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Lời giải:
$a+b+c=0\Rightarrow a+b=-c$
Ta có:
$a^3+b^3+c^3=(a+b)^3-3a^2b-3ab^2+c^3$
$=(a+b)^3-3ab(a+b)+c^3=(-c)^3-3ab(-c)+c^3=(-c)^3+3abc+c^3=3abc$ chứ không phải bằng $0$ nhé.
bài 5 nhé:
a) (a+1)2>=4a
<=>a2+2a+1>=4a
<=>a2-2a+1.>=0
<=>(a-1)2>=0 (luôn đúng)
vậy......
b) áp dụng bất dẳng thức cô si cho 2 số dương 1 và a ta có:
a+1>=\(2\sqrt{a}\)
tương tự ta có:
b+1>=\(2\sqrt{b}\)
c+1>=\(2\sqrt{c}\)
nhân vế với vế ta có:
(a+1)(b+1)(c+1)>=\(2\sqrt{a}.2\sqrt{b}.2\sqrt{c}\)
<=>(a+1)(b+1)(c+1)>=\(8\sqrt{abc}\)
<=>(a+)(b+1)(c+1)>=8 (vì abc=1)
vậy....
1.
\(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
Ta có:
\(\dfrac{\left(a+2b\right)^2+\left(b+2c\right)^2+\left(c+2a\right)^2}{\left(a-2b\right)^2+\left(b-2c\right)^2+\left(c-2a\right)^2}\)
\(=\dfrac{a^2+4b^2+4ab+b^2+4c^2+4bc+c^2+4a^2+4ca}{a^2+4b^2-4ab+b^2+4c^2-4bc+c^2+4a^2-4ca}\)
\(=\dfrac{5\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)}{5\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)}\)
\(=\dfrac{-10\left(ab+bc+ca\right)+4\left(ab+bc+ca\right)}{-10\left(ab+bc+ca\right)-4\left(ab+bc+ca\right)}\)
\(=\dfrac{-6}{-14}=\dfrac{3}{7}\)
b.
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-3abc\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c\)
\(\Rightarrow\dfrac{ab+2bc+3ca}{3a^2+4b^2+5c^2}=\dfrac{a^2+2a^2+3a^2}{3a^2+4a^2+5a^2}=\dfrac{6}{12}=\dfrac{1}{2}\)
\(\left(a^3+b^3\right)\left(a+b\right)=ab\left(1-a\right)\left(1-b\right)\)
\(\Leftrightarrow\left(1-a\right)\left(1-b\right)=\left(\dfrac{a^2}{b}+\dfrac{b^2}{a}\right)\left(a+b\right)\ge\left(a+b\right)^2\ge4ab\)
\(\Rightarrow1+ab-4ab\ge a+b\ge2\sqrt{ab}\)
\(\Rightarrow3ab+2\sqrt{ab}-1\le0\)
\(\Leftrightarrow\left(\sqrt{ab}+1\right)\left(3\sqrt{ab}-1\right)\le0\)
\(\Leftrightarrow ab\le\dfrac{1}{9}\)
BĐT \(\Leftrightarrow a^3-b^3+a^2b-ab^2\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2\right)+ab\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)^2\ge0\) (luôn đúng do \(a\geq b\)).
1. b3+b= 3
(b3+b)=3
b.(3+1)=3
b. 4= 3
b=\(\dfrac{3}{4}\)
a3+a= 3 b3
(a3+a)=3
a.(3+1)=3
a. 4= 3
a=\(\dfrac{3}{4}\)
2