1
ĐK: \(x\ne\dfrac{4}{3};y\ge-1\)
Đặt \(a=\dfrac{1}{3x-4};b=\sqrt{y+1}\) ta có:
\(\left\{{}\begin{matrix}a+3b=2\\3a+5b=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a+9b=6\\3a+5b=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4b=2\\a+3b=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{2}\\a+\dfrac{3}{2}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{2}\\a=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{3x-4}=\dfrac{1}{2}\\\sqrt{y+1}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-4=2\\y+1=\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{3}{4}\end{matrix}\right.\)(T/m)
2
Xét PT hoành độ giao điểm (d) và (P) có:
\(x^2=\left(m+2\right)x+3\Leftrightarrow x^2-\left(m+2\right)x-3\left(1\right)\)
a)
(d) cắt (P) tại 2 điểm phân biệt\(\Leftrightarrow PT\left(1\right)\) có 2 nghiệm phân biệt
\(\Delta=\left[-\left(m+2\right)\right]^2-4\cdot1\cdot\left(-3\right)=\left(m+2\right)^2+12>0\forall m\)
\(\Rightarrow\) PT luôn có 2 nghiệm phân biệt
\(\Rightarrow\) (d) luôn cắt (P) tại 2 điểm phân biệt
b)
Theo Vi-ét ta có:\(\left\{{}\begin{matrix}x_1+x_2=m+2\\x_1\cdot x_2=-3\end{matrix}\right.\)
Theo đề bài ta có: \(x_1^2+x_2^2=12-x_1x_2\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=12-x_1x_2\)
\(\Leftrightarrow\left(x_1+x_2\right)^2=12+x_1x_2\)
\(\Leftrightarrow\left(m+2\right)^2=12-3\)
\(\Leftrightarrow\left(m+2\right)^2=9\)
\(\Leftrightarrow m+2=\pm3\)
\(\Leftrightarrow\left\{{}\begin{matrix}m=1\\m=-5\end{matrix}\right.\)
