\(\frac{x\sqrt{x}-1}{x+\sqrt{x}+1}+\frac{x-1}{\sqrt{x}-1}\)
ĐK x >= 0, x#1
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6 tháng 9 2017
\(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
TP
21 tháng 6 2019
\(B=\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\)(đk: x ≥ 0 và x ≠ 9)
\(B=\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\sqrt{x}+3}-\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(B=\left(3-\sqrt{x}\right)-\left(\sqrt{x}-3\right)-6\)
\(B=3-\sqrt{x}-\sqrt{x}+3-6\)
\(B=-2\sqrt{x}\)
TP
21 tháng 6 2019
\(A=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\)(đk: x ≥ 0 và x ≠ 36)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+6\right)-3\left(\sqrt{x-6}\right)-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3(\sqrt{x}+6)}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3}{\sqrt{x}-6}\)
WJ
9 tháng 8 2016
Q=\(\frac{\sqrt{x}-1}{x-\sqrt{x}+1}+\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\) điều kiện x>=0
=\(\frac{x-1+x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
ta thấy cả tử và mẫu đề >=0=> Q>=0
dấu = xảy ra khi x=0
=> Q=0 khi x=0
TH
8 tháng 10 2015
\(=-1+\frac{1}{2\sqrt{x}-1}+\frac{x\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=\frac{-2\sqrt{x}+1+1+2\text{x}\sqrt{x}-x}{2\sqrt{x}-1}\)
\(=\frac{-2\sqrt{x}+2+2\text{x}\sqrt{x}-x}{2\sqrt{x}-1}\)
tick cho mình nha
VH
26 tháng 11 2019
a.
\(A=\left[\frac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}\right]:\left[\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)
\(=\frac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\)
\(=\frac{-4\sqrt{x}.\sqrt{x}}{-\sqrt{x}+3}=\frac{4x}{\sqrt{x}-3}\)
b.
\(A=-1\Leftrightarrow\frac{4x}{\sqrt{x}-3}=-1\)
\(\Leftrightarrow4x=-\sqrt{x}+3\)
\(\Leftrightarrow4x+\sqrt{x}-3=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(l\right)\\x=\frac{3}{4}\end{matrix}\right.\)
Vậy \(A=-1\Leftrightarrow x=\frac{3}{4}\)
T
14 tháng 10 2019
Sửa đề: \(P=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
Bài này chắc là quy đồng full quá nhỉ?
a)P\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
b) P =1/4 \(\Leftrightarrow\frac{\sqrt{x}-2}{3\sqrt{x}}=\frac{1}{4}\Leftrightarrow\sqrt{x}-2=\frac{3}{4}\sqrt{x}\)
\(\Leftrightarrow\frac{1}{4}\sqrt{x}=2\Leftrightarrow\sqrt{x}=8\Rightarrow x=64\left(TM\right)\)
P/s: Ko chắc..
14 tháng 10 2019
Sửa đề :
a) \(P=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
b) Để P \(=\frac{1}{4}\Leftrightarrow\frac{\sqrt{x}-2}{3\sqrt{x}}=\frac{1}{4}\Leftrightarrow4\left(\sqrt{x}-2\right)=3\sqrt{x}\)
\(\Leftrightarrow4\sqrt{x}-8=3\sqrt{x}\Leftrightarrow\sqrt{x}=8\Leftrightarrow x=64\left(TM\right)\)
10 tháng 8 2019
A=\(\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
A= \(\frac{2x-3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)=\(\frac{2x-2\sqrt{x}-\sqrt{x}+1}{x-1}=\frac{2\sqrt{x}-1}{x+1}\)
Để A=1/2 thì
\(\frac{2\sqrt{x}-1}{x+1}=\frac{1}{2}\)
nhân chéo ta đc pt \(x-4\sqrt{x}+3=0\)
giải pt ta đc x=1 (loại) hoặc x= 9
vậy x=9 TM
Để A<1 thì \(\frac{2\sqrt{x}-1}{\sqrt{x}+1}< 1\Leftrightarrow2\sqrt{x}-1< \sqrt{x}+1\Leftrightarrow\sqrt{x}< 2\)
=> x<4
vậy vs 0\(\le x< 4\) và x khác 1 TM
10 tháng 8 2019
Mình nghĩ thế này ạ
a) Với \(x\ge0,x\ne1\)ta có: \(\frac{\sqrt{x}+1}{\sqrt{x}-1x}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{3\sqrt{x}+1}{x-1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x-1}\right)}-\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x-\sqrt{x}-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(2\sqrt{x}-1\right)-\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)
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