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a: \(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2\left(x-1\right)\left(x+1\right)}{5}\cdot2\)
\(=\dfrac{10}{5}\cdot2=4\)
b: \(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}\cdot\dfrac{x^2+6x+9-x^2}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x}{x-3}-\dfrac{3}{x-3}=1\)
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
\(\dfrac{4}{5}x^2\left(\dfrac{1}{3}x-\dfrac{1}{2}\right)-\left(\dfrac{1}{5}x-\dfrac{2}{3}\right)\left(\dfrac{4}{3}x^2+1\right)=\dfrac{22}{45}x^2\)
\(\Leftrightarrow\dfrac{4}{15}x^3-\dfrac{2}{5}x^2-\left(\dfrac{4}{15}x^3+\dfrac{1}{5}x-\dfrac{8}{9}x^2-\dfrac{2}{3}\right)=\dfrac{22}{45}x^2\)
\(\Leftrightarrow\dfrac{4}{15}x^3-\dfrac{2}{5}x^2-\dfrac{4}{15}x^3-\dfrac{1}{5}x+\dfrac{8}{9}x^2+\dfrac{2}{3}-\dfrac{22}{45}x^2=0\)
\(\Leftrightarrow-\dfrac{1}{5}x=-\dfrac{2}{3}\)
hay x=10/3
a: \(=6x^4-9x^3+3x^2-4x^3+6x^2-2x+10x^2-15x+5\)
\(=6x^4-13x^3+19x^2-17x+5\)
b: \(=6x^4-\dfrac{9}{4}x^3-\dfrac{9}{2}x^2-\dfrac{8}{3}x^3+x^2+2x-\dfrac{20}{3}x^2+\dfrac{5}{2}x+5\)
\(=6x^4-\dfrac{59}{12}x^3-\dfrac{67}{6}x^2+\dfrac{9}{2}x+5\)
c: \(=3x^4-\dfrac{9}{8}x^3-\dfrac{3}{4}x^2+8x^3-3x^2-6x-\dfrac{4}{3}x^2+\dfrac{1}{2}x+1\)
\(=3x^4-\dfrac{55}{8}x^3-\dfrac{25}{12}x^2-\dfrac{11}{2}x+1\)
`P=((3+x)/(3-x)-(3-x)/(3+x)+(4x^2)/(x^2-9)):((2x+1)/(x+3)-1)`
`=((4x^2-(3-x)^2-(3+x)^2)/(x^2-9)):((2x+1-x-3)/(x+3))`
`=((4x^2-x^2+6x-9-x^2-6x-9)/(x^2-9)):((x-2)/(x+3))`
`=((2x^2-18)/(x^2-9))*(x+3)/(x-2)`
`=((2(x^2-9))/(x^2-9))*(x+3)/(x-2)`
`=(2x+6)/(x-2)`
ĐKXĐ: \(x\ne\pm3;x\ne-\dfrac{1}{2};x\ne2\)
\(P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{4x^2}{\left(3-x\right)\left(3+x\right)}\right):\dfrac{2x+1-x-3}{x+3}\)
\(=\dfrac{\left(3+x\right)^2-\left(3-x\right)^2-4x^2}{\left(3+x\right)\left(3-x\right)}:\dfrac{x-2}{x+3}\)
\(=\dfrac{\left(3+x-3+x\right)\left(3+x+3-x\right)-4x^2}{\left(x+3\right)\left(3-x\right)}.\dfrac{x+3}{x-2}\)
\(=\dfrac{12x-4x^2}{3-x}\cdot\dfrac{1}{x-2}\)
\(=\dfrac{4x\left(3-x\right)}{3-x}\cdot\dfrac{1}{x-2}\) \(=\dfrac{4x}{x-2}\)