\(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
help me:((
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AH
Akai Haruma
Giáo viên
30 tháng 6 2020
Lời giải:
Sửa lại đề, chỗ $\sqrt{x}-2}$ chuyển thành $\sqrt{x}+2$ mới đúng.
a)
\(B=\frac{3(x+\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+2)}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{\sqrt{x}}.\frac{\sqrt{x}}{1-\sqrt{x}}=\frac{3(x+\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+2)}-\frac{\sqrt{x}+1}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\frac{3(x+\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+2)}-\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}+2)(\sqrt{x}-1)}-\frac{(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}\)
\(=\frac{x+3\sqrt{x}+2}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{(\sqrt{x}+1)(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b)
Để $B$ âm thì $\frac{\sqrt{x}-1}{\sqrt{x}+1}< 0$
Mà $\sqrt{x}+1>0$ nên $\sqrt{x}-1< 0$
$\Leftrightarrow 0\leq x< 1$
Kết hợp với đkxđ suy ra $0< x< 1$ thì $B$ âm.
24 tháng 7 2020
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne1\\x\ge0\\x\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có : \(B=\left(\frac{1}{1-\sqrt{x}}+\frac{1}{1+\sqrt{x}}\right):\left(\frac{1}{1-\sqrt{x}}-\frac{1}{1+\sqrt{x}}\right)+\frac{1}{2\sqrt{x}}\)
=> \(B=\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right):\left(\frac{1+\sqrt{x}-1+\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right)+\frac{1}{2\sqrt{x}}\)
=> \(B=\left(\frac{2}{1-x}\right):\left(\frac{2\sqrt{x}}{1-x}\right)+\frac{1}{2\sqrt{x}}=\frac{2\left(1-x\right)}{2\sqrt{x}\left(1-x\right)}+\frac{1}{2\sqrt{x}}\)
=> \(B=\frac{1}{\sqrt{x}}+\frac{1}{2\sqrt{x}}=\frac{2}{2\sqrt{x}}+\frac{1}{2\sqrt{x}}=\frac{3}{2\sqrt{x}}\)
Vậy ....
17 tháng 8 2019
a, ĐKXĐ: x>0 (1)
b,T= (\(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)+6\sqrt{x}-9\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}\left(\sqrt{x}+1\right)}\)).(\(\frac{\sqrt{x}+1}{2-4\sqrt{x}}\))+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\)
= \(\left(\frac{x+3\sqrt{x}+2+6\sqrt{x}-9x-9\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}+1\right)}\right)\).\(\left(\frac{\sqrt{x}+1}{2-4\sqrt{x}}\right)\)+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\)
= \(\left(\frac{2-8x}{3\sqrt{x}\left(\sqrt{x}+1\right)}\right)\).\(\left(\frac{\sqrt{x}+1}{2-4\sqrt{x}}\right)\)+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\)
= \(\left(\frac{2\left(1-2\sqrt{x}\right)\left(1+2\sqrt{x}\right)}{3\sqrt{x}\left(\sqrt{x}+1\right)}\right)\).\(\left(\frac{\sqrt{x}+1}{2\left(1-2\sqrt{x}\right)}\right)\)+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\)
= \(\frac{1+2\sqrt{x}}{3\sqrt{x}}\)+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\) = \(\frac{x-\sqrt{x}}{3\sqrt{x}}\)=\(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{3\sqrt{x}}\)=\(\frac{\sqrt{x}-1}{3}\)
c, Để T<0 \(\Leftrightarrow\)\(\frac{\sqrt{x}-1}{3}\) <0 \(\Leftrightarrow\) \(\sqrt{x}\)-1<0 \(\Leftrightarrow\) \(\sqrt{x}\)<1\(\Leftrightarrow\) x<1 mà do ĐK (1)
=> Để T<0 \(\Leftrightarrow\) 0<x<1
17 tháng 8 2019
Cho mk hỏi là bước t2 từ dưới lên phần b thì \(\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)\) sao lại khai triển đc như vậy
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2019
ĐKXĐ: \(x>0\)
\(S=\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\div\left(\frac{\sqrt{x}}{x+\sqrt{x}}\right)=\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\right)\)
\(=\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\left(\sqrt{x}+1\right)=\frac{\sqrt{x}+1}{\sqrt{x}}+\sqrt{x}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)
\(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
ĐKXĐ : \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(=\frac{\left|x^2\right|\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left|x^2\right|\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\sqrt{x\cdot x^2}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x\cdot x^2}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\sqrt{x^3}-1^3}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x^3}+1^3}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)
\(=\frac{x+2\sqrt{x}+1}{\sqrt{x}}\)
\(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)
\(=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)