1, Đặt \(\dfrac{1}{\sqrt{x}-4}=t_1\left(t_1>0\right)\\ \dfrac{1}{\sqrt{y}+3}=t_2\left(t_2>0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}6t_1-10t_2=1\\2t_1+15t_2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6t_1-10t_2=1\\6t_1+45t_2=12\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-55t_2=-11\\2t_1+15t_2=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}t_2=\dfrac{1}{5}\\t_1=\dfrac{1}{2}\end{matrix}\right.\left(t/m\right)\)
\(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}-4}=\dfrac{1}{2}\\\dfrac{1}{\sqrt{y}+3}=\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-4=2\\\sqrt{y}+3=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=6\\\sqrt{y}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=36\\y=4\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(36;4\right)\)
1: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0;x< >16\\y>=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{6}{\sqrt{x}-4}-\dfrac{10}{\sqrt{y}+3}=1\\\dfrac{2}{\sqrt{x}-4}+\dfrac{15}{\sqrt{y}+3}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{6}{\sqrt{x}-4}-\dfrac{10}{\sqrt{y}+3}=1\\\dfrac{6}{\sqrt{x}-4}+\dfrac{45}{\sqrt{y}+3}=12\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{55}{\sqrt{y}+3}=-11\\\dfrac{6}{\sqrt{x}-4}-\dfrac{10}{\sqrt{y}+3}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{y}+3=5\\\dfrac{6}{\sqrt{x}-4}=1+\dfrac{10}{5}=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{y}=2\\\sqrt{x}-4=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{y}=2\\\sqrt{x}=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=36\\y=4\end{matrix}\right.\left(nhận\right)\)
2:
a: Khi m=3 thì phương trình sẽ là:
\(x^2+4x-3^2+4=0\)
=>\(x^2+4x-5=0\)
=>(x+5)(x-1)=0
=>\(\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
b: \(\text{Δ}=4^2-4\cdot1\left(-m^2+4\right)\)
\(=16+4m^2-16=4m^2>=0\forall m\)
Để Phương trình có hai nghiệm phân biệt thì Δ>0
=>4m^2>0
=>m<>0
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-4\\x_1x_2=\dfrac{c}{a}=-m^2+4\end{matrix}\right.\)
\(x_2=x_1^3+4x_1^2\)
=>\(x_2=x_1^3+x_1^2\left(-x_1-x_2\right)\)
=>\(x_2=x_1^3-x_1^3-x_2\cdot x_1^2\)
=>\(x_2+x_2\cdot x_1^2=0\)
=>\(x_2\left(1+x_1^2\right)=0\)
=>\(x_2=0\)
=>\(x_1=-4\)
\(x_1x_2=-m^2+4\)
=>\(4-m^2=-4\)
=>m2=8
=>\(m=\pm2\sqrt{2}\)
