2 câu đầu của bạn như bạn này mik giải rồi nhá

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+) \(P=\dfrac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(c+ab\right)\left(a+bc\right)\left(b+ca\right)}\)

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

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

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

+) \(M=\dfrac{1}{ab+a+1}+\dfrac{1}{bc+b+1}+\dfrac{1}{ca+c+1}\)

\(=\dfrac{1}{ab+a+1}+\dfrac{abc}{bc+ab^2c+abc}+\dfrac{abc}{ca+c+abc}\left(vìabc=1\right)\)

\(=\dfrac{1}{ab+a+1}+\dfrac{abc}{bc\left(ab+a+1\right)}+\dfrac{abc}{c\left(ab+a+1\right)}\)

\(=\dfrac{ab+a+1}{ab+a+1}=1\)

\(\left(x^2-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)

\(=\left(x^2\right)^2-2.x^2.2y+\left(2y\right)^2+\left(x^2-\left(\dfrac{1}{2}y\right)^2\right)\)

\(=x^4-4x^2y+4y^2+x^2-\dfrac{y^2}{4}\)

\(=x^4-4x^2y+x^2+\dfrac{15y^2}{4}\)

x² + 8 - xy + 6x - 5y = 0

<=> x² + 6x + 8 - y(x + 5) = 0

<=> x² + 6x + 8 = y(x + 5) 

<=> \(y=\dfrac{x^2+6x+8}{x+5}\)

Có \(y=\dfrac{x^2+6x+8}{x+5}=x+1+\dfrac{3}{x+5}\)

Để \(y\inℤ\Rightarrow3⋮x+5\Rightarrow x+5\inƯ\left(3\right)\)

\(\Rightarrow x+5\in\left\{1;3;-1;-3\right\}\Leftrightarrow x\in\left\{-4;-2;-6-8\right\}\)

Với x = -2 => y  = 0 

Với x = -4 => y = -1

Với x = - 6 => y = -8

Với x = -8 => y = -8

Vậy (x;y) = (-2 ; 0) ; (-4 ; -1) ; (-6 ; -8) ; (-8 ; -8) 

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