Bài 8: Phép chia các phân thức đại số

b: 

ĐKXĐ: \(x\notin\left\{0;2;-2\right\}\)

\(\left(\dfrac{4}{x^3-4x}+\dfrac{1}{x+2}\right):\left(\dfrac{x-2}{x^2+2x}-\dfrac{x}{2x+4}\right)\)

\(=\left(\dfrac{4}{x\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right):\left(\dfrac{x-2}{x\left(x+2\right)}-\dfrac{x}{2\left(x+2\right)}\right)\)

\(=\dfrac{4+x\left(x-2\right)}{x\left(x-2\right)\cdot\left(x+2\right)}:\dfrac{2\left(x-2\right)-x^2}{x\left(x+2\right)\cdot2}\)

\(=\dfrac{x^2-2x+4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{2x\left(x+2\right)}{-\left(x^2-2x+4\right)}\)

\(=\dfrac{-2}{x-2}\)

c:ĐKXĐ: x<>0

\(\left(x-\dfrac{3}{x}\right):\left(\dfrac{x^2+2x+1}{x}-\dfrac{2x+4}{x}\right)\)

\(=\dfrac{x^2-3}{x}:\dfrac{x^2+2x+1-2x-4}{x}\)

\(=\dfrac{x^2-3}{x}\cdot\dfrac{x}{x^2-3}\)

=1

Cầngấpquá,mngiúpmìnhvớiạainhanhmìnhtickchoạ :))

a) \(\left(x^3-1\right):\left(x-1\right)=\left(x-1\right)\left(x^2+x+1\right):\left(x-1\right)=x^2+x+1\)

a: \(=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x-1}=x^2+x+1\)

b: \(=\dfrac{x^3+2x^2-x^2-2x+x+2-1}{x+2}=x^2-x+1+\dfrac{-1}{x+2}\)

\(3^{n+3}-2.3^n+2^{n+5}-7.2^n=3^n.3^3-2.3^n+2^n.2^5-7.2^n=3^n.\left(27-2\right)+2^n.\left(32-7\right)=3^n.25+2^n.25=\left(3^n+2^n\right).25⋮25\)

bạn ơi có thiếu +3 hay j ko

\(a^3+b^3+c^3=3abc\)

\(=> (a^3+b^3) + c^3 - 3abc = 0\)

\(=> (a+b)^3 - 3ab(a+b) + c^3 - 3abc=0\)

\(=> [(a+b)^3+c^3] - 3ab(a+b+c) = 0\)

\(=> (a+b+c)[(a+b)^2-c(a+b)+c^2]-3ab(a+b+c)=0\)

\(=> (a+b+c)(a^2+b^2+c^2+2ab-bc-ca)-3ab(a+b+c)=0\)

\(=> (a+b+c)(a^2+b^2+c^2+2ab-bc-ca-3ab)=0\)

\(=> (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0\)

\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

a: \(A=\left(\dfrac{x}{x+2}+\dfrac{4x-12}{5x^2-15x}-\dfrac{8}{5x^2+10x}\right):\dfrac{x^2-2x+2}{x^2-x-6}\)

\(=\left(\dfrac{x}{x+2}+\dfrac{4x-12}{5x\left(x-3\right)}-\dfrac{8}{5x\left(x+2\right)}\right)\cdot\dfrac{\left(x-3\right)\left(x+2\right)}{x^2-2x+2}\)

\(=\left(\dfrac{x}{x+2}+\dfrac{4}{5x}-\dfrac{8}{5x\left(x+2\right)}\right)\cdot\dfrac{\left(x-3\right)\left(x+2\right)}{x^2-2x+2}\)

\(=\dfrac{5x^2+4x+8-8}{5x\left(x+2\right)}\cdot\dfrac{\left(x-3\right)\left(x+2\right)}{x^2-2x+2}\)

\(=\dfrac{5x^2+4x}{5x}\cdot\dfrac{x-3}{x^2-2x+2}=\dfrac{\left(5x+4\right)\left(x-3\right)}{5\left(x^2-2x+2\right)}\)

b: Khi x=1 thì \(A=\dfrac{\left(5+4\right)\left(1-3\right)}{5\left(1-2+2\right)}=\dfrac{9\cdot\left(-2\right)}{5}=\dfrac{-18}{5}\)

Khi x=3 thì \(A=\dfrac{\left(5\cdot3+4\right)\left(3-3\right)}{A}=0\)

a ) \(\left(\dfrac{20x}{3y^2}\right):\left(\dfrac{4x^3}{5y}\right)=\dfrac{20x}{3y^2}.\dfrac{5y}{4x^3}=\dfrac{100xy}{12x^3y^2}=\dfrac{25}{3x^2y}\)

b ) Đ/k : \(x\ne-4\)

Ta có : \(\dfrac{4x+12}{\left(x+4\right)^2}:\dfrac{3\left(x+3\right)}{x+4}\)

\(=\dfrac{4\left(x+3\right)}{\left(x+4\right)^2}.\dfrac{x+4}{3\left(x+3\right)}\)

\(=\dfrac{4\left(x+3\right)\left(x+4\right)}{3\left(x+3\right)\left(x+4\right)^2}\)

\(=\dfrac{4}{3\left(x+4\right)}\)

\(=\dfrac{4}{3x+12}\)

\(=\dfrac{a^2+b^2}{a}+b:\dfrac{a^2+b^2}{a^2b^2}\cdot\dfrac{a^2+b^2}{a^3+b^3}\)

\(=\dfrac{a^2+b^2}{a}+b\cdot\dfrac{a^2b^2}{a^2+b^2}\cdot\dfrac{a^2+b^2}{a^3+b^3}\)

\(=\dfrac{a^2+b^2}{a}+\dfrac{a^2b^3}{a^3+b^3}\)

\(=\dfrac{a^5+a^2b^3+a^3b^2+b^5+a^3b^3}{a\left(a^3+b^3\right)}\)