Ta có:\(\left(a^3-3ab^2\right)^2=5^2\)

\(\Leftrightarrow a^6-6a^4b^2+9a^2b^4=25\)

\(\left(b^3-3a^2b\right)^2=10^2\)

\(\Leftrightarrow b^6-6a^2b^4+9a^4b^2=100\)

\(\left(a^3-3ab^2\right)^2+\left(b^3-3a^2b\right)^2=25+100\)

\(\Leftrightarrow a^6-6a^4b^2+9a^2b^4+a^6-6a^2b^4+9a^4b^2=125\)

\(\Leftrightarrow a^6+3a^4b^2+3a^2b^4+b^6=125\)

\(\Leftrightarrow\left(a^2+b^2\right)^3=5^3\)

\(\Leftrightarrow a^2+b^2=5\)

\(\)

Ta có:\(a^2-5a+2=0\Rightarrow a^2=5a-2\)

\(P=a^5-a^4-18a^3+9a^2-5a+2017+\frac{a^4-40a^2+4}{a^2}\)

\(=a^5-a^4-18a^3+9a^2-5a+2017+\frac{\left(a^2-2\right)^2-36a^2}{a^2}\)

\(=a^5-a^4-18a^3+9a^2-5a+2015+2+\frac{\left(a^2-2\right)^2-\left(6a\right)^2}{a^2}\)

\(=\left(a^2-5a+2\right)\left(a^3+4a^2+1\right)+2015+\frac{\left(a^2-2+6a\right)\left(a^2-2-6a\right)}{a^2}\)

\(=0\times\left(a^3+4a^2+1\right)+2015+\frac{\left(a^2-2+6a\right)\left(a^2-2-6a\right)}{a^2}\)

\(=0+2015+\frac{\left(a^2-2+6a\right)\left(a^2-2-6a\right)}{a^2}\)

\(=2015+\frac{\left(5a-2-6a-2\right)\left(5a-2+6a-2\right)}{a^2}\)Vì \(a^2=5a-2\)

\(=2015+\frac{-\left(a+4\right)\left(11a-4\right)}{a^2}\)

\(=2015+\frac{-\left(a^2+40a-16\right)}{a^2}\)

\(=2015+\frac{-\left[a^2+8\left(5a-2\right)\right]}{a^2}\)Vì \(a^2=5a-2\)

\(=2015+\frac{-\left(a^2+8a^2\right)}{a^2}\)

\(=2015+\frac{-9a^2}{a^2}\)

\(=2015+\frac{-9}{1}\)

\(=2015-9\)

\(=2006\)

Cre:hoidap247

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

Để đa thức chia hết \(\Leftrightarrow a+2=0\Leftrightarrow a=-2\)

a) Điều kiện: \(x\ne\pm1\)

 \(B=\frac{x-1}{x+1}-\frac{x+1}{x-1}-\frac{4}{1-x^2}\)

\(B=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}-\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{-4}{\left(x-1\right).\left(x+1\right)}\)

\(B=\frac{x^2-x-x+1-x^2-x-x-1+4}{\left(x-1\right).\left(x+1\right)}\)

\(B=\frac{-4x+4}{\left(x-1\right).\left(x+1\right)}=\frac{-4.\left(x-1\right)}{\left(x-1\right).\left(x+1\right)}=\frac{-4}{x+1}\)

b) \(x^2-x=0\Leftrightarrow x.\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Khi  \(x=0\Leftrightarrow\frac{-4}{0-1}=\frac{-4}{-1}=4\)

Khi \(x=1\Leftrightarrow\frac{-4}{1-1}=0\)

c) \(\frac{-4}{x+1}=-3\Leftrightarrow-3.\left(x+1\right)=-4\Leftrightarrow x+1=\frac{4}{3}\Leftrightarrow x=\frac{1}{3}\)