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\(\left\{{}\begin{matrix}a^2+b^2=32\\a+b+2ab=40\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2+2ab+a+b=72\\a+b+2ab=40\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^2+\left(a+b\right)-72=0\\a+b+2ab=40\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a+b=8\\a+b=-9\end{matrix}\right.\\a+b+2ab=40\end{matrix}\right.\)
TH1: \(a+b=8\Rightarrow ab=16\)
\(\Rightarrow a\left(8-a\right)=16\Leftrightarrow a^2-8a+16=0\)
\(\Leftrightarrow\left(a-4\right)^2=0\Rightarrow a=4\Rightarrow b=4\)
TH2: \(a+b=-9\Rightarrow ab=\dfrac{49}{2}\)
\(\Rightarrow a\left(-9-a\right)=\dfrac{49}{2}\) \(\Leftrightarrow2a^2+18a+49=0\)
\(\Leftrightarrow2\left(a+\dfrac{9}{2}\right)^2+\dfrac{17}{2}=0\) (ko tồn tại a thỏa mãn)
Vậy \(\left\{{}\begin{matrix}a=4\\b=4\end{matrix}\right.\)
Cách 2:
Với mọi số thực a; b ta luôn có:
\(\left(a-4\right)^2+8\left(a-b\right)^2+\left(b-4\right)^2\ge0\)
\(\Leftrightarrow a^2-8a+16+8\left(a^2-2ab+b^2\right)+b^2-8a+16\ge0\)
\(\Leftrightarrow9\left(a^2+b^2\right)\ge8\left(a+b+2ab\right)-32\)
\(\Leftrightarrow9\left(a^2+b^2\right)\ge288\)
\(\Leftrightarrow a^2+b^2\ge32\)
Đẳng thức xảy ra khi và chỉ khi \(a=b=4\)
\(a\ge2b\Rightarrow\dfrac{a}{b}\ge2\)
\(P=2\left(\dfrac{a}{b}\right)+\left(\dfrac{b}{a}\right)-2=\dfrac{a}{4b}+\dfrac{b}{a}+\dfrac{7}{4}\left(\dfrac{a}{b}\right)-2\ge2\sqrt{\dfrac{ab}{4ab}}+\dfrac{7}{4}.2-2=\dfrac{5}{2}\)
\(P_{min}=\dfrac{5}{2}\) khi \(a=2b\)
a,= a\(^2\)+2a+b\(^2\)-2b-2ab+37
=a\(^2\)-2ab+b\(^2\)+2a-2b+37
=(a-b)\(^2\)+2(a-b)+37
⇒5\(^2\)+2.5+37= 25+10+37= 72
b,= a\(^3\)+a\(^2\)-b\(^3\)+b\(^2\)+ab-3a\(^2\)b+3ab\(^2\)-3ab-95
=a\(^3\)-3a\(^2\)b+3ab\(^2\)-b\(^3\)+a\(^2\)-2ab+b\(^2\)-95
=(a-b)\(^3\)+(a-b)\(^2\)-95
⇒5\(^3\)+5\(^2\)-95= 125+25-95= 60
Bài 2:
b: Ta có: \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x^3-4x-x^4+1\)
\(=-x^4+x^3-4x+1\)
c: Ta có: \(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2ab\)
\(=\left(a+b-c-a+c\right)\left(a+b-c+a-c\right)\)
\(=b\left(2a+b-2c\right)\)
\(=2ab+b^2-2bc\)
sai ở chỗ ab + a^2 - 2ab