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a: \(=\dfrac{2\left(x+2\right)\left(x-1\right)}{x+2}=2x-2\)
b: \(=\dfrac{2x^3+x^2-6x^2-3x+2x+1}{2x+1}=x^2-3x+1\)
c: \(=\dfrac{x^3+2x^2-2x^2-4x+2x+4}{x+2}=x^2-2x+2\)
d: \(=\dfrac{x^2\left(x-3\right)}{x-3}=x^2\)
\(\dfrac{6}{x^2+4x}+\dfrac{3}{2x+8}\\ =\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\\ =\dfrac{6.2}{2x\left(x+4\right)}+\dfrac{3x}{2x\left(x+4\right)}\\ =\dfrac{12+3x}{2x\left(x+4\right)}\\ =\dfrac{3\left(4+x\right)}{2x\left(x+4\right)}\\ =\dfrac{3}{2x}\)
________
\(\dfrac{x+1}{x-2}+\dfrac{x-2}{x+2}+\dfrac{x-14}{x^2-4}\\ \left(\text{đ}k\text{x}\text{đ}:x\ne\pm2\right)\\ =\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}+\dfrac{x-14}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2+2x+x+2+x^2-4x+4+x-14}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2x^2-8}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2\left(x^2-4\right)}{x^2-4}\\ =2\)
a: \(=\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\)
\(=\dfrac{12+3x}{2x\left(x+4\right)}=\dfrac{3\left(x+4\right)}{2x\left(x+4\right)}=\dfrac{3}{2x}\)
b: \(=\dfrac{\left(x+1\right)\left(x+2\right)+\left(x-2\right)^2+x-14}{x^2-4}\)
\(=\dfrac{x^2+3x+2+x^2-4x+4+x-14}{x^2-4}=\dfrac{2x^2-8}{x^2-4}=2\)
a) Ta có: \(\left(x-2\right)^3-\left(3+x^2\right)\left(3-x\right)\)
\(=x^3-6x^2+12x-8+\left(x-3\right)\left(x^2+3\right)\)
\(=x^3-6x^2+12x-8+x^3+3x-3x^2-9\)
\(=2x^3-9x^2+15x-17\)
b) Ta có: \(x\left(x-14\right)-10\left(x-1\right)^2\)
\(=x^2-14x-10\left(x^2-2x+1\right)\)
\(=x^2-14x-10x^2+20x-10\)
\(=-9x^2+6x-10\)
c) Ta có: \(2x\left(x+2\right)-\left(x+2\right)\left(x-2\right)\)
\(=2x^2+4x-\left(x^2-4\right)\)
\(=2x^2+4x-x^2+4\)
\(=x^2+4x+4\)
d) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x^3-27\right)\)
\(=x^3-27-x^3+27\)
=0
\(68^2+64.68+32^2\)
\(=68^2+2.32.68+32^2\)
\(=\left(68+32\right)^2\)
\(=100^2\)
\(=10000\)
\(=\dfrac{3x^2+5x+14+x^2-1-4x^2+4x-4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{9x+9}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{9}{x^2-x+1}\)
a) \(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}=\dfrac{x^2+2+2\left(x-1\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
b) \(=\dfrac{1}{x+2}+\dfrac{3}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x+2\right)\left(x-2\right)+3\left(x+2\right)+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\)
c) \(=\dfrac{x^2+xy+y^2-3xy+\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{x^2-2xy+y^2+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x^2-2xy+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
`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))`
Chọn đáp án A