Bài 8: Phép chia các phân thức đại số

Cầngấpquá,mngiúpmìnhvớiạainhanhmìnhtickchoạ :))

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

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

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

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

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

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

\(=\dfrac{-x-2}{5x}\)

b: Để P=x² thì \(5x^3+x+2=0\)

hay \(x\in\left\{-0.65\right\}\)

c: Để |P|<P thì \(x\in\varnothing\)

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

b: \(=1:\dfrac{a-1}{a}=\dfrac{a}{a-1}\)

c: \(=\dfrac{a+6-3}{3\left(a+3\right)}\cdot\dfrac{27a}{a+2}=\dfrac{a+3}{3\left(a+3\right)}\cdot\dfrac{27a}{a+2}\)

\(=\dfrac{9a}{a+2}\)

\(A=\left(\dfrac{x}{y^2-xy}+\dfrac{y}{x^2-xy}\right):\left(\dfrac{x^2+y^2}{xy^2+x^2y}\right)\)

\(A=\left(\dfrac{x}{y\left(y-x\right)}+\dfrac{y}{x\left(x-y\right)}\right):\left(\dfrac{x^2+y^2}{xy\left(x+y\right)}\right)\\ \)

\(x,y\ne0;\left|y\right|\ne\left|x\right|\)

\(\)\(A=\left(\dfrac{x}{y\left(y-x\right)}+\dfrac{y}{x\left(x-y\right)}\right).\dfrac{xy\left(x+y\right)}{x^2+y^2}\)

\(A=\left(\dfrac{x}{y\left(y-x\right)}.\dfrac{xy}{x^2+y^2}+\dfrac{y}{x\left(x-y\right)}.\dfrac{xy}{x^2+y^2}\right)\left(x+y\right)\)

\(A=\left(\dfrac{x^2}{\left(y-x\right)\left(x^2+y^2\right)}+\dfrac{y^2}{\left(x-y\right)\left(x^2+y^2\right)}\right)\left(x+y\right)\)

\(A=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(y-x\right)\left(x^2+y^2\right)}\left(x+y\right)=\dfrac{-\left(x+y\right)^2}{x^2+y^2}\)

\(\)

\(\left(\dfrac{x}{-y\left(x-y\right)}+\dfrac{y}{x\left(x-y\right)}\right).\dfrac{xy\left(x+y\right)}{x^2+y^2}\)

\(=\dfrac{x^2-y^2}{-xy\left(x-y\right)}.\dfrac{xy\left(x+y\right)}{x^2+y^2}\)

\(=\dfrac{\left(x-y\right)\left(x+y\right)}{-xy\left(x-y\right)}.\dfrac{xy\left(x+y\right)}{x^2+y^2}\)

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