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\(a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Đặt \(\frac{1}{a}=x,\frac{1}{b}=y,\frac{1}{c}=z\)
\(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)
mà \(a,b,c\)dương nên \(x=y=z\Rightarrow a=b=c\).
\(A=\left(2+\frac{a}{b}\right)\left(2+\frac{b}{c}\right)\left(2+\frac{c}{a}\right)=3^3=27\).
\(3a^2\)\(b^2\)\(c^2\)
\(=>ab+bc+ca=0\)
\(=>ab^2\)\(+bc^2\)\(+ca^2\)\(=0\)
\(TH1:ab+bc+ca=0\)
\(ab+bc=-ca\)
\(=>a+c=-\frac{ac}{b}\)
\(=>a+b=-\frac{ab}{c}\)
\(b+c=-\frac{bc}{a}\)
\(Thay\)\(A\)
\(=>A=-3\)
\(\left(ab-bc\right)^2\)\(+\left(bc-ca\right)^2\)\(+\left(ca-ab\right)^2\)\(=0\)
\(=>ab-bc=0\)
\(bc-ca=0\)
\(ca-ab=0\)
\(=>ab=bc=ca\)
\(=>a=b=c\)
\(Thay\)\(A\)
\(=>A=-24\)
\(=>A=\left(-3;-24\right)\)
Em làm sai mong anh thông cảm cho ạ
Ai biết cách làm thì nhanh tay giải giùm mình nhé!!!!!!!!!!!!
mk đang cần gấp....<3<3<3<3<3<3
Cho a,b,c > 0 thỏa mãn: a2+b2+c2=1
Tìm GTNN của C= \(\frac{bc}{a}\)+\(\frac{ac}{b}\)+\(\frac{ab}{c}\)
\(sigma\frac{a^2+b^2}{ab\left(a+b\right)^3}\ge sigma\frac{\frac{\left(a+b\right)^2}{2}}{\left(a+b\right)^2\left(a^3+b^3\right)}=sigma\frac{1}{2\left(a^3+b^3\right)}\ge\frac{9}{4\left(a^3+b^3+c^3\right)}=\frac{9}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt[3]{3}}\)
\(a\left(a^2-bc\right)+b\left(b^2-ca\right)+c\left(c^2-ab\right)=0\)
\(\Rightarrow a^3-abc+b^3-abc+c^3-abc=0\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
Mà \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-ac-bc=0\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}\Rightarrow}a=b=c\)
Vậy \(P=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=1+1+1=3\)
\(P\ge\frac{\left(a^2+b^2+c^2\right)^2}{a+b+c+6}=\frac{9}{a+b+c+6}\)(1)
lại có: \(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\Leftrightarrow a+b+c\le3\)
Vậy: \(\left(1\right)\ge\frac{9}{6+3}=1\)
Dấu = xảy ra khi a=b=c=1/căn 3
\(9=3a^2+2b^2+2c^2+2bc\)
\(\Leftrightarrow9=\left(a+b+c\right)^2+2a^2+b^2+c^2-2a\left(b+c\right)\)
\(\Leftrightarrow9\ge\left(a+b+c\right)^2+2a^2+\frac{1}{2}\left(b+c\right)^2-2a\left(b+c\right)\)
\(\Leftrightarrow9\ge\left(a+b+c\right)^2+\frac{1}{2}\left(2a-b-c\right)^2\ge\left(a+b+c\right)^2\)
\(\Rightarrow a+b+c\le3\)
Ta có:
\(P=a+b+c+\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\ge a+b+c+\frac{18}{a+b+c}\)
\(P\ge a+b+c+\frac{9}{a+b+c}+\frac{9}{a+b+c}\)
\(P\ge2\sqrt{\frac{9\left(a+b+c\right)}{a+b+c}}+\frac{9}{3}=9\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Thanks bn