NT
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
27 tháng 9 2018
\(\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\frac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{x-1-\left(x-4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{3}\)
\(=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
NA
27 tháng 9 2018
\(\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)Đkxđ : x>2
=(\(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)) \(:\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)-\left(x-4\right)}\)
\(=\frac{1}{\sqrt{x}}.\frac{\sqrt{x}-2}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
8 tháng 9 2020
Ta có: \(\frac{2x+2}{\sqrt{x}}+\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x^2+\sqrt{x}}{x\sqrt{x}+x}\)
\(=\frac{2\left(x+1\right)}{\sqrt{x}}+\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}\)
\(=\frac{2\left(x+1\right)}{\sqrt{x}}+\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)
\(=\frac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
TM
16 tháng 7 2016
\(D=\frac{3\sqrt{1-4x+4x^2}}{2x-1}=\frac{3\sqrt{\left(2x\right)^2-2.2x.1+1^2}}{2x-1}=\frac{3\sqrt{\left(2x-1\right)^2}}{2x-1}=\frac{3.\left(2x-1\right)}{2x-1}=3\)
mình làm lại nè, bài kia mình hơi nhầm 1 chút
AH
Akai Haruma
Giáo viên
4 tháng 7 2020
Lời giải:
a) \(A=\frac{\sqrt{x}-2+\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2}{\sqrt{x}+2}\)
b)
\(A>\frac{1}{2}\Leftrightarrow \frac{2}{\sqrt{x}+2}>\frac{1}{2}\Leftrightarrow 4> \sqrt{x}+2\Leftrightarrow 4> x\geq 0\)
Kết hợp với ĐKXĐ suy ra $4>x>0$
24 tháng 8 2019
P=\(\frac{2\sqrt{x}+\left|\sqrt{x}-1\right|}{3x+2\sqrt{x}-1}\)(đk :\(x\ge0,x\ne\frac{1}{9},x\ne1\))
=\(\frac{2\sqrt{x}+\left|\sqrt{x}-1\right|}{3x+3\sqrt{x}-\sqrt{x}-1}=\frac{2\sqrt{x}+\left|\sqrt{x}-1\right|}{\left(\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\)(1)
TH1 : \(0\le\sqrt{x}\le1\)
Từ (1)=> \(P=\frac{2\sqrt{x}+1-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}=\frac{1}{3\sqrt{x}-1}\)
TH2: x>1
Từ (1) => \(P=\frac{2\sqrt{x}+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}+1}\)
Vậy với \(0\le x\le1\) => \(P=\frac{1}{3\sqrt{x}-1}\)
x>1=> P=\(\frac{1}{\sqrt{x}+1}\)
KH
24 tháng 8 2019
\(P=\frac{2\sqrt{x}+\left|\sqrt{x}-1\right|}{3x+2\sqrt{x}-1}\)
ĐK: \(x\ge0;x\ne\frac{1}{9}\)
\(TH_1:\sqrt{x}-1\ge0\Leftrightarrow x\ge1\)
\(P=\frac{2\sqrt{x}+\sqrt{x}-1}{3x+2\sqrt{x}-1}\\ =\frac{3\sqrt{x}-1}{3x+3\sqrt{x}-\sqrt{x}-1}\\ =\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}\\ =\frac{1}{1+\sqrt{x}}=\frac{1-\sqrt{x}}{1-x}\)
\(TH_2:\sqrt{x}-1< 0\Leftrightarrow x< 1\)
\(P=\frac{2\sqrt{x}+1-\sqrt{x}}{3x+2\sqrt{x}-1}\\ =\frac{\sqrt{x}+1}{3x+3\sqrt{x}-\sqrt{x}-1}\\ =\frac{\sqrt{x}+1}{\left(3\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}\\ =\frac{1}{3\sqrt{x}-1}\\ =\frac{3\sqrt{x}+1}{9x-1}\)
19 tháng 6 2015
\(P=\frac{\sqrt{x}\left(\sqrt{x^3}+1\right)}{\left(x-\sqrt{x}+1\right)}+1-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)
\(P=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-2\sqrt{x}-1\)
\(P=\sqrt{x}\left(\sqrt{x}+1\right)-2\sqrt{x}=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\)