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Đề sai

hoặc thiếu dữ kiện

Ta có: \(F=\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\)

\(\Leftrightarrow F=\frac{a^2}{ab+ac}+\frac{b^2}{bc+bd}+\frac{c^2}{cd+ca}+\frac{d^2}{da+db}\)  

\(\Leftrightarrow F\ge\frac{\left(a+b+c+d\right)^2}{2ac+2bd+\left(a+c\right)\left(b+d\right)}=P\)

\(\Leftrightarrow P=\frac{a^2+b^2+c^2+2ab+2bc+2cd+2ad+2ac+2bd}{ab+ac+bc+bd+cd+ac+ad+bd}\)

\(\Leftrightarrow P=\frac{\left(a^2+c^2\right)+\left(b^2+d^2\right)+2ab+2bc+2cd+2ad+2ac+2bd}{2ac+2bd+ab+bc+cd+ad}\)

(Vì \(a^2+c^2\ge2ac\Leftrightarrow\left(a-c\right)^2\ge0\)luôn đúng; \(b^2+d^2\ge2bd\Leftrightarrow\left(b-d\right)^2\ge0\)luôn đúng)

\(\Leftrightarrow P\ge\frac{2ac+2bd+2ab+2bc+2cd+2ad+2ac+2bd}{2ac+2bd+ab+cd+ad+ac+bd}\)

\(\Leftrightarrow P\ge\frac{4ac+4bd+2ab+2bc+2cd+2ad}{2ac+2bd+ab+bc+cd+ad}=2\)

\(\Leftrightarrow F\ge P\ge2\)

\(\LeftrightarrowĐPCM\)

Ta có:

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

=> \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=2^2\)

=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=4\)

=> \(2+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)

=> \(2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=2\)

=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\)

=> \(abc.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=abc\)

=> \(c+a+b=abc\) (đpcm)

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{ac}\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)

\(\Rightarrow2^2=2+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)

\(\Leftrightarrow2=2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)

\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\)

\(\Leftrightarrow a+b+c=abc\)

đpcm

\(\frac{\Leftrightarrow c}{abc}+\frac{a}{abc}+\frac{b}{abc}=\frac{abc}{abc}\)

Ta có:

\(\frac{x-1}{x+1}+\frac{x+1}{x-1}+\frac{x^2-3}{x^2-1}\)

\(=\frac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x^2-3}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2-2x+1+x^2+2x+1+x^2-3}{\left(x+1\right)\left(x-1\right)}\)

\(=\frac{3x^2-1}{\left(x+1\right)\left(x-1\right)}\)

bn nan đúng nhưng cho sửa lại đề

a) Ta có: A = x² + y² - xy - 2x - 2y + 9

2A = 2x² + 2y² - 2xy - 4x - 4y + 18

2A = (x² + y² - 2xy) + (x² - 4x + 4) + (x² - 4y + 4) + 10

2A = (x - y)² + (x - 2)² + (y - 2)² + 10 \(\ge\)10 \(\forall\)x

=>A \(\ge\)5 \(\forall\)x

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y=0\\x-2=0\\y-2=0\end{cases}}\) <=> \(\hept{\begin{cases}x=y\\x=2\\y=2\end{cases}}\) <=> x = y = 2

Vậy MinA = 5 <=> x = y = 2

b) Ta có: 3x² + 3y² + 4xy + 2x - 2y + 2 = 0

=> (2x² + 2y² + 4xy) + (x² + 2x + 1) + (y² - 2y + 1) = 0

=> 2(x + y)² + (x + 1)² + (y - 1)² = 0

<=> \(\hept{\begin{cases}x+y=0\\x+1=0\\y-1=0\end{cases}}\) 

<=> \(\hept{\begin{cases}x=-y\\x=-1\\y=1\end{cases}}\)

<=> \(\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

chắc là có

có vì 2y-2y=0

9x-(3^2)x=9x-9x=0