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27 tháng 2 2022

Trả lời:

 \(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\left(ĐK:x\ge0;x\ne16\right)\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\frac{\sqrt{x}+2}{x+16}\)

\(=\frac{x-4\sqrt{x}+4\sqrt{x}+16}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\frac{\sqrt{x}+2}{x+16}\)

\(=\frac{x+16}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\frac{\sqrt{x}+2}{x+16}=\frac{\sqrt{x}+2}{x-16}\)

8 tháng 3 2022

a, Với x >= 0 ; x khác 16 

\(A=\left(\frac{x+5\sqrt{x}-27+\left(3-\sqrt{x}\right)\left(\sqrt{x}+4\right)}{x-16}\right):\frac{1}{\sqrt{x}+4}\)

\(=\left(\frac{x+5\sqrt{x}-27+3\sqrt{x}+12-x-4\sqrt{x}}{x-16}\right):\frac{1}{\sqrt{x}+4}\)

\(=\left(\frac{4\sqrt{x}-15}{x-16}\right):\frac{1}{\sqrt{x}+4}=\frac{4\sqrt{x}-15}{\sqrt{x}-4}\)

b, Ta có \(B=-2A\Rightarrow\sqrt{x}-4=-\frac{8\sqrt{x}-30}{\sqrt{x}-4}\)

\(\Leftrightarrow x-8\sqrt{x}+16=-8\sqrt{x}+30\Leftrightarrow x-14=0\Leftrightarrow x=14\left(tm\right)\)

6 tháng 7 2019

\(a,ĐKXĐ:x\ge0;x\ne4\)

Ta có: \(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}-\frac{5\sqrt{x}+2}{x-4}\)

\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x+2\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2x-4\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

\(=\frac{3\sqrt{x}}{\sqrt{x}+2}\)

Vậy....

\(b,ĐKXĐ:x\ge0;x\ne4\)

\(ĐểP=2\Rightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}=2\)

\(\Leftrightarrow2\left(\sqrt{x}+2\right)=3\sqrt{x}\)

\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)

\(\Leftrightarrow3\sqrt{x}-2\sqrt{x}=4\)

\(\Leftrightarrow\sqrt{x}=4\)

\(\Leftrightarrow x=16\text{(Thỏa mãn ĐKXĐ)}\)

Vậy...

13 tháng 12 2020

a) 

\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}-\frac{5\sqrt{x}+2}{x-4}\)

\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3\sqrt{x}}{\sqrt{x}+2}\)

b) Thay P = 2 vào , ta được :

\(2=\frac{3\sqrt{x}}{\sqrt{x}+2}\Leftrightarrow2\sqrt{x}+4=3\sqrt{x}\)

\(\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\)

Vậy x = 16 thì P = 2

6 tháng 7 2019

\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}\ne1\end{cases}\Rightarrow}\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}}\)

\(M=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}.\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}+\frac{3\left(\sqrt{x}-1\right)}{x-1}-\frac{6\sqrt{x}-4}{x-1}\)

\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

\(b,M< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}< \frac{1}{2}\)

\(\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}< 0\)\(\Rightarrow\frac{2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{2\left(\sqrt{x}+1\right)}< 0\)

\(\Rightarrow\frac{2\sqrt{x}-1-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)\(\Rightarrow\frac{\sqrt{x}-2}{2\left(\sqrt{x}+1\right)}< 0\)

Vì \(2\left(\sqrt{x}+1\right)>0\Rightarrow\sqrt{x}-2>0\Rightarrow\sqrt{x}>2\)

\(\Rightarrow\sqrt{x}>\sqrt{4}\Leftrightarrow x>4\)

6 tháng 7 2019

\(M=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\left(x\ge0;x\ne1\right)\)

\(M=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{6\sqrt{x}-4}{x-1}\)

\(M=\frac{x+\sqrt{x}+3\sqrt{x}-3}{\left(\sqrt{x}\right)^2-1^2}-\frac{6\sqrt{x}-4}{x-1}\)

\(M=\frac{x-2\sqrt{x}+1}{x-1}\)

\(M=\frac{\left(\sqrt{x}-1\right)^2}{x-1}\)

31 tháng 7 2017

\(A=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}+2\right)+8\sqrt{x}}{x-4}:\frac{2\left(\sqrt{x}+2\right)-2\sqrt{x}-3}{\sqrt{x}+2}\)

\(A=\frac{2x}{x-4}.\left(\sqrt{x}+2\right)=\frac{2x\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(A=\frac{2x}{\sqrt{x}-2}\)