cho bt A \(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
a) rut gon bt A
b) cm A < 1/3
K
Khách
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8 tháng 7 2019
\(ĐKXĐ:x\ge0\)
Đề sai???
Sửa lại
\(a,P=\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}-1}{x-\sqrt{x}+1}\)
\(=\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{x+2-x+\sqrt{x}-1+x-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
KD
28 tháng 1 2020
\(a,Đkxđ:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x+1}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x^3}-1\right)}{x+\sqrt{x}+1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}-1\right)\)
\(=x-\sqrt{x}\)
\(b,P=x-\sqrt{x}=x-\sqrt{x}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\)
Ta có: \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\forall x\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall x\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow x=\frac{1}{4}\)
\(Min_P=-\frac{1}{4}\Leftrightarrow x=\frac{1}{4}\)
c, Đề thiếu không bạn?
10 tháng 8 2020
\(A=\frac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}.\frac{\sqrt{x}\left(\sqrt{x^3}-1\right)}{1}\)
\(A=\frac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}.\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)
\(A=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(A=x-1\)
(ĐKXĐ là: \(x>0;x\ne1\))
AH
Akai Haruma
Giáo viên
13 tháng 7 2020
Lời giải:
ĐKXĐ: \(x\geq 0; x\neq 1\)
Ta có:
\(A=\frac{x+\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+2)}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}=\frac{x+\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+2)}+\frac{\sqrt{x}+2}{(\sqrt{x}-1)(\sqrt{x}+2)}+\frac{\sqrt{x}-1}{(\sqrt{x}+2)(\sqrt{x}-1)}\)
\(=\frac{x+\sqrt{x}+1+\sqrt{x}+2+\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{x+3\sqrt{x}+2}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{(\sqrt{x}+1)(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
15 tháng 10 2019
a) \(A=\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}-1}{2\sqrt{x}}\)
\(A=\left(\frac{x+2}{\sqrt{x^3}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}-1}{2\sqrt{x}}\)
\(A=\frac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2\sqrt{x}}{\sqrt{x}-1}\)
\(A=\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2\sqrt{x}}{\sqrt{x}-1}\)
\(A=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2\sqrt{x}}{\sqrt{x}-1}\)
\(A=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2\sqrt{x}}{\sqrt{x}-1}=\frac{2\sqrt{x}}{x+\sqrt{x}+1}\)
CC
15 tháng 10 2019
Ta có: A-\(\frac{2}{3}\)= \(\frac{\sqrt{x}}{x+\sqrt[]{x}+1}-\frac{2}{3}\)=\(\frac{6\sqrt{x}-2x-2\sqrt{x}-2}{3\left(x+\sqrt{x}+1\right)}\)
=\(\frac{-2\left(x-2\sqrt{x}+1\right)}{3\left(x+\sqrt{x}+1\right)}\)=\(\frac{-2}{3}.\frac{\left(\sqrt{x}-1\right)^2}{x+\sqrt{x}+1}\)<0
hay A\(-\frac{2}{3}\)<0
=>A<\(\frac{2}{3}\)
N
24 tháng 3 2019
ĐK: \(x>0,x\ne4\)
1, \(A=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\frac{\left(x-4\right)^2}{\sqrt{x}^3}\)
\(A=\frac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\frac{\left(x-4\right)^2}{x}\)
\(A=\frac{2}{\sqrt{x}+2}.\frac{x-4}{x}\)
\(A=\frac{2\sqrt{x}-4}{x}\)
2, \(x=\left(\sqrt{3}+1\right)^2\)
Thay \(x=\left(\sqrt{3}+1\right)^2\):
\(A=\frac{2\sqrt{3}-2}{\left(\sqrt{3}+1\right)^2}\)
3, \(A\ge\frac{1}{4}\Rightarrow\)\(\frac{2\sqrt{x}-4}{x}-\frac{1}{4}\ge0\)
\(\Leftrightarrow\frac{8\sqrt{x}-16-x}{4x}\ge0\)
\(\Rightarrow x-8\sqrt{x}+16\le0\)
\(\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\left(TM\right)\)
BT
12 tháng 5 2017
a/ \(P=\left(\frac{3}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{x-1}\right):\left(\frac{x+2}{x+\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)
\(P=\left(\frac{3}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2}{\sqrt{x^2}-1+\sqrt{x}-1}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)
\(P=\left(\frac{3\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{1}{\left(\sqrt{x}-1\right)}\right)\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{\sqrt{x}-1}{1}\right)\)
=> \(P=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)}\)
b/ \(P=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)}=\sqrt{x}-1\)
<=> \(4\sqrt{x}=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
<=> \(4\sqrt{x}=x-1\). Bình phương 2 vế, ta được:
<=> 16x=(x-1)2
<=> 16x=x2-2x+1
<=> x2-18x+1=0
\(\Delta'=81-1=80=>\sqrt{\Delta'}=4\sqrt{5}\)
=> \(x_1=9-4\sqrt{5}\)
\(x_2=9+4\sqrt{5}\)
