\(B=\frac{7y^2-4xy}{x^2-2xy+2y^2}=\frac{\left(-x^2+2xy-2y^2\right)+\left(x^2-6xy+9y^2\right)}{x^2-2xy+2y^2}=\frac{\left(x-3y\right)^2}{x^2-2xy+2y^2}-1\ge-1\)

Dấu =  xảy ra khi x = 3y

Ta có:

\(\left(\frac{1}{a+3b}+\frac{1}{a+b+2c}\right)+\left(\frac{1}{b+3c}+\frac{1}{2a+b+c}\right)+\left(\frac{1}{c+3a}+\frac{1}{a+2b+c}\right)\)

\(\ge\frac{4}{2a+4b+2c}+\frac{4}{2a+2b+4c}+\frac{4}{4a+2b+2c}\)

\(\ge\frac{2}{a+2b+c}+\frac{2}{a+b+2c}+\frac{2}{2a+b+c}\)

Vậy

\(\frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a}\ge\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}+\)

Đặt \(\hept{\begin{cases}a=3x^2-4x+1\\b=3x^2+2x+1\end{cases}}\)

\(\Rightarrow b-a=6x\)

Thế lại bài toán được

\(\frac{2}{a}+\frac{13}{b}=\frac{36}{b-a}\)

\(\Leftrightarrow2b\left(b-a\right)+13a\left(b-a\right)-36ab=0\)

\(\Leftrightarrow2b^2-25ab-13a^2=0\)

\(\Leftrightarrow\left(ab+2b^2\right)-\left(13a^2+26ab\right)=0\)

\(\Leftrightarrow\left(a+2b\right)\left(b-13a\right)=0\)

Làm nốt nha

accept đi mình ở An Quý nè

Xét \(x=0\Rightarrow B=0\left(1\right)\)

Xét \(x\ne0\) thì ta có:

\(\frac{1}{B}=\frac{x^2+20x+100}{x}=x+\frac{100}{x}+20\ge2\sqrt{x.\frac{100}{x}}+20=40\)

\(\Rightarrow B\le\frac{1}{40}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\)GTLN là \(\frac{1}{40}\)

\(\frac{a}{-a+2b+2c}+\frac{b}{2a-b+2c}+\frac{2}{2a+2b-c}\)

\(=\frac{a^2}{-a^2+2ab+2ca}+\frac{b^2}{2ab-b^2+2bc}+\frac{c^2}{2ac+2bc-c^2}\)

\(\ge\frac{\left(a+b+c\right)^2}{-\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)}\)

\(=\frac{\left(a+b+c\right)^2}{6\left(ab+bc+ca\right)-\left(a+b+c\right)^2}\)

\(\ge\frac{\left(a+b+c\right)^2}{\frac{6\left(a+b+c\right)^2}{3}-\left(a+b+c\right)^2}=1\)

Rút y ra ta có

\(\)\(y=\frac{x^2+2x-1}{x^2-2x-1}=1+\frac{4x}{x^2-2x-1}\) là số nguyên 

với x=0 thì y=1

với x khác 0 ta có  \(\frac{x^2-2x-1}{x}\text{ là ước của 4}\)hay \(x-2-\frac{1}{x}\in\left\{\pm1;\pm2\right\}\)

\(\Rightarrow x=\pm1\Rightarrow y=-2\)

vậy ta có 3 nghiệm nguyên \(\left(x,y\right)\in\left\{\left(0,1\right),\left(1,-2\right),\left(-1,-2\right)\right\}\)

Nghiệm nguyên hả?

Đáp án đây nhé

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