1 Rút gọn
\(\dfrac{x^3-3x^2+3x-1}{1-x+x^2y-xy}\)
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22 tháng 8 2021
a: Ta có: \(\left(8x^3-4x^2\right):4x-\left(4x^2-5x\right):2x+\left(2x\right)^2\)
\(=2x^2-x-2x+\dfrac{5}{2}+4x^2\)
\(=6x^2-3x+\dfrac{5}{2}\)
b: Ta có: \(\left(3x^3-x^2y\right):x^2-\left(xy^2+x^2y\right):xy+2x\left(x-1\right)\)
\(=3x-y-y-x+2x^2-2x\)
\(=2x^2-2y\)
19 tháng 12 2021
c: \(=\dfrac{3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}=\dfrac{3x}{x^2+1}\)
21 tháng 10 2021
a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)
16 tháng 12 2022
f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)
g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)
h: \(=\dfrac{4-2+x}{x-1}=\dfrac{x+2}{x-1}\)
n: \(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{x\left(x+3\right)}=\dfrac{2}{x}\)
p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)
k: \(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-6}{x^2-4}\)
m: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
17 tháng 10 2021
a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)
\(=\left(3x+2+1-2y\right)^2\)
\(=\left(3x-2y+3\right)^2\)
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31 tháng 12 2021
\(A=\dfrac{3x}{x-1}+\dfrac{2}{x+1}+\dfrac{3-3x-2x^2}{x^2-1}.\) \(\left(ĐKXĐ:x\ne1;x\ne-1\right).\)
\(A=\dfrac{3x\left(x+1\right)+2\left(x-1\right)+3-3x-2x^2}{\left(x-1\right)\left(x+1\right)}.\)
\(A=\dfrac{3x^2+3x+2x-2+3-3x-2x^2}{\left(x-1\right)\left(x+1\right)}.\)
\(A=\dfrac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x-1}.\)
27 tháng 12 2020
a) Ta có: \(\dfrac{2x^2-2x}{x-1}\)
\(=\dfrac{2x\left(x-1\right)}{x-1}\)
=2x
b) Ta có: \(\dfrac{x^2+2x+1}{3x^2+3x}\)
\(=\dfrac{\left(x+1\right)^2}{3x\left(x+1\right)}\)
\(=\dfrac{x+1}{3x}\)
c) Ta có: \(\dfrac{x}{3x-3}+\dfrac{1}{x^2-1}\)
\(=\dfrac{x}{3\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+1+3}{3\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+4}{3x^2-3}\)
25 tháng 1 2022
\(Q=x^2+2xy+\left(-3x^3+3x^3\right)+\left(2y^3-y^3\right)=x^2+2xy+y^3\)
\(P=\left(\dfrac{1}{3}x^2y-\dfrac{1}{3}x^2y\right)+\left(xy^2+\dfrac{1}{2}xy^2\right)-\left(xy+5xy\right)=\dfrac{3}{2}xy^2-6xy\)
\(=\dfrac{\left(x-1\right)^3}{xy\left(x-1\right)+\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{xy+1}\)
\(\dfrac{x^3-3x^2+3x-1}{1-x+x^2y-xy}=\dfrac{\left(x-1\right)^3}{\left(xy-1\right)\left(x-a\right)}=\dfrac{\left(x-1\right)^2}{xy-1}\)