![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(\Leftrightarrow6\sqrt{x-2}-15\cdot\dfrac{\sqrt{x-2}}{5}=20+4\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{x-2}\cdot6-3\sqrt{x-2}-4\sqrt{x-2}=20\)
\(\Leftrightarrow-\sqrt{x-2}=20\)(vô lý)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
\(\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x-1}\right]:\dfrac{2\sqrt{3x}}{x-1}\)
\(=\left(\dfrac{x+\sqrt{x}+x-\sqrt{x}}{x-1}\right).\dfrac{x-1}{2\sqrt{3x}}\)
\(=\dfrac{2x}{x-1}.\dfrac{x-1}{2\sqrt{3x}}=\dfrac{\sqrt{x}}{\sqrt{3}}=\dfrac{\sqrt{3x}}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3+\sqrt{2x-3}=x\) (ĐKXĐ: x \(\ge\)1,5)
\(\Leftrightarrow\sqrt{2x-3}=x-3\)
\(\Leftrightarrow2x-3=x^2-6x+9\)
\(\Leftrightarrow-x^2+8x-12=0\)
\(\Leftrightarrow-\left(x^2-8x+12\right)=0\)
\(\Leftrightarrow x^2-6x-2x+12=0\)
\(\Leftrightarrow x.\left(x-6\right)-2.\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-2\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=6\\x=2\end{cases}\left(\text{TMĐK}\right)}\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
1:
1: =2căn 3-3căn 3+4căn 3=3căn 3
2: =(3căn 5+2căn 5-4căn 5)/căn 5=1
3: =6căn 3-4/3*căn 3-4*căn 3-5/3*căn 3
=-căn 3
4: =(2căn 7-2căn 14+căn 7)*căn 7+14căn 2
=21
5:
\(=\left(3\sqrt{5}-2\sqrt{3}\right)\left(3\sqrt{5}+2\sqrt{3}\right)=45-12=33\)
6: \(=\left(6\sqrt{5}-5\sqrt{5}-3\sqrt{5}\right)\cdot\sqrt{5}\)
=-2căn 5*căn 5=-10
7:
\(=\dfrac{6\cdot8\sqrt{2}-\dfrac{3}{5}\cdot5\sqrt{2}+14\sqrt{2}}{3\sqrt{2}}=16-1+\dfrac{14}{3}=15+\dfrac{14}{3}=\dfrac{59}{3}\)
8: \(=\left(8\sqrt{3}-\sqrt{3}+3\sqrt{3}\right)\cdot2\sqrt{3}\)
=60
11:
=2+căn 3-2+căn 3=2*căn 3
![](https://rs.olm.vn/images/avt/0.png?1311)
a:
b: Tọa độ A là:
x=-x+3 và y=x
=>x=3/2 và y=3/2
=>A(1,5;1,5)
Tọa độ B là:
2x=-x+3 và y=2x
=>3x=3 và y=2x
=>x=1 và y=2
=>B(1;2)
O(0;0): A(1,5;1,5); B(1;2)
\(OA=\sqrt{1.5^2+1.5^2}=\dfrac{3}{2}\sqrt{2}\)
\(OB=\sqrt{1^2+2^2}=\sqrt{5}\)
\(AB=\sqrt{\left(1-1.5\right)^2+\left(2-1.5\right)^2}=\dfrac{\sqrt{2}}{2}\)
\(cosAOB=\dfrac{OA^2+OB^2-AB^2}{2\cdot OA\cdot OB}=\dfrac{3}{\sqrt{10}}\)
=>\(sinAOB=\dfrac{1}{\sqrt{10}}\)
\(S_{OAB}=\dfrac{1}{2}\cdot OA\cdot OB\cdot sinAOB=\dfrac{1}{2}\cdot\dfrac{1}{\sqrt{10}}\cdot\dfrac{3}{\sqrt{10}}=\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(Q=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{x-1}\cdot\dfrac{1}{\sqrt{x}}=\dfrac{2}{x-1}\)
b: |Q|>-Q
=>|Q|+Q>0
=>Q>0
=>\(\dfrac{2}{x-1}>0\)
=>x-1>0
=>x>1
c: Để Q nguyên thì \(2⋮x-1\)
=>\(x-1\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{2;0;3;-1\right\}\)
mà x>0 và x<>1
nên \(x\in\left\{2;3\right\}\)