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Ta có (a + b + c)(ab + bc + ca) = 2020
<=> \(\left(a+b+c\right)abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2020\)
<=> \(\left(a+b+c\right).2020.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2020\)
<=> \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)
<=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
<=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
<=> \(\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)
<=> \(\left(a+b\right)\left(\frac{1}{ab}+\frac{1}{c\left(a+b+c\right)}\right)=0\)
=> (a + b)(a + c)(b + c) = 0
=> a = -b hoặc a = -c ; b = -c (1)
Khi đó P = (b2c + 2020)(c2a + 2020)(a2b + 2020)
= (b2c + abc)(c2a + abc)(a2b + abc)
= bc(b + a)ac(b + c)ab(a + c)
= (abc)2.0 = 0 (Theo (1))
T cx trl đầy câu ra đấy thôi, có đc tick cái nào đâu, nhiều lần rồi, ngồi cả buổi trưa để lm bài mà cx = 0
\(x^2+y^2+1\ge xy+x+y\)
\(\Leftrightarrow2\left(x^2+y^2+1\right)\ge2\left(xy+x+y\right)\)
\(\Leftrightarrow x^2-2xy+y^2+y^2-2y+1+x^2-2x+1\ge0\)\(\Leftrightarrow\left(x-y\right)^2-\left(y-1\right)^2-\left(x-1\right)^2\ge0\)
Đúng với mọi x , y
Đẳng thức xảy ra khi \(\left[{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-y=0\\y-1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\y=1\\x=1\end{matrix}\right.\Rightarrow x=y=1\)
b, \(A=\dfrac{x-2}{x^3-x^2-x-2}=\dfrac{x-2}{x^3-2x^2+x^2-2x+x-2}\)
\(=\dfrac{x-2}{x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)}\)
\(=\dfrac{x-2}{\left(x^2+x+1\right)\left(x-2\right)}=\dfrac{1}{x^2+x+1}\)
\(=\dfrac{1}{x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)
Ta có: \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
Dấu " = " xảy ra khi \(\left(x+\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{-1}{2}\)
Vậy \(MAX_A=\dfrac{4}{3}\) khi \(x=\dfrac{-1}{2}\)
