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Ta có
a^3+b^3+3ab(a^2+b^2)+6ab(a+b)=a^3+b^3+3ab.a^2+3ab.b^2+6ab=a^3+b^3+3(a^2)b+3(b^2)a+3a(b-1)b^2+3b(a-1)a^2+6ab
=(a+b)^3+3ab((b-1).b+(a-1).a)+6ab=(a+b)^3+3ab((1-b).(-b)+(1-a)(-a))+6ab=(a+b)^3+3ab(-2ab)+6ab
=(a+b)^3+(-6ab)ab+6ab
=>(a+b)^3+6ab(-ab-1)=6ab(-ab-1)+1 Vậy M=6ab(-ab-1)+1
k cho mình nhá
Bài 2:
a: Ta có: \(M=\left(x+y\right)^3+2x^2+4xy+2y^2\)
\(=\left(x+y\right)^3+2\cdot\left(x+y\right)^2\)
\(=7^3+2\cdot7^2=441\)
do a>0, b>0 nên 1=a+b+3ab\(\ge3\sqrt[3]{3\left(ab\right)^2}\Leftrightarrow\frac{1}{3}\ge\sqrt[3]{3\left(ab\right)^2}\)
\(\Leftrightarrow\frac{1}{27}\ge3\left(ab\right)^2\Leftrightarrow\frac{1}{81}\ge\left(ab\right)^2\Leftrightarrow\frac{1}{9}\ge ab\Leftrightarrow\frac{1}{3}\ge\sqrt{ab}\)do đó
P=\(\frac{6ab}{a+b}-a^2-b^2=\frac{6ab}{a+b}-\left(a^2+b^2\right)\le\frac{6ab}{2\sqrt{ab}}-2ab=-2ab+3\sqrt{ab}=-2\left(ab-\frac{3}{2}\sqrt{ab}\right)\)
\(=-2\left[ab-2\sqrt{ab}\cdot\frac{1}{3}+\left(\frac{1}{3}\right)^2-\left(\frac{1}{3}\right)^2-\frac{5}{6}\sqrt{ab}\right]\)
\(=-2\left(\sqrt{ab}-\frac{1}{3}\right)^2+\frac{2}{9}+\frac{5}{3}\sqrt{ab}\le\frac{2}{9}+\frac{5}{3}\cdot\frac{1}{3}=\frac{7}{9}\)
vậy maxP=\(\frac{7}{9}\Leftrightarrow\hept{\begin{cases}a=b>0\\a+b+3ab=1\end{cases}\Leftrightarrow a=b=\frac{1}{3}}\)
1/ \(\left(x-y\right)^2+\left(x+y\right)^2-2\left(x^2-y^2\right)-4y^2+10\)
\(=x^2-2xy+y^2+x^2+2xy+y^2-2x^2+2y^2-4y^2+10\)
\(=10\)
2/ \(5a^2+b^2=6ab\Leftrightarrow\left(5a^2-5ab\right)+\left(b^2-ab\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(5a-b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=b\\5a=b\end{cases}}\)
Với a = b thì
\(M=\frac{a-b}{a+b}=\frac{a-a}{a+a}=0\)
Với 5a = b thì
\(M=\frac{a-b}{a+b}=\frac{a-5a}{a+5a}=\frac{-4}{6}=\frac{-2}{3}\)
1.(x-y)2+(x+y)2-2(x2-y2)-4y2+10
=x2-2xy+y2+x2+2xy+y2-2x2+2y2-4y2+10
=x2+x-2x2-2xy+2xy+y2+y2+2y2-4y2+10
=10
=>dpcm
2.Ta co : 5a2+b2=6ab
5a2+b2-6ab=0
5a2+b2-5ab-ab=0
5a2-5ab+b2-ab=0
5a(a-b)+b(b-a)=0
5a(a-b)-b(a-b)=0
(a-b)(5a-b)=0
Ta lai co : a-b=0 \(\Rightarrow\)a=b
Va : 5a-b=0 \(\Rightarrow\)5a=b
Thay : a=b vao M
\(\Rightarrow M=\frac{a-b}{a+b}=\frac{b-b}{b+b}=\frac{0}{2b}=0\)
Thay : 5a=b vao M
\(\Rightarrow M=\frac{a-b}{a+b}=\frac{a-5a}{a+5a}=-\frac{4a}{6a}=-\frac{4}{6}=-\frac{2}{3}\)
\(M=\left(3a\right)^2+b^2+2.3a.b+\left(2^2\right)^b-2.2^b.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+5+1\)
\(=\left(3a+b\right)^2+\left(2^b-\frac{1}{2}\right)^2+\frac{23}{4}\ge\frac{23}{4}\)
min M=23/4 <=>\(\hept{\begin{cases}3a+b=0\\2^b-\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}3a=-b\\2^b=\frac{1}{2}=2^{-1}\end{cases}\Leftrightarrow}\hept{\begin{cases}a=\frac{1}{3}\\b=-1\end{cases}}}\)