Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
ĐK : \(\hept{\begin{cases}x,y>0\\x\ne y\end{cases}}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\frac{x+2\sqrt{xy}+y}{x-y}-\frac{x-2\sqrt{xy}+y}{x-y}\)
\(=\frac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y}{x-y}=\frac{4\sqrt{xy}}{x-y}\)
Với \(\hept{\begin{cases}x=7+2\sqrt{3}\\y=7-2\sqrt{3}\end{cases}}\)( tmđk )
=> \(A=\frac{4\sqrt{\left(7+2\sqrt{3}\right)\left(7-2\sqrt{3}\right)}}{7+2\sqrt{3}-\left(7-2\sqrt{3}\right)}\)
\(=\frac{4\sqrt{7^2-\left(2\sqrt{3}\right)^2}}{7+2\sqrt{3}-7+2\sqrt{3}}\)
\(=\frac{4\sqrt{49-12}}{4\sqrt{3}}\)
\(=\frac{4\sqrt{37}}{4\sqrt{3}}=\frac{\sqrt{37}}{\sqrt{3}}=\frac{\sqrt{37}\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{\sqrt{111}}{3}\)
\(a,C=\frac{7}{\sqrt{x}+3}< 1\)
\(C=\frac{7}{\sqrt{x}+3}-1< 0\)
\(C=\frac{7-\sqrt{x}-3}{\sqrt{x}+3}< 0\)
\(C=\frac{4-\sqrt{x}}{\sqrt{x}+3}< 0\)
\(\sqrt{x}+3>0< =>4-\sqrt{x}< 0\)
\(\sqrt{x}>4\)
\(x>16\)
\(b,\sqrt{x}+2C=\sqrt{x}+\frac{14}{\sqrt{x}+3}\)
\(=\sqrt{x}+3+\frac{14}{\sqrt{x}+3}-3\)
\(\sqrt{x}+3+\frac{14}{\sqrt{x}+3}\ge2\sqrt{\left(\sqrt{x}+3\right).\frac{14}{\sqrt{x}+3}}=2\sqrt{14}\)
\(\sqrt{x}+2C\ge2\sqrt{14}-3\)dấu "=" xảy ra khi \(\sqrt{x}+3=\frac{14}{\sqrt{x}+3}\)
\(x+6\sqrt{x}+9=14\)
\(x+6\sqrt{x}-5=0\)
rồi bạn giải pt bậc 2
\(< =>MIN=2\sqrt{14}-3\)
a) \(M=\frac{x+1+\sqrt{x}}{x+1}:\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\sqrt{x}\left(x+1\right)-\left(x+1\right)}\right)\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\frac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\)\(=\frac{x+\sqrt{x}+1}{x+1}.\frac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
b) \(M>3\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}-1}>3\Leftrightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}-1}-3>0\)
\(\Leftrightarrow\frac{x+\sqrt{x}+1-3\left(\sqrt{x}-1\right)}{\sqrt{x}-1}>0\Leftrightarrow\frac{x+\sqrt{x}+1-3\sqrt{x}+3}{\sqrt{x}-1}>0\)\(\Leftrightarrow\frac{x-2\sqrt{x}+4}{\sqrt{x}-1}>0\)
Ta có: \(x-2\sqrt{x}+4=x-2\sqrt{x}+1+3=\left(\sqrt{x}-1\right)+3>0\)\(\Rightarrow\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)
Vậy x>1
c) \(M=7\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}-1}=7\Rightarrow x+\sqrt{x}+1=7\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow x+\sqrt{x}+1=7\sqrt{x}-7\Leftrightarrow x-6\sqrt{x}+8=0\)\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\\sqrt{x}-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=16\end{cases}\left(tm\right)}}\)
Vậy \(x\in\text{{}4;16\)
Đáp án là D