\(\hept{\begin{cases}x+y=-5\\xy=6\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-5-x\\x.\left(-5-x\right)=6\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-5-x\\x^2+5x+6=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=-5-x\\\left(x+2\right)\left(x+3\right)=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2,y=-3\\x=-3,y=-2\end{cases}}\)

Vì xy = 6

=> \(x=\frac{6}{y}\)

Khi đó x + y = -5

<=> \(\frac{6}{y}+y=-5\)

=> \(\frac{y^2+6}{y}=-5\)

=> y² + 6 = -5y

=> y² + 5y + 6 = 0

=> y² + 2y + 3y + 6 = 0

=> y(y + 2) + 3(y + 2) = 0

=> (y + 3)(y + 2) = 0 

=> \(\orbr{\begin{cases}y+3=0\\y+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}y=-3\\y=-2\end{cases}}\)

Khi y = -3 => x = -2

Khi y = -2 => x = -3

Vậy các cặp (x;y) thỏa mãn là (-2 ; - 3) ; (-3 ; -2)

(a^4+b^4+c^4)^3.(1+1+1)≥(a^3+b^3+c^3)^4≥(a^3+b^3+c^3)3.\(\frac{\text{(a+b+c)^3 }}{9}\)

=3(a^3+b^3+c^3)^3
⇒a^4+b^4+c^4≥a^3+b^3+c^3

\(\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}-\frac{2x}{x-1}\)(ĐK: \(x\ne1,x\ne-3\))

\(\Leftrightarrow\frac{4}{x^2+2x-3}=\frac{\left(2x-5\right)\left(x-1\right)-2x\left(x+3\right)}{\left(x+3\right)\left(x-1\right)}\)

\(\Leftrightarrow\frac{-13x+1}{\left(x+3\right)\left(x-1\right)}=0\)

\(\Rightarrow-13x+1=0\Leftrightarrow x=\frac{1}{13}\)(tm).

\(\frac{3}{x^2+x-2}-\frac{1}{x-1}=\frac{-7}{x+2}\)

ĐKXĐ : x ≠ 1 , x ≠ -2

pt <=> \(\frac{3}{\left(x-1\right)\left(x+2\right)}-\frac{x+2}{\left(x-1\right)\left(x+2\right)}+\frac{7\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=0\)

<=> \(\frac{3-x-2+7x-7}{\left(x-1\right)\left(x+2\right)}=0\)

<=> \(\frac{6x-6}{\left(x-1\right)\left(x+2\right)}=0\)

=> 6x - 6 = 0

<=> x = 1 ( ktm )

Vậy phương trình vô nghiệm 

\(\frac{12}{x^2-4}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\)

ĐKXĐ : x ≠ ±2

pt <=> \(\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x+7\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=0\)

<=> \(\frac{12-x^2-3x-2+x^2+5x-14}{\left(x-2\right)\left(x+2\right)}=0\)

<=> \(\frac{2x-4}{\left(x-2\right)\left(x+2\right)}=0\)

=> 2x - 4 = 0

<=> x = 2 ( ktm )

Vậy phương trình vô nghiệm

\(\frac{4}{x^2+2x-3}-\frac{2x-5}{x+3}-\frac{2x}{x-1}ĐK:x\ne1;-3\)

\(=\frac{4}{\left(x+3\right)\left(x-1\right)}-\frac{\left(2x-5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}\)

\(=\frac{4-2x^2+2x+5x-5-2x^2-6x}{\left(x-1\right)\left(x+3\right)}\)

\(=\frac{-1-4x^2+x}{\left(x-1\right)\left(x+3\right)}\)