tìm x,y biết x+y=-5 và xy=6
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\(\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}-\frac{2x}{x-1}\)(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
\(\frac{4}{x^2+2x-3}-\frac{2x-5}{x+3}-\frac{2x}{x-1}ĐK:x\ne1;-3\)
\(=\frac{4}{\left(x+3\right)\left(x-1\right)}-\frac{\left(2x-5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}\)
\(=\frac{4-2x^2+2x+5x-5-2x^2-6x}{\left(x-1\right)\left(x+3\right)}\)
\(=\frac{-1-4x^2+x}{\left(x-1\right)\left(x+3\right)}\)
\(\hept{\begin{cases}x+y=-5\\xy=6\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-5-x\\x.\left(-5-x\right)=6\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-5-x\\x^2+5x+6=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y=-5-x\\\left(x+2\right)\left(x+3\right)=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2,y=-3\\x=-3,y=-2\end{cases}}\)
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)