Bài 8:
a) Ta có:
\(a+b=S\)
\(\Rightarrow\left(a+b\right)^2=S^2\)
\(\Rightarrow a^2+2ab+b^2=S^2\)
\(\Rightarrow a^2+2P+b^2=S^2\)
\(\Rightarrow a^2+b^2=S^2-2P\)
\(\Rightarrow A=a^2+b^2=S^2-2P\)
b) Ta có:
\(a+b=S\)
\(\Rightarrow\left(a+b\right)^3=S^3\)
\(\Rightarrow a^3+3ab\left(a+b\right)+b^3=S^3\)
\(\Rightarrow a^3+3SP+b^3=S^3\)
\(\Rightarrow a^3+b^3=S^3-3SP\)
\(\Rightarrow B=a^3+b^3=S^3-3SP\)
c) Ta có:
\(C=a^4+b^4\)
\(C=a^4+b^4+2a^2b^2-2a^2b^2\)
\(C=\left(a^2+b^2\right)^2-2a^2b^2\)
\(C=\left(S^2-2P\right)^2-2\cdot\left(ab\right)^2\) (Theo câu a \(a^2+b^2=S^2-2P\))
\(C=\left(S^2-2P\right)^2-2P^2\)
\(C=S^4-4S^2P+4P^2-2P^2\)
\(C=S^4-4S^2P+2P^2\)
6:
a: \(A=-x^2+6x-11\)
\(=-\left(x^2-6x+11\right)\)
\(=-\left(x^2-6x+9+2\right)\)
\(=-\left(x-3\right)^2-2< =-2\)
Dấu = xảy ra khi x=3
b: \(B=-x^2+x-4\)
\(=-\left(x^2-x+4\right)\)
\(=-\left(x^2-x+\dfrac{1}{4}+\dfrac{15}{4}\right)\)
\(=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{15}{4}< =-\dfrac{15}{4}\)
Dấu = xảy ra khi x=1/2
c: \(C=-x^2-3x\)
\(=-\left(x^2+3x\right)\)
\(=-\left(x^2+3x+\dfrac{9}{4}-\dfrac{9}{4}\right)\)
\(=-\left(x+\dfrac{3}{2}\right)^2+\dfrac{9}{4}< =\dfrac{9}{4}\)
Dấu = xảy ra khi x=-3/2
