\(\left(a+b\right)^2=\left(-1\right)^2=1\\ =>\left(a^2+b^2\right)+2ab=1\\ =>5+2ab=1\\ =>ab=\dfrac{1-5}{2}=-2\)
Ta có: `a^4+b^4=(a^4+2a^2b^2+b^4)-2a^2b^2`
`=(a^2+b^2)^2-2(ab)^2=5^2-2*(-2)^2=25-8=17`
\(M=a^6+b^6=\left(a^2\right)^3+\left(b^2\right)^3\\ =\left(a^2+b^2\right)\left(a^4-a^2b^2+b^4\right)\\ =\left(a^2+b^2\right)\left[\left(a^4+b^4\right)-\left(ab\right)^2\right]\\ =5\cdot\left(17-\left(-2\right)^2\right)=5\cdot\left(17-4\right)=5\cdot13=65\)
\(a=-1-b\)
Thay \(a\) vào \(a^2+b^2=5\)
\(\Leftrightarrow\left(-1-b\right)^2+b^2=5\)
\(\Leftrightarrow1+2b+b^2+b^2=5\)
\(\Leftrightarrow2b^2+2b-4=0\)
\(\Leftrightarrow\left(2b^2-2b\right)+\left(4b-4\right)=0\)
\(\Leftrightarrow2b\left(b-1\right)+4\left(b-1\right)=0\)
\(\Leftrightarrow\left(2b+4\right)\left(b-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2b+4=0\\b-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}b=2\\b=1\end{matrix}\right.\)
+) Nếu \(b=2\)
\(\Rightarrow a=-1-2=-3\)
Thay \(a,b\) vào \(M=a^6+b^6\)
\(\Rightarrow M=\left(-3\right)^6+2^6\)
\(\Rightarrow M=729+64=793\)
+) Nếu \(b=1\)
\(\Rightarrow a=-1-1=-2\)
Thay \(a,b\) vào \(M=a^6+b^6\)
\(\Rightarrow M=\left(-2\right)^6+1\)
\(\Rightarrow M=65\)
a+b=-1<0 không thể bình phương hai vế được.
\(a^2+b^2=5\)
\(\Rightarrow\left(a+b\right)^2-2ab=5\)
\(\Rightarrow\left(-1\right)^2-2ab=5\) Vì \(a+b=-1\)
\(\Rightarrow ab=-2\)
\(a^6+b^6=\left(a^2\right)^3+\left(b^2\right)^3=\left(a^2+b^2\right)\left(a^4+b^4-a^2b^2\right)=\left(a^2+b^2\right)\left[\left(a^2+b^2\right)^2-3a^2b^2\right]=5.\left(5^2-3.\left(-2\right)^2\right)=65\)
