a) \(18x^4y^3:12\left(-x\right)^3y\)
\(=\left(18:-12\right)\left(x^4:x^3\right)\left(y^3:y\right)\)
\(=-\dfrac{3}{2}xy^2\)
b) \(x^2y^2-2xy^3:\dfrac{1}{2}xy^2\)
\(=\dfrac{xy^2\left(x-2y\right)}{\dfrac{1}{2}xy^2}\)
\(=\dfrac{x-2y}{\dfrac{1}{2}}\)
\(=2x-4y\)
1) \(A=\left[x^4-\left(x-1\right)^2\right]:\left(x^2+x-1\right)-x^2+x=\left[\left(x^2-x+1\right)\left(x^2+x-1\right)\right]:\left(x^2+x-1\right)-x^2+x=x^2-x+1-x^2+x=1\)
2) \(B=\dfrac{\left(x+1\right)\left(x+2\right)+4\left(x-2\right)+2-7x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4}{x^2-4}=1\)
A= ( x^4-x^2-2x-1)