Câu 2,cho biểu thức
A=\(\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)
a,Rút gọn A
b,tìm x để A=1/2
K
Khách
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Những câu hỏi liên quan
14 tháng 9 2021
\(a,A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(x>0;x\ne1\right)\\ A=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ A=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(b,\dfrac{P}{A}\left(x-1\right)=0\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\cdot\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow x=0\left(\sqrt{x}+1>0\right)\)
14 tháng 9 2021
a) \(A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(đk:x>0,x\ne1\right)\)
\(=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b) \(\dfrac{P}{A}\left(x-1\right)=0\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}:\dfrac{\sqrt{x}+1}{\sqrt{x}}.\left(x-1\right)=0\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-1}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow x=0\)( do \(\sqrt{x}+1\ge1>0\))(không thỏa đk)
Vậy \(S=\varnothing\)
AT
30 tháng 7 2021
a) \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{x}{x-1}\right):\left(\dfrac{2x}{x-1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\left(x\ge0,x\ne1\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2x-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\sqrt{x}}=-\dfrac{1}{\sqrt{x}-1}\)
b) \(A=2\Rightarrow\dfrac{-1}{\sqrt{x}-1}=2\Rightarrow-1=2\sqrt{x}-2\Rightarrow2\sqrt{x}=1\Rightarrow\sqrt{x}=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{4}\)
AH
Akai Haruma
Giáo viên
30 tháng 7 2021
Lời giải:
ĐK: $x\geq 0; x\neq 1$
a.
\(A=\frac{\sqrt{x}(\sqrt{x}-1)-x}{(\sqrt{x}-1)(\sqrt{x}+1)}:\frac{2x-\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}\)
\(=\frac{-\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}:\frac{x-\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{-\sqrt{x}}{x-\sqrt{x}}=\frac{-\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}=\frac{1}{1-\sqrt{x}}\)
b.
$A=2\Leftrightarrow 1-\sqrt{x}=\frac{1}{2}$
$\Leftrightarrow \sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}$ (tm)
T
29 tháng 12 2023
a) ĐKXĐ: \(x>0;x\ne4\)
\(Q=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right):\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right]:\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}:\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)
b) Để biểu thức \(Q\) có giá trị âm thì \(\dfrac{3\sqrt{x}}{\sqrt{x}-2}< 0\)
\(\Rightarrow\sqrt{x}-2< 0\) (vì \(3\sqrt{x}>0\forall x>0;x\ne4\))
\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0\le x< 4\)
Kết hợp với điều kiện xác định của \(x\), ta được: \(0< x< 4\)
\(\text{#}\mathit{Toru}\)
24 tháng 11 2021
\(a,ĐK:x>0;x\ne9\\ b,A=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ A=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\\ c,A>\dfrac{2}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{2}{5}>0\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{5}>0\\ \Leftrightarrow\dfrac{2-\sqrt{x}}{5\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow2-\sqrt{x}>0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)
5 tháng 4 2022
a: \(A=\dfrac{x-1}{2\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1\right)-\sqrt{x}\left(x+2\sqrt{x}+1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{2\sqrt{x}}=\dfrac{-4x}{2\sqrt{x}}=-2\sqrt{x}\)
b: Để A>-6 thì \(2\sqrt{x}< 6\)
=>0<x<9
Kết hợp ĐKXĐ, ta được:
0<x<9 và x<>1
Y
5 tháng 4 2022
\(a,\)
\(=\dfrac{x-1}{2\sqrt{x}}.\dfrac{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{2\sqrt{x}}.\dfrac{x\sqrt{x}-x-x+\sqrt{x}-x\sqrt{x}-x-x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{2\sqrt{x}}.\dfrac{-4x}{x-1}\)
\(=-2\sqrt{x}\)
Vậy \(A=-2\sqrt{x}\)
\(b,\)Đề \(A\ge-6\) thì \(-2\sqrt{x}\ge-6\) \(\Leftrightarrow\sqrt{x}\le3\) \(\Leftrightarrow x\le3^2\Leftrightarrow x\le9\)
Vậy \(x\le9\) thì \(A\ge-6\)
AH
Akai Haruma
Giáo viên
12 tháng 6 2021
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1$
a)
\(A=\frac{x+\sqrt{x}+1}{x+1}:\left[\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{(\sqrt{x}-1)(x+1)}\right]\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\frac{x+1-2\sqrt{x}}{(\sqrt{x}-1)(x+1)}=\frac{x+\sqrt{x}+1}{x+1}.\frac{(\sqrt{x}-1)(x+1)}{(\sqrt{x}-1)^2}=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
b)
\(A=7\Leftrightarrow x+\sqrt{x}+1=7(\sqrt{x}-1)\)
\(\Leftrightarrow x-6\sqrt{x}+8=0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}-4)=0\)
\(\Leftrightarrow \left[\begin{matrix} x=4\\ x=16\end{matrix}\right.\) (đều thỏa mãn)
c)
\(x=2(2+\sqrt{3})=4+2\sqrt{3}=3+1+2\sqrt{3.1}=(\sqrt{3}+1)^2\Rightarrow \sqrt{x}=\sqrt{3}+1\)
\(\Rightarrow A=\frac{4+2\sqrt{3}+\sqrt{3}+1+1}{\sqrt{3}}=\frac{6+3\sqrt{3}}{\sqrt{3}}=3+2\sqrt{3}\)
d)
\(A< 1\Leftrightarrow \frac{x+\sqrt{x}+1}{\sqrt{x}-1}-1<0\Leftrightarrow \frac{x-2\sqrt{x}+2}{\sqrt{x}-1}<0\)
\(\Leftrightarrow \frac{(\sqrt{x}-1)^2+1}{\sqrt{x}-1}<0\Leftrightarrow \sqrt{x}-1< 0\Leftrightarrow 0\leq x< 1\)
16 tháng 11 2021
a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
7 tháng 11 2021
ĐKXĐ: x>0; x≠1
A = 2(1√x−1−1√x):(√x+1x2−x)=2[√x−√x+1(√x−1)√x]:√x+1x(√x−1)(√x+1)2(1x−1−1x):(x+1x2−x)=2[x−x+1(x−1)x]:x+1x(x−1)(x+1)
=2√x.(√x−1):1x(√x−1)=2√x=2x.(x−1):1x(x−1)=2x
b) Để A = √20122012 thì 2√x=√20122x=2012
⇔2√x=2√503⇔2x=2503
⇔x=503⇔x=503 (thỏa mãn điều kiện)
Vậy x=503
21 tháng 12 2021
a: \(A=\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}-1}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-\left(\sqrt{x}-2\right)^2}{3}\)
22 tháng 12 2021
Đề bạn gõ sai, mình có sửa lại r nha
\(a,A=\dfrac{1-\sqrt{x}+1}{\sqrt{x}-1}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3}\\ x=5\Leftrightarrow A=\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{3}=\dfrac{5-2\sqrt{5}}{3}\\ c,A=-\dfrac{1}{3}\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=-1\Leftrightarrow x-2\sqrt{x}+1=0\\ \Leftrightarrow\left(\sqrt{x}-1\right)^2=0\Leftrightarrow x=1\left(ktm\right)\Leftrightarrow x\in\varnothing\)
a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\cdot\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)