c) Để Q=3 thì \(\sqrt{x}+1=3\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow\sqrt{x}+1-3\sqrt{x}+3=0\)
\(\Leftrightarrow4-2\sqrt{x}=0\)
\(\Leftrightarrow2\sqrt{x}=4\)
\(\Leftrightarrow\sqrt{x}=2\)
hay x=4
d) Để \(Q>\dfrac{1}{2}\) thì \(Q-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+2-\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}>0\)
\(\Leftrightarrow\sqrt{x}-1>0\)
\(\Leftrightarrow\sqrt{x}>1\)
hay x>1
Kết hợp ĐKXĐ, ta được: x>1
e) Để Q nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)
\(\Leftrightarrow2⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{2;0;3\right\}\)
hay \(x\in\left\{0;4;9\right\}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
a) Ta có: \(Q=\dfrac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
\(=\dfrac{3x+3\sqrt{x}-3-\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) Ta có: |2x-5|=3
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=3\\2x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\2x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)
Thay x=4 vào Q, ta được:
\(Q=\dfrac{2+1}{2-1}=3\)
còn phần c, d, e nữa cậu giải giúp mình đc ko
