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J
2
26 tháng 11 2018
1) \(\dfrac{x^2-18x-19}{x^2-1}=\dfrac{x^2-19x+x-19}{\left(x-1\right)\left(x+1\right)}=\dfrac{x\left(x-19\right)+x-19}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-19\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-19}{x-1}\)
2) \(\dfrac{x\left(4x^2-8x+4\right)}{2x^3-2x^2}=\dfrac{4x\left(x^2-2x+1\right)}{2x^2\left(x-1\right)}=\dfrac{4x\left(x-1\right)^2}{2x^2\left(x-1\right)}=\dfrac{2\left(x-1\right)}{x}\)
6 tháng 8 2016
\(\left(x+2\right)\left(x-2\right)-\left(x+2\right)^2\)
\(=\left(x+2\right)\left(x-2-x-2\right)\)
\(=\left(-4\right)\left(x+2\right)\)
N
2
VM
22 tháng 12 2019
\(A=\frac{x^2-2x+1}{x^2-x}-\frac{2x^3-x^2}{x^4-x^3}\)
\(A=\frac{x^2-2x+1}{x.\left(x-1\right)}-\frac{2x^3-x^2}{x^3.\left(x-1\right)}\)
\(A=\frac{x^2.\left(x^2-2x+1\right)}{x^2.x.\left(x-1\right)}-\frac{2x^3-x^2}{x^3.\left(x-1\right)}\)
\(A=\frac{x^4-2x^3+x^2}{x^3.\left(x-1\right)}+\frac{-\left(2x^3-x^2\right)}{x^3.\left(x-1\right)}\)
\(A=\frac{x^4-2x^3+x^2-2x^3+x^2}{x^3.\left(x-1\right)}\)
\(A=\frac{x^4-4x^3+2x^2}{x^3.\left(x-1\right)}\)
\(A=\frac{\left(x^4-4x^3\right)+2x^2}{x^3.\left(x-1\right)}\)
\(A=\frac{x^3.\left(x-4\right)+2x^2}{x^3.\left(x-1\right)}\)
\(A=\frac{x-4+2x^2}{x-1}.\)
Không chắc nha.
Chúc bạn học tốt!
\(\)
DH
22 tháng 12 2019
\(\frac{x^2-2x+1}{x^2-x}-\frac{2x^3-x^2}{x^4-x^3}=\frac{\left(x-1\right)^2}{x\left(x-1\right)}-\frac{x^2\left(2x-1\right)}{x^3\left(x-1\right)}=\frac{\left(x-1\right)^2}{x^2-x}-\frac{2x-1}{x^3-x}=\frac{x^2-2x+2x+1}{x^2-x}=\frac{x^2-4x+2}{x^2-x}\)
19 tháng 6 2019
\(\frac{1}{2}x^2\left(6x-3\right)-x\left(x^2+\frac{1}{2}\right)+\frac{1}{2}\left(x+4\right)\)
\(=\frac{1}{2}x^2.6x+\frac{1}{2}x^2.\left(-3\right)+\left(-x\right).x^2+\left(-x\right).\frac{1}{2}+\frac{1}{2}.x+\frac{1}{2}.4\)
\(=3x^3-\frac{3}{2}x^2-x^3-\frac{1}{2}x+\frac{1}{2}x+2\)
\(=\left(3x^3-x^3\right)-\frac{3}{2}x^2+\left(-\frac{1}{2}x+\frac{1}{2}x\right)+2\)
\(=2x^3-\frac{3}{2}x^2+2\)
19 tháng 6 2019
\(a,\)\(\frac{1}{2}x^2\left(6x-3\right)-x\left(x^2+\frac{1}{2}\right)+\frac{1}{2}\left(x+4\right)\)
\(=3x^3-\frac{3}{2}x^2-x^3-\frac{1}{2}x+\frac{1}{2}x+2\)
\(=2x^3-\frac{3}{2}x^2+2\)
\(b,\)\(2x\left(3x^3-x\right)-4x^2\left(x-x^2+1\right)+\left(x-3x^2\right)x\)
\(=6x^4-2x^2-4x^3+4x^4-4x^2+x^2-3x^3\)
\(=10x^4-7x^3-5x^2\)
6 tháng 4 2018
Bài 2:
a, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}-\dfrac{3x+1}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}+\dfrac{3x+1}{x^2-1}\right).\dfrac{x^2-1}{2x+1}\)
\(P=\dfrac{\left(x-1\right)^2-x\left(x+1\right)+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{x^2-2x+1-x^2-x+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{2}{2x+1}\)
b, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P=\dfrac{3}{x-1}\Leftrightarrow\dfrac{2}{2x+1}=\dfrac{3}{x-1}\Leftrightarrow2\left(x-1\right)=3\left(2x+1\right)\)
\(\Leftrightarrow2x-2=6x+3\)\(\Leftrightarrow-4x=5\Leftrightarrow x=\dfrac{-5}{4}\)(TMĐK)
c, \(ĐKXĐ:x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P\in Z\Leftrightarrow\dfrac{2}{2x+1}\in Z\Leftrightarrow2x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
+) Với \(2x+1=1\Leftrightarrow x=0\left(TMĐK\right)\)
+) Với \(2x+1=-1\Leftrightarrow x=-1\left(KTMĐK\right)\)
+) Với \(2x+1=2\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)\)
+) Với \(2x+1=-2\Leftrightarrow x=\dfrac{-3}{2}\left(TMĐK\right)\)
Vậy để \(P\in Z\Leftrightarrow x\in\left\{0;\dfrac{1}{2};\dfrac{-3}{2}\right\}\)
\(x^2-x+2x-2=\)\(\left(x^2-x\right)+\left(2x-2\right)\)
\(=x\left(x-1\right)+2\left(x-1\right)\)
\(=\left(x-1\right)\left(x+2\right)\)
\(x^2-x+2x-2=\left(x^2-x\right)+\left(2x-2\right)\)
\(=x\left(x-1\right)+2\left(x-1\right)\)
\(=\left(x-1\right)\left(x+2\right)\)