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)
a: ĐKXĐ: x>=0; x<>4
\(Q=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+2\sqrt{x}\left(\sqrt{x}+2\right)-3x-4}{x-4}\cdot\dfrac{\sqrt{x}-2+2}{2}\)
\(=\dfrac{x-2\sqrt{x}+2x+4\sqrt{x}-3x-4}{x-4}\cdot\dfrac{\sqrt{x}}{2}\)
\(=\dfrac{2\sqrt{x}-4}{x-4}\cdot\dfrac{\sqrt{x}}{2}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
b: \(M=P\cdot Q=\dfrac{\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{1-5\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)
\(M\left(M-1\right)=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-5x-x-3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)\left(-6x-2\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)^2\cdot\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{\sqrt{x}\left(5\sqrt{x}-1\right)\left(6x+2\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}+1\right)^2}\)
TH1: M>=căn M
=>M^2>=M
=>M^2-M>=0
=>5*căn x-1>=0
=>x>=1/25 và x<>4
TH2: M<căn M
=>5căn x-1<0
=>x<1/25
Kết hợp ĐKXĐ, ta được: 0<=x<1/25
![](https://rs.olm.vn/images/avt/0.png?1311)
Với \(x\ge0;x\ne4\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{x-4}{\sqrt{x}-2}\)
\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}-\sqrt{x}-2-3\sqrt{x}+2}{x-4}.\dfrac{x-4}{\sqrt{x}-2}\)
\(=\dfrac{2x-4\sqrt{x}}{x-4}.\dfrac{x-4}{\sqrt{x}-2}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=2\sqrt{x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-3\sqrt{x}-2}{x-4}\right):\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\\ =\dfrac{x+2\sqrt{x}+x-\sqrt{x}-2\sqrt{x}+2-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\times\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\\ =\dfrac{2x-4\sqrt{x}}{\sqrt{x}-2}\times\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\times\dfrac{1}{\sqrt{x}-2}=\dfrac{2\sqrt{x}}{\sqrt{x}-2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(M=1:\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{3x}{2\left(x-4\right)}+\dfrac{1}{2\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{4-2\sqrt{x}}{1}\)
\(=1:\left(\dfrac{2\sqrt{x}-4-3x+\sqrt{x}+2}{2\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\cdot\dfrac{-2\left(\sqrt{x}-2\right)}{1}\)
\(=\dfrac{2\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\cdot\left(-2\right)\cdot\left(\sqrt{x}-2\right)}{-3x+3\sqrt{x}-2}\)
\(=\dfrac{-4\left(\sqrt{x}-2\right)^2\left(\sqrt{x}+2\right)}{-3x+3\sqrt{x}-2}\)
b: M=20
=>\(-4\left(x-4\right)\left(\sqrt{x}-2\right)=-60x+60\sqrt{x}-40\)
=>\(x\sqrt{x}-2x-4\sqrt{x}+8=-15x+15\sqrt{x}-10\)
=>\(x\sqrt{x}+13x-19\sqrt{x}+18=0\)
=>\(x\in\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(P=\left(\dfrac{x-2\sqrt{x}+4}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{x+4}{x-4}\right)\)
\(=\dfrac{x-2\sqrt{x}+4}{\sqrt{x}-2}:\dfrac{x+4\sqrt{x}+4+x-2\sqrt{x}-x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-2\sqrt{x}+4}{1}\cdot\dfrac{\sqrt{x}+2}{x+2\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+4}{\sqrt{x}}\)
b: \(P-2=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}}>0\forall x\) thỏa mãn ĐKXĐ
nên P>2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(P=\left(\dfrac{\sqrt{x}}{x-4}-\dfrac{2}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\left(\dfrac{\sqrt{x}-2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\dfrac{-\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\dfrac{-\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+4}\)
\(=\dfrac{-1}{\sqrt{x}-2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{2}{\sqrt{x^2}-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right):\left(\sqrt{x}-\dfrac{x-4}{\sqrt{x}+2}\right)\) (ĐK: \(x>0;x\ne4\))
\(=\left[\dfrac{2}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]:\left[\sqrt{x}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\right]\)
\(=\dfrac{2-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\left(\sqrt{x}-\sqrt{x}+2\right)\)
\(=-\dfrac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-2\right)}:2\)
\(=-\dfrac{1}{\sqrt{x}}:2\)
\(=-\dfrac{1}{\sqrt{x}}\cdot\dfrac{1}{2}\)
\(=-\dfrac{1}{2\sqrt{x}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{x-4}=\dfrac{x-2\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
M= \(\dfrac{x+2+\sqrt{x^2-4}}{x+2-\sqrt{x^2-4}}+\dfrac{x+2-\sqrt{x^2-4}}{x+2+\sqrt{x^2-4}}\)
M=\(\dfrac{(x+2+\sqrt{x^2-4})^2}{(x+2)^2+\sqrt{x^2-4}^2}+\dfrac{(x+2-\sqrt{x^2-4})^2}{(x+2)^2+\sqrt{x^2-4}^2}\)
M=\(\dfrac{\left(x+2\right)^2+2.\left(x+2\right).\sqrt{x^2-4}+x^2-4}{x^2+4x+4-x^2+4}+\dfrac{\left(x+2\right)^2-2.\left(x+2\right).\sqrt{x^2-4}+x^2-4}{x^2+4x+4-x^2+4}\)
M=\(\dfrac{\left(x+2\right)^2+x^2-4+\left(x+2\right)^2+x^2-4}{x^2+4x+4-x^2+4}\)
M=\(\dfrac{x^2+4x+4+x^2-4+x^2+4x+4+x^2-4}{4x+8}\)
M= \(\dfrac{4x^2+8}{4x+8}\)
M= \(\dfrac{x.\left(4x+8\right)}{4x+8}\)
M= x
OK nha bn !!