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điều kiện xác định : \(x>0;x\ne4\)
a) ta có : \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4\sqrt{x}-3}{2\sqrt{x}-x}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\right)\)
\(\Leftrightarrow P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\right)\) \(\Leftrightarrow P=\left(\dfrac{x-4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\left(\sqrt{x}-2\right)^2-\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\left(\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\left(\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{4}\right)\) \(\Leftrightarrow P=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{4}\)b) để \(P>0\) \(\Leftrightarrow\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{4}>0\) \(\Leftrightarrow\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-3>0\\\sqrt{x}-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-3< 0\\\sqrt{x}-1< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>9\\x>1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 9\\x< 1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>9\\x< 1\end{matrix}\right.\)
kết hợp với điều kiện xác định ta có : \(0< x< 1\) hoặc \(x>9\)
c) ta có : \(\sqrt{P}=\sqrt{\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{4}}\ge0\forall x\)
dấu "=" xảy ra khi \(\left[{}\begin{matrix}\sqrt{x}-3=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
vậy ....................................................................................................
d) ta có : \(m\left(\sqrt{x}-3\right)P=12m\sqrt{x}-4\)
\(\Leftrightarrow m\left(\sqrt{x}-3\right)\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{4}=12m\sqrt{x}-4\)
\(\Leftrightarrow m\left(x-6\sqrt{x}+9\right)\left(\sqrt{x}-1\right)=48m\sqrt{x}-4\)
nhân tung ra giải bình thường ............(mk nghỉ có vấn đề ở câu d này nha )
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4\sqrt{x}-3}{2\sqrt{x}-x}\right):\)\(\left(\frac{\sqrt{x}+2}{\sqrt{x}}-\frac{\sqrt{x}-4}{\sqrt{x}-2}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)\(:\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}:\frac{x-4-x+4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-3}{4}\)
\(b,\)Để \(P>0\Rightarrow\frac{\sqrt{x}-3}{4}>0\)
Mà \(4>0\Rightarrow\sqrt{x}-3>0\Rightarrow\sqrt{x}>3\Rightarrow x>9\)
\(c,\sqrt{P}_{min}=0\Rightarrow\frac{\sqrt{x}-3}{4}=0\)
\(\Leftrightarrow\sqrt{x}-3=0\Rightarrow\sqrt{x}=3\Rightarrow x=9\)
Phần 3:
Ta đã rút gọn được \(P=\frac{4x}{\sqrt{x}-3}\)
Ta có: \(m(\sqrt{x}-3)P> x+1\) với mọi \(x>4\)
\(\Leftrightarrow m(\sqrt{x}-3).\frac{4x}{\sqrt{x}-3}> x+1\) với mọi \(x>4\)
\(\Leftrightarrow 4mx> x+1\) với mọi \(x>4\)
\(\Leftrightarrow m> \frac{x+1}{4x}\) với mọi \(x>4\)
Điều này xảy ra khi mà \(m> max \left(\frac{x+1}{4x}\right)\)
Ta có: \(\frac{x+1}{4x}=\frac{1}{4}+\frac{1}{4x}<\frac{1}{4}+\frac{1}{4.4}\Leftrightarrow \frac{x+1}{4x}< \frac{5}{16}\) (do \(x>4\) )
\(\Rightarrow max\left(\frac{x+1}{4x}\right)< \frac{5}{16}\)
Do đó \(m\geq \frac{5}{16}\) thỏa mãn điều kiện đã cho.
Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
a) P=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right).\dfrac{x-4}{\sqrt{4x}}\) với x > 0 và x≠4
=\(\left(\dfrac{\sqrt{x}.\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{x-4}\right).\dfrac{x-4}{2\sqrt{x}}\)
=\(\left(\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\right)\).\(\dfrac{x-4}{2\sqrt{x}}\)
=\(\dfrac{2x}{x-4}.\dfrac{x-4}{2\sqrt{x}}\)
=\(\dfrac{x}{\sqrt{x}}\)
b) \(\dfrac{x}{\sqrt{x}}\) >3
<=> x> \(3\sqrt{x}\)
<=> x>9
a: \(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\cdot\dfrac{x-4}{2\sqrt{x}}\)
\(=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)
b: Để P>3 thì \(\sqrt{x}>3\)
hay x>9
1. ĐK \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
a. Ta có \(R=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}.\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
b. Với \(x=4+2\sqrt{3}\Rightarrow R=\frac{\sqrt{4+2\sqrt{3}}+2}{\sqrt{4+2\sqrt{3}}\left(\sqrt{4+2\sqrt{3}}-2\right)}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}-2\right)}\)
\(=\frac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+3}{3-1}=\frac{\sqrt{3}+3}{2}\)
c. Để \(R>0\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
Vậy \(x>4\)thì \(R>0\)
2. Ta có \(A=6+2\sqrt{2}=6+\sqrt{8};B=9=6+3=6+\sqrt{9}\)
Vì \(\sqrt{8}< \sqrt{9}\Rightarrow A< B\)
3. a. \(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}\right)=a-b=VP\left(đpcm\right)\)
b. Ta có \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)
\(b,M=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\\ x=3+2\sqrt{2}\Leftrightarrow\sqrt{x}=\sqrt{2}+1\\ \Leftrightarrow M=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}=\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)=1\\ c,M>0\Leftrightarrow\sqrt{x}-2>0\left(\sqrt{x}>0\right)\\ \Leftrightarrow x>4\)