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1. Ta có : \(\left(\frac{1}{a}-\frac{1}{b}\right)^2\ge0\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\)
Tương tự : \(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\); \(\frac{1}{a^2}+\frac{1}{c^2}\ge\frac{2}{ac}\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\). Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=9\)
\(9\le3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)
Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c = 1
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=7\)\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=49\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}=49\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=49\)
Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
=\(2\left(a+b+c\right)+\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}-\frac{ab}{c}-\frac{bc}{a}-\frac{ca}{b}=2\left(a+b+c\right)\)
\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)
=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)
2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)
a) \(a+\frac{1}{a}=3\)
\(\Leftrightarrow\)\(\left(a+\frac{1}{a}\right)^2=9\)
\(\Leftrightarrow\)\(a^2+2+\frac{1}{a^2}=9\)
\(\Leftrightarrow\)\(a^2+\frac{1}{a^2}=7\)
Ta có: \(\left(a+\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}\right)=3.7\)
\(\Leftrightarrow\)\(a^3+\frac{1}{a}+a+\frac{1}{a^3}=21\)
\(\Leftrightarrow\)\(a^3+\frac{1}{a^3}=21-3=18\)
Ta lại có: \(\left(a^2+\frac{1}{a^2}\right)\left(a^3+\frac{1}{a^3}\right)=7.18\)
\(\Leftrightarrow\)\(a^5+\frac{1}{a}+a+\frac{1}{a^5}=126\)
\(\Leftrightarrow\)\(a^5+\frac{1}{a^5}=126-3=123\)
Bài làm:
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)
\(\Leftrightarrow a^2b+ab^2+c^2a+ca^2+b^2c+bc^2+2abc=0\)
\(\Leftrightarrow\left(a^2+2ab+b^2\right)c+ab\left(a+b\right)+c^2\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca+c^2\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
=> Hoặc a+b=0 hoặc b+c=0 hoặc c+a=0
=> Hoặc a=-b hoặc b=-c hoặc c=-a
Ko mất tổng quát, g/s a=-b
a) Ta có: vì a=-b thay vào ta được:
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{1}{b^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{c^3}\)
\(\frac{1}{a^3+b^3+c^3}=\frac{1}{-b^3+b^3+c^3}=\frac{1}{c^3}\)
=> đpcm
b) Ta có: \(a+b+c=1\Leftrightarrow-b+b+c=1\Rightarrow c=1\)
=> \(P=-\frac{1}{b^{2021}}+\frac{1}{b^{2021}}+\frac{1}{c^{2021}}=\frac{1}{1^{2021}}=1\)
\(abc=a+b+c\Leftrightarrow\frac{abc}{abc}=\frac{a+b+c}{abc}\)
\(\Leftrightarrow1=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=Q\)
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(\Rightarrow P=3^2-2Q=9-2=7\)
** Bạn lưu ý lần sau viết đề bằng công thức toán để được hỗ trợ tốt hơn.
Lời giải:
$\frac{a+b}{c}+\frac{a+c}{b}+\frac{b+c}{a}=-2$
$\Leftrightarrow \frac{a+b}{c}+1+\frac{a+c}{b}+1+\frac{b+c}{a}=0$
$\Leftrightarrow (a+b+c)(\frac{1}{c}+\frac{1}{b})+\frac{b+c}{a}=0$
$\Leftrightarrow \frac{(a+b+c)(b+c)}{bc}+\frac{b+c}{a}=0$
$\Leftrightarrow (b+c)(\frac{a+b+c}{bc}+\frac{1}{a})=0$
$\Leftrightarrow (b+c).\frac{a(a+b+c)+bc}{abc}=0$
$\Leftrightarrow \frac{(b+c)(a+b)(a+c)}{abc}=0$
$\Rightarrow (a+b)(b+c)(c+a)=0$
$\Rightarrow a+b=0$ hoặc $b+c=0$ hoặc $c+a=0$
Không mất tổng quát giả sử $a+b=0\Rightarrow a=-b$
$1=a^3+b^3+c^3=(-b)^3+b^3+c^3=c^3\Rightarrow c=1$
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{-1}{b}+\frac{1}{b}+\frac{1}{1}=1$
Vậy..........