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Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
a) đã rút gọn
b) (x-3)(x+3)-(x-3)(x+1)
= (x-3)(x+3-x-1)
= (x-3)2
\(\left(x-2\right)\left(x-3\right)-2x\left(1-x\right)\)
\(=x^2-3x-2x+6-2x+2x^2\)
\(=x^2-5x+6-2x+2x^2\)
\(=3x^2-7x+6\)
_______________
\(\left(x+5\right)^2-\left(x+3\right)\left(x-2\right)\)
\(=\left(x^2+10x+25\right)-\left(x^2-2x+3x-6\right)\)
\(=x^2+10x+25-x^2-x+6\)
\(=9x+31\)
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: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)
\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)
\(=-5\left(x^2-16\right)=-5x^2+80\)
\(-3x\left(x+2\right)^2+\left(x+3\right)\left(x-1\right)\left(x+1\right)-\left(2x-3\right)^2\\ =-3x\left(x^2+4x+4\right)+\left(x+3\right)\left(x^2-1\right)-\left(4x^2-12x+9\right)\\ =-3x^3-12x^2-12x+x^3-x+3x^2-3-4x^2+12x-9\\ =-2x^3-13x^2-x-12\)
\(=x^2+2x+1-2x^2+18+x^2-4x+4\\ =-2x+23=-2\cdot12+23=-24+23=-1\)
\(\dfrac{x^5+x^3+x^2+1}{x^3+x^2+x+1}=\dfrac{x^3\left(x^2+1\right)+\left(x^2+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}\)
= \(\dfrac{\left(x^3+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{x+1}=x^2-x+1\)
\(\dfrac{x^5+x^3+x^2+1}{x^3+x^2+x+1}=\dfrac{x^3.\left(x^2+1\right)+\left(x^2+1\right)}{x.\left(x^2+1\right)+\left(x^2+1\right)}\) \(=\dfrac{\left(x^3+1\right).\left(x^2+1\right)}{\left(x+1\right).\left(x^2+1\right)}=\dfrac{x^3+1}{x+1}=\dfrac{\left(x+1\right).\left(x^2-x+1\right)}{x+1}\) \(=x^2-x+1\)
a: =x^3+6x^2+12x+8-(x^3+3x^2+3x+1)
=x^3+6x^2+12x+8-x^3-3x^2-3x-1
=3x^2+9x+7
b: =x^3-9x^2+27x-27-x(x^2-6x+9)
=x^3-9x^2+27x-27-x^3+6x^2-9x
=-3x^2+18x-27
c: =x^3+3x^2+3x+1-x^3-x^2-2x^2-4x
=-x+1
\(a,\left(x+2\right)^3-\left(x+1\right)^3\\ =\left(x+2-x-1\right)\left(x^2+4x+4+x^2+3x+2+x^2+2x+1\right)\\ =3x^2+9x+7\\ b,\left(x-3\right)^3-x\left(x-3\right)^2\\ =x^3-6x^2+9x-27-x^3+6x^2-9x\\ =-27\)
\(\left(2x-1\right)\left(x+2\right)-\left(x-3\right)\left(x+3\right)\)
\(=2x^2+3x-2-x^2+9\)
\(=x^2+3x+7\)
\(\left(2x-1\right)\left(x+2\right)-\left(x-3\right)\left(x+3\right)\)
\(=2x^2+4x-x-2-x^2+9\)
\(=x^2+3x+7\)
\(\left(x+3\right)^2-\left(x+1\right).\left(x-3\right)\)
\(=x^3+3x^2y+3xy^2+9-\text{[}x^2-3x+x-3\text{]}\)
\(=x^3+3x^2y+3xy^2+9-x^2+3x-x+3\)
\(=x^3+3x^2y+3xy^2+12-x^2+2x\)
\(\left(x+3\right)^2-\left(x+1\right)\left(x-3\right)\)
\(=\left(x^2+6x+9\right)-\left(x^2+x-3x-3\right)\)
\(=x^2+6x+9-x^2-x+3x+3\)
\(=8x+12=4\left(2x+3\right)\)