rút gọn:
(-x+3)(2x-1)+(x+1)(x-2)
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a) Ta có: \(\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=x^3+1-x^3+1\)
=2
b) Ta có: \(\left(2x+3\right)\left(2x-3\right)-\left(2x+1\right)^2\)
\(=4x^2-9-4x^2-4x-1\)
\(=-4x-10\)
Ta có: \(2x\left(3x-1\right)-\left(2x+1\right)\left(x-3\right)\)
\(=6x^2-2x-\left(2x^2-6x+x-3\right)\)
\(=6x^2-2x-2x^2+5x+3\)
\(=4x^2+3x+3\)
Ta có: \(3\left(x^2-2x\right)-\left(4x+2\right)\left(x-1\right)\)
\(=3x^2-6x-\left(4x^2-4x+2x-2\right)\)
\(=3x^2-6x-4x^2+2x+2\)
\(=-x^2-4x+2\)
\(2x\left(3x-1\right)-\left(2x+1\right)\left(x-3\right)=6x^2-2x-2x^2+5x+3=4x^2+3x+3\)
\(3\left(x^2-2x\right)-\left(4x+2\right)\left(x-1\right)=3x^2-6x-4x^2+2x-2=-x^2-4x-2\)
ĐKXĐ: \(x\notin\left\{1;\dfrac{1}{2}\right\}\)
\(\left(\dfrac{1}{x-1}+2+\dfrac{2x^3+x^2-x}{1-x^3}\right):\dfrac{1-2x}{x^3+x-2}\)
\(=\left(\dfrac{1}{x-1}+2-\dfrac{2x^3+x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right)\cdot\dfrac{x^3+x-2}{1-2x}\)
\(=\dfrac{x^2+x+1+2\left(x^3-1\right)-2x^3-x^2+x}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^3-x^2+x^2-x+2x-2}{-\left(2x-1\right)}\)
\(=\dfrac{2x+1+2x^3-2-2x^3}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x^2+x+2\right)}{-\left(2x-1\right)}\)
\(=\dfrac{2x-1}{x^2+x+1}\cdot\dfrac{-\left(x^2+x+2\right)}{2x-1}=\dfrac{-x^2-x-2}{x^2+x+1}\)
\(a,=x^2-6x+9-x^2+6x=9\\ b,=4x^2+4x+1-4x^2+9-4x-8=2\\ c,=\left(2x^2-2x-x+1\right):\left(x-1\right)\\ =\left(x-1\right)\left(2x-1\right):\left(x-1\right)=2x-1\)
`a)(x-3)^2-x(x-6)`
`=x^2-6x+9-x^2+6x=9`
`b)(2x+1)^2-(3+2x)(2x-3)-4(x+2)`
`=4x^2+4x+1-(4x^2-9)-4x-8`
`=2`
`c)(2x^2-3x+1):(x-1)`
`=(2x^2-2x-x+1):(x-1)`
`=[2x(x-1)-(x-1)]:(x-1)`
`=2x-1`
1) `2x(3x-1)-(2x+1)(x-3)`
`=6x^2-2x-2x^2+6x-x+3`
`=4x^2+3x+3`
2) `3(x^2-3x)-(4x+2)(x-1)`
`=3x^2-9x-4x^2+4x-2x+2`
`=-x^2-7x+2`
3) `3x(x-5)-(x-2)^2-(2x+3)(2x-3)`
`=3x^2-15x-(x^2-4x+4)-(4x^2-9)`
`=3x^2-15x-x^2+4x-4-4x^2+9`
`=-2x^2-11x+5`
4) `(2x-3)^2+(2x-1)(x+4)`
`=4x^2-12x+9+2x^2+8x-x-4`
`=6x^2-5x+5`
Lời giải:
$(2x+1)^2-3(x-1)^2-(x+1)(x-1)$
$=(4x^2+4x+1)-3(x^2-2x+1)-(x^2-1)$
$=4x^2+4x+1-3x^2+6x-3-x^2+1$
$=(4x^2-3x^2-x^2)+(4x+6x)+(1-3+1)$
$=10x-1$
1:
a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)
\(=4x^2-20x+25-4x^2-12x\)
=-32x+25
b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)
\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)
c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)
\(=\left(x-1-x-2\right)^2+5\left(2x-3\right)\)
\(=\left(-3\right)^2+5\left(2x-3\right)\)
\(=9+10x-15=10x-6\)
2:
a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)
\(=9x^2-12x+4-5x^2+20x+4x-4\)
\(=4x^2+12x\)
b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)
\(=27-x^3+x^3-9x^2+27x-27\)
\(=-9x^2+27x\)
c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)
\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)
\(=-5\left(x^2-16\right)=-5x^2+80\)
a: Ta có: \(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)
\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)
\(=x^2+10x+25-16x^3-48x^2-36x-2x^3+18x+x^2-9\)
\(=-18x^3-46x^2-8x+16\)
\(\left(x-2\right)^3+\left(2x+1\right)^2+2\left(x+2\right)\left(1-x\right)-9x^3+2x\)
\(=x^3-6x^2+12x-8+8x^3+12x^2+6x+1+2\left(x+2\right)\left(1-x\right)-9x^3+2x\)
\(=9x^3+6x^2+18x-7+2\left(x-x^2+2-2x\right)-9x^3+2x\)
\(=6x^2+20x-7-2x^2-2x+4=4x^2+18x-3\)
(-x+3)(2x-1)+(x+1)(x-2)
= -2x\(^2\)+x+6x+x\(^2\)-2x+x-2
= -x\(^2\)+6x-2