Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(H=\frac{2x^2+2x}{x^2-1}+\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(\Leftrightarrow H=\frac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(\Leftrightarrow H=\frac{2x}{x-1}+\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(\Leftrightarrow H=\frac{2x+\sqrt{x}-1-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow H=\frac{2x-2}{x-1}\)
\(\Leftrightarrow H=2\)
b) Để \(\sqrt{x}< H\)
\(\Leftrightarrow\sqrt{x}< 2\)
\(\Leftrightarrow x< 4\)
Mà \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}0\le x< 1\\1< x< 4\end{cases}}\)
p/s : vì đề bài không yêu cầu \(x\)nguyên nên mình làm như vậy !
\(a,x>0;x\ne4,9\)
\(b,Q=\left(\frac{1}{\sqrt{x}-3}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right)\)
\(Q=\left(\frac{\sqrt{x}-\sqrt{x}+3}{x-3\sqrt{x}}\right):\left(\frac{x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(Q=\frac{3}{x-3\sqrt{x}}:\frac{-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(Q=\frac{3}{\sqrt{x}\left(\sqrt{x}-3\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{-5}\)
\(Q=\frac{3\sqrt{x}-6}{-5\sqrt{x}}\)
\(c,Q< 0< =>\frac{3\sqrt{x}-6}{-5\sqrt{x}}\)
\(-5\sqrt{x}< 0\)
\(< =>3\sqrt{x}-6>0\)
\(\sqrt{x}>2\)
\(x>4\)
a) ĐKXĐ: \(x\ge0\); \(1-4x\ne\)0; \(2\sqrt{x}-1\ne0\); \(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\ne\)0
<=> \(x\ge0\); x \(\ne\)1/4
Ta có: \(A=\left(\frac{\sqrt{x}-4x}{1-4x}-1\right):\left(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\right)\)
\(A=\left(\frac{\sqrt{x}-4x-1+4x}{1-4x}\right):\left(\frac{1+2x+2\sqrt{x}\left(2\sqrt{x}+1\right)-1+4x}{\left(1-2\sqrt{x}\right)\left(1+2\sqrt{x}\right)}\right)\)
\(A=\frac{\sqrt{x}-1}{1-4x}\cdot\frac{1-4x}{6x+4x+2\sqrt{x}}\)
\(A=\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\)
b)Với x \(\ge\)0 và x \(\ne\)1/4
Ta có: A > A2 <=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}>\left(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\right)^2\)
<=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\cdot\left(1-\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\right)>0\)
<=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\cdot\frac{10x+2\sqrt{x}-\sqrt{x}+1}{10x+2\sqrt{x}}>0\)
<=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\cdot\frac{10+\sqrt{x}+1}{10x+2\sqrt{x}}>0\)
<=> \(\sqrt{x}-1>0\) <=> \(x>1\)
c) Với x\(\ge\)0 và x \(\ne\)1/4 (1)
Ta có: \(\left|A\right|>\frac{1}{4}\) <=> \(\orbr{\begin{cases}A>\frac{1}{4}\\A< -\frac{1}{4}\end{cases}}\)
TH1: \(A>\frac{1}{4}\) <=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}>\frac{1}{4}\)
<=> \(4\left(\sqrt{x}-1\right)>10x+2\sqrt{x}\)
<=> \(4\sqrt{x}-4>10x+2\sqrt{x}\)
<=> \(10x-2\sqrt{x}+4< 0\)(vô liia vì \(10x-2\sqrt{x}+4>0\))
TH2: \(A< -\frac{1}{4}\) <=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}< -\frac{1}{4}\)
<=> \(4\left(\sqrt{x}-1\right)< -10x-2\sqrt{x}\)
<=> \(4\sqrt{x}-4+10x+2\sqrt{x}< 0\)
<=> \(10x+6\sqrt{x}-4< 0\)
<=> \(5x+3\sqrt{x}-2< 0\)
<=> \(\left(5\sqrt{x}-2\right)\left(\sqrt{x}+1\right)< 0\)
<=> \(x< \frac{4}{25}\) (2)
Từ (1) và (2) => \(0\le x< \frac{4}{25}\)
a, Ta có: H = \(\frac{2x.\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}\) + \(\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\) - \(\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)
= \(\frac{2x}{x-1}+\frac{\sqrt{x}-1}{x-1}-\frac{\sqrt{x}+1}{x-1}\)
= \(\frac{2x+\sqrt{x}-1-\sqrt{x}-1}{x-1}\)
= \(\frac{2x-2}{x-1}\)
= 2
b, Ta có: \(\sqrt{x}\) < H <=> \(\sqrt{x}\) < 2
<=> x < 4
Vậy x = 4 thì \(\sqrt{x}\) < H