a. Phương trình hoành độ giao điểm:

\(3x-5=-2x\)

\(\Leftrightarrow5x=5\)

\(\Rightarrow x=1\)

Thế vào \(y=3x-5\Rightarrow y=3.1-5=-2\)

Vậy \(A\left(1;-2\right)\)

b. Gọi phương trình d có dạng \(y=ax+b\)

Do d song song \(d_1\Rightarrow a=1\Rightarrow y=x+b\)

Do d qua A nên: \(y_A=x_A+b\Leftrightarrow-2=1+b\Rightarrow b=-3\)

Vậy pt d có dạng: \(y=x-3\)

giúp gì vậy ạ??

ta có sinB=\(\dfrac{AH}{AB}\)\(\Rightarrow\)AH=AB.sinB=3,6.sin62=3,18

BH=\(\sqrt{AB^2-AH^2}\)(pytago)=\(\sqrt{3,6^2-3,18^2}\)=1,69

\(_{\widehat{C}}\)=90-\(\widehat{B}\)=90-62=28\(^0\)

sinC=\(\dfrac{AB}{BC}\)\(\Rightarrow\)BC=\(\dfrac{AB}{sinC}\)=\(\dfrac{3,6}{sin28}\)=7,67

mà:CH=BC-BH=7,67-1,69=5,98

AC=\(\sqrt{BC^2-AB^2}\)(pytago)=\(\sqrt{7,67^2-3,6^2}\)=6.77

Bài 2: 

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{4;9\right\}\end{matrix}\right.\)

b) Ta có: \(B=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}-9-\left(x-9\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

c) Để B>1 thì B-1>0

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}>0\)

\(\Leftrightarrow\dfrac{4}{\sqrt{x}-3}>0\)

\(\Leftrightarrow\sqrt{x}>3\)

hay x>9

Bài 2: 

d) Để B nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)

\(\Leftrightarrow4⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{-2;-1;1;2;4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{1;2;4;5;7\right\}\)

hay \(x\in\left\{1;16;25;49\right\}\)

a, Ta có : \(A=\dfrac{1}{x+\sqrt{x}}+\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(1+\sqrt{x}\right)}=\dfrac{1}{\sqrt{x}}\)

\(\Rightarrow P=\dfrac{A}{B}=\dfrac{\dfrac{1}{\sqrt{x}}}{\dfrac{2}{\sqrt{x}+1}}=\dfrac{1}{\sqrt{x}}.\dfrac{\sqrt{x}+1}{2}=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)

b, Ta có : \(P=\dfrac{\sqrt{x}+1}{2\sqrt{x}}=\dfrac{1}{2}+\dfrac{1}{2\sqrt{x}}\)

Mà \(x>0\)

\(\Rightarrow\dfrac{1}{2\sqrt{x}}>0\)

\(\Rightarrow P>\dfrac{1}{2}\)

Vậy ...

\(\begin{cases}\dfrac{1}{x+3}-\dfrac{2}{y-1}=9\\\dfrac{3}{x+3}+\dfrac{1}{y-1}=6\\\end{cases}\)

`<=>` \(\begin{cases}\dfrac{1}{x+3}-\dfrac{2}{y-1}=9\\\dfrac{6}{x+3}+\dfrac{2}{y-1}=12\\\end{cases}\)

`<=>` \(\begin{cases}\dfrac{7}{x+3}=21\\\dfrac{3}{x+3}+\dfrac{1}{y-1}=6\\\end{cases}\)

`<=>` \(\begin{cases}\dfrac{1}{x+3}=3\\\dfrac{3}{x+3}+\dfrac{1}{y-1}=6\\\end{cases}\)

`<=>` \(\begin{cases}x+3=\dfrac13\\\dfrac{1}{y-1}=6-9=-3\\\end{cases}\)

`<=>` \(\begin{cases}x=\dfrac{-8}{3}\\y=\dfrac23\\\end{cases}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne-3\\y\ne1\end{matrix}\right.\)

- Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x+3}=a\\\dfrac{1}{y-1}=b\end{matrix}\right.\)

HPTTT : \(\left\{{}\begin{matrix}a-2b=9\\3a+b=6\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a-2b=9\\6a+2b=12\end{matrix}\right.\)

- Cộng hai phương trình ta được : 7a = 21

=> a = 3

=> b = -3

- Thay lại hệ phương trình ta được ; \(\left\{{}\begin{matrix}\dfrac{1}{x+3}=3\\\dfrac{1}{y-1}=-3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{8}{3}\\y=\dfrac{2}{3}\end{matrix}\right.\) ( TM )

Vậy ..