a/ ĐKXĐ: x khác 0
Đặt \(\dfrac{1}{x}=a\)
hpt <=> \(\left\{{}\begin{matrix}\dfrac{3}{2}a-y=6\left(1\right)\\a+2y=-4\left(2\right)\end{matrix}\right.\)
pt (2) : a + 2y = -4
<=> a = -2y - 4
Thay a = -2y - 4 vào pt (1) ta có:
\(\dfrac{3}{2}\cdot\left(-2y-4\right)-y=6\)
<=> -3y - 6 - y = 6
<=> -4y = 12 <=> y = -3
=> a = -2y - 4 = -2 . (-3) - 4 = 2
hay \(\dfrac{1}{x}=2\Rightarrow x=\dfrac{1}{2}\)
Vậy nghiệm của hệ là:
(x;y) = (\(\dfrac{1}{2};-3\))
b/ \(\dfrac{2x+1}{x-1}+\dfrac{3\left(x-1\right)}{x+1}=6\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{6\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)+3\left(x-1\right)^2=6\left(x^2-1\right)\)
\(\Leftrightarrow2x^2+3x+1+3x^2-6x+3=6x^2-6\)
\(\Leftrightarrow2x^2+3x^2-6x^2+3x-6x=-6-3-1\)
\(\Leftrightarrow-x^2-3x+10=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Vậy............