Hoktot

\(tanB=\frac{3}{4}\)

\(\Rightarrow\frac{AB}{BC}=\frac{3}{4}\)

Ta có:

\(AC^2+AB^2=BC^2\)

\(\Rightarrow AB^2=BC^2-AC^2=\frac{16}{9}AC^2-AC^2=\frac{7}{9}AC^2=144\)

\(\Rightarrow AC=13,6\)

\(\Rightarrow BC=18,1\)

a, \(x-3\sqrt{x}+2=0\Leftrightarrow x-2\sqrt{x}-\sqrt{x}+2=0\)đk : x >= 0 

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)=0\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\Leftrightarrow x=1;x=4\)

b, \(\sqrt{x^2-1}-\sqrt{x+1}=0\Leftrightarrow\sqrt{\left(x-1\right)\left(x+1\right)}-\sqrt{x+1}=0\)đk : \(x\ge1\)

\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x-1}-1\right)=0\)

TH1 : \(x=-1\)( loại )

TH2 : \(\sqrt{x-1}=1\Leftrightarrow x-1=1\Leftrightarrow x=2\)

c, \(x^2+4x+4-\sqrt{2x+1}-\left(x-1\right)^2=0\)đk : x>= -1/2 

\(\Leftrightarrow\left(x+2\right)^2-\left(x-1\right)^2-\sqrt{2x+1}=0\)

\(\Leftrightarrow3\left(2x+1\right)-\sqrt{2x+1}=0\Leftrightarrow\sqrt{2x+1}\left(3\sqrt{2x+1}-1\right)=0\)

TH1 : \(x=-\frac{1}{2}\)

TH2 : \(\sqrt{2x+1}=\frac{1}{3}\Leftrightarrow2x+1=\frac{1}{9}\Leftrightarrow x=\frac{\frac{1}{9}-1}{2}=\frac{-\frac{8}{9}}{2}=-\frac{4}{9}\)

a) ĐK : x \(\ge0\) 

\(x-3\sqrt{x}+2=0\)

<=> \(\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)

<=> \(\orbr{\begin{cases}\sqrt{x}-1=0\\\sqrt{x}-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=4\end{cases}}\)(tm) 

b) ĐK \(\hept{\begin{cases}x\ge-1\\x\notin\left\{x\in R|-1< x< 0\right\}\end{cases}}\)

\(\sqrt{x^2-1}-\sqrt{x+1}=0\)

<=> \(\sqrt{x-1}\sqrt{x+1}-\sqrt{x+1}=0\)

<=> \(\sqrt{x-1}\left(\sqrt{x+1}-1\right)=0\)

<=> \(\orbr{\begin{cases}\sqrt{x+1}=0\\\sqrt{x-1}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-1=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)(tm) 

c) ĐK : \(x\ge-\frac{1}{2}\)

 \(x^2+4x+4-\sqrt{2x+1}-\left(x-1\right)^2=0\)

<=> \(6x+3-\sqrt{2x+1}=0\)

<=> \(\sqrt{2x+1}\left(3\sqrt{2x+1}-1\right)=0\)

<=> \(\orbr{\begin{cases}\sqrt{2x+1}=0\\3\sqrt{2x+1}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-\frac{4}{9}\end{cases}}\)(tm) 

a, \(\frac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}=\frac{\sqrt{10}+\sqrt{6}}{\sqrt{10.3}+\sqrt{6.3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)

b, Với a;b > 0 

\(\frac{a+\sqrt{ab}}{b+\sqrt{ab}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}=\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{ab}}{b}\)

c, Với x >= 0 

\(\frac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}=\frac{\left(\sqrt{x}-1\right)\left(4\sqrt{x}+7\right)}{4\sqrt{x}+7}=\sqrt{x}-1\)

d, Với x >= 0 ; x khác 14

\(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)

a) \(\frac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}=\frac{\sqrt{10}+\sqrt{6}}{\sqrt{3}\left(\sqrt{10}+\sqrt{6}\right)}=\frac{1}{\sqrt{3}}\)

b) \(\frac{a+\sqrt{ab}}{b+\sqrt{ab}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}=\frac{\sqrt{a}}{\sqrt{b}}\)

c) \(\frac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}=\frac{\left(\sqrt{x}-1\right)\left(4\sqrt{x}+7\right)}{\left(4\sqrt{x}+7\right)}=\sqrt{x}-1\)

d) \(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\frac{x+\sqrt{x}-4\sqrt{x}-4}{x-4\sqrt{x}+3\sqrt{x}-12}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)

a) \(3a-2\sqrt{ab}-b=3a-3\sqrt{ab}+\sqrt{ab}-b\)

\(=3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)=\left(3\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)

b) \(5a+3\sqrt{ab}-8b=5a-5\sqrt{ab}+8\sqrt{ab}-8b\)

\(=5\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)+8\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\left(5\sqrt{a}+8\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)

a) (\(\sqrt{a}-\sqrt{b}\))(3\(\sqrt{a}+b\))

b) \(\left(\sqrt{a}-\sqrt{b}\right)\left(5\sqrt{a}+8\sqrt{b}\right)\)

ta có :

Hay quá, Minh Quang không bị lừa :)))

tui lại 

úi sao bạn cũng là quản lý giống mình à, mình trả lời câu hỏi của bạn có được không nhỉ 

a) \(x-2\sqrt{x-1}-4=\left(x-1\right)-2\sqrt{x-1}+1-4\)

\(=\left(\sqrt{x-1}-1\right)^2-4=\left(\sqrt{x-1}-3\right)\left(\sqrt{x-1}+1\right)\)

b) \(x-2\sqrt{x-6}-5-y^2=\left(x-6\right)-2\sqrt{x-6}+1-y^2\)

\(=\left(\sqrt{x-6}-1\right)^2-y^2=\left(\sqrt{x-6}-1+y\right)\left(\sqrt{x-6}-1-y\right)\)

c) \(x-2\sqrt{x-8}-7-a^2=\left(x-8\right)-2\sqrt{x-8}+1-a^2\)

\(=\left(\sqrt{x-8}-1\right)^2-a^2=\left(\sqrt{x-8}+a-1\right)\left(\sqrt{x-8}-a-1\right)\)

a) \(\left(\sqrt{x-1}-3\right)\left(\sqrt{x-1}+1\right)\)

b) \(\left(\sqrt{x-6}-1-y\right)\left(\sqrt{x-6}-1+y\right)\)

c) \(\left(\sqrt{x-8}-1-a\right)\left(\sqrt{x-8}-1+a\right)\)

lll

a) (\(\sqrt{x}\)-1)(\(\sqrt{x}\)+7)

b) (\(\sqrt{x}\)-4)(\(\sqrt{x}\)-2)

c) (\(\sqrt{x}\)+1)(3\(\sqrt{x}\)+2)