Question

Answer 1

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-2\\y\ge0\\y\ne1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{1}{x+2}+\dfrac{1}{\sqrt{y}-1}=1\\\dfrac{3}{x+2}+\dfrac{2}{\sqrt{y}-1}=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+2}+\dfrac{2}{\sqrt{y}-1}=2\\\dfrac{3}{x+2}+\dfrac{2}{\sqrt{y}-1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-1}{x+2}=1\\\dfrac{3}{x+2}+\dfrac{2}{\sqrt{y}-1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\\dfrac{3}{-3+2}+\dfrac{2}{\sqrt{y}-1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\-3+\dfrac{2}{\sqrt{y}-1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\\dfrac{2}{\sqrt{y}-1}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\\sqrt{y}-1=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\\sqrt{y}=\dfrac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=\dfrac{9}{4}\left(tm\right)\end{matrix}\right.\)

 

Answer 2

a: Khi m=1 thì (d1): y=-2x+2

Tọa độ giao điểm là:

\(\left\{{}\begin{matrix}-2x+2=2x-3\\y=2x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{4}\\y=\dfrac{5}{2}-3=-\dfrac{1}{2}\end{matrix}\right.\)

Answer 3

Bài 1:
Đặt $\frac{1}{x+2}=a; \frac{1}{\sqrt{y}-1}=b$ thì hpt trở thành:
\(\left\{\begin{matrix} a+b=1\\ 3a+2b=1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 2a+2b=2\\ 3a+2b=1\end{matrix}\right.\Rightarrow (3a+2b)-(2a+2b)=1-2\)

$\Leftrightarrow a=-1$

$\Leftrightarrow \frac{1}{x+2}=-1\Leftrightarrow x+2=-1$

$\Leftrightarrow x=-3$

Khi $a=-1$ thì $b=1-a=1-(-1)=2$

$\Leftrightarrow \frac{1}{\sqrt{y}-1}=2\Leftrightarrow \sqrt{y}=\frac{3}{2}$

$\Leftrightarrow y=\frac{9}{4}$ (tm)

Answer 4

Bài 2:
a. Khi $m=1$ thì $(d_1): y=-2x+2$ và $(d_2): y=2x-3$
PT hoành độ giao điểm:

$-2x+2=2x-3$

$\Leftrightarrow 5=4x\Leftrightarrow x=\frac{5}{4}$

$y=2x-3=2.\frac{5}{4}-3=\frac{-1}{2}$

Vậy tọa độ giao điểm là $(\frac{5}{4}, \frac{-1}{2})$

b.Gọi $(x_0, y_0)$ là điểm cố định luôn đi qua $(d_1)$ với mọi $m$ 

Tức là:
$y_0=(m-3)x_0+m+1, \forall m\in\mathbb{R}$

$\Leftrightarrow m(x_0+1)+(1-3x_0-y_0)=0, \forall m\in\mathbb{R}$

$\Leftrightarrow x_0+1=1-3x_0-y_0=0$

$\Leftrightarrow y_0=-3x_0+1$

$\Rightarrow$ điểm cố định $(x_0, y_0)$ luôn đi qua đt có pt $y=-3x+1$

$\Rightarrow x_0=-4$