Câu hỏi của oops

1: \(\dfrac{3a^3-2a^2-3a+2}{a^2-1}\)\(=\dfrac{3a^3-3a-2a^2+2}{a^2-1}\)\(=\dfrac{3a\cdot\left(a^2-1\right)-2\left(a^2-1\right)}{a^2-1}=3a-2\)2: \(\dfrac{3}{4}\left(-x\right)^3:\dfrac{1}{8}x=\left(\dfrac{3}{4}:\dfrac{1}{8}\right)\cdot\left(-x^3:x\right)=-6x^2\)3: \(x^4-9x^3+21x^2+x+k⋮x-2\)=>\(x^4-2x^3-7x^3+14x^2+7x^2-14x+15x-30+k+30⋮x-2\)=>k+30=0=>k=-30Câu trả lời: 1: \(\dfrac{3a^3-2a^2-3a+2}{a^2-1}\)\(=\dfrac{3a^3-3a-2a^2+2}{a^2-1}\)\(=\dfrac{3a\cdot\left(a^2-1\right)-2\left(a^2-1\right)}{a^2-1}=3a-2\)2: \(\dfrac{3}{4}\left(-x\right)^3:\dfrac{1}{8}x=\left(\dfrac{3}{4}:\dfrac{1}{8}\right)\cdot\left(-x^3:x\right)=-6x^2\)3: \(x^4-9x^3+21x^2+x+k⋮x-2\)=>\(x^4-2x^3-7x^3+14x^2+7x^2-14x+15x-30+k+30⋮x-2\)=>k+30=0=>k=-30