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Ta có:
\(\dfrac{1}{a^2}+\dfrac{1}{ab}+\dfrac{1}{ac}=\dfrac{2}{a}\)
\(\dfrac{1}{ab}+\dfrac{1}{b^2}+\dfrac{1}{bc}=\dfrac{2}{b}\)
\(\dfrac{1}{ac}+\dfrac{1}{bc}+\dfrac{1}{c^2}=\dfrac{2}{c}\)
Cộng vế với vế ta được:
\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\)\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\dfrac{c+a+b}{abc}=2.2\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\left(đpcm\right)\)
Ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)
=> \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\) = 4
=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\) = 4
=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\) + \(2\left(\dfrac{c}{abc}+\dfrac{b}{abc}+\dfrac{a}{abc}\right)\) = 4
=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\dfrac{a+b+c}{abc}\) = 4
=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\) + \(2.\dfrac{abc}{abc}\) = 4 ( vì a+b + c = abc)
=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\) => đpcm
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=4\)
Thay \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\) vào, ta được:
\(2+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=4\)
\(\Rightarrow\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=4-2=2\)
\(\Rightarrow2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=2\)
\(\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}=1\)
\(\Rightarrow abc\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=abc\)
\(\Rightarrow\dfrac{abc}{ab}+\dfrac{abc}{bc}+\dfrac{abc}{ac}=abc\)
\(\Rightarrow a+b+c=abc\)
Có \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=2^2\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{c+a+b}{abc}\right)=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\) (do \(a+b+c=abc\))
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\). (đpcm).
3.
\(A=\dfrac{2x+1}{x^2+2}=\dfrac{x^2+2-x^2+2x-1}{x^2+2}=\dfrac{\left(x^2+2\right)-\left(x^2-2x+1\right)}{x^2+2}=1-\dfrac{\left(x-1\right)^2}{x^2+2}\)
Ta có: \(\dfrac{\left(x-1\right)^2}{x^2+2}\ge0\forall x\in R\)
⇒ \(A=1-\dfrac{\left(x-1\right)^2}{x^2+2}\le1\)
Vậy: \(Max_A=1\Leftrightarrow x=1\)
* \(A=\dfrac{2x+1}{x^2+2}=\dfrac{2\left(2x+1\right)}{2\left(x^2+2\right)}=\dfrac{4x+2}{2\left(x^2+2\right)}=\dfrac{-x^2-2+x^2+4x+4}{2\left(x^2+2\right)}\)
\(=-\dfrac{1}{2}+\dfrac{x^2+4x+4}{x^2+2}=-\dfrac{1}{2}+\dfrac{\left(x+2\right)^2}{x^2+2}\ge-\dfrac{1}{2}\)
Vậy: \(Min_A=-\dfrac{1}{2}\Leftrightarrow x=-2\)
* \(B=\dfrac{4x+3}{x^2+1}\) ( 1 cách khác)
\(\Rightarrow B\left(x^2+1\right)=4x+3\)
\(\Rightarrow Bx^2-4x+B-3=0\) (1) \(\left(a=B;b=-4,c=B-3\right)\)
* Với B = 0, pt (1) có nghiệm x = \(-\dfrac{3}{4}\)
* Với B ≠ 0, pt (1) có nghiệm khi và chỉ khi:
\(\Delta=b^2-4ac\ge0\)
\(\Rightarrow\left(-4\right)^2-4.B.\left(B-3\right)\ge0\)
\(\Rightarrow16-4B^2+12B\ge0\)
\(\Rightarrow\left(B-4\right)\left(B+1\right)\ge0\)
\(\Rightarrow-1\le B\le4\)
Suy ra: \(Min_B=-1\Leftrightarrow x=\dfrac{-b}{2a}=\dfrac{4}{2.\left(-1\right)}=-2\)
\(Max_B=4\Leftrightarrow x=\dfrac{-b}{2a}=\dfrac{4}{2.4}=\dfrac{1}{2}\)
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{ac}+\dfrac{2}{bc}=4\)
<=>\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\) +\(2\left(\dfrac{c}{abc}+\dfrac{b}{abc}+\dfrac{a}{abc}\right)=4\)
<=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{a+b+c}{abc}\right)=4\)
<=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{abc}{abc}\right)=4\) (vì a+b+c =abc)
<=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\left(đpcm\right)\)
\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac})=4 \\<=>\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\dfrac{a+b+c}{abc}=4 \\<=>\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4(do\ a+b+c=abc) \\<=>\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2 (đpcm)\)