cho A=\(\dfrac{1}{\sqrt{x}+\sqrt{x-1}}-\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x\sqrt{x-x}}{1-\sqrt{x}}\)
rút gọn biểu thức A
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15 tháng 12 2022
a: \(C=\dfrac{3x+3\sqrt{x}-3-x+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{2x+3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để \(C=\sqrt{x}\) thì \(x-\sqrt{x}=\sqrt{x}+1\)
=>\(x-2\sqrt{x}-1=0\)
=>\(\Leftrightarrow x=3+2\sqrt{2}\)
c: |2x-5|=3
=>2x-5=3 hoặc 2x-5=-3
=>2x=2 hoặc 2x=8
=>x=4(nhận) hoặc x=1(loại)
Khi x=4 thì \(C=\dfrac{2+1}{2-1}=3\)
Y
4 tháng 4 2022
\(a,\)
\(=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3}{3\sqrt{x}+1}\right)\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+3x}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
Vậy \(P=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
\(b,\)Thay \(P=\dfrac{6}{5}\) vào pt, ta có :
\(\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}=\dfrac{6}{5}\)
\(\Leftrightarrow5\left(3\sqrt{x}+1\right)=6\left(3\sqrt{x}-1\right)\)
\(\Leftrightarrow15\sqrt{x}+5-18\sqrt{x}+6=0\)
\(\Leftrightarrow-3\sqrt{x}+11=0\)
\(\Leftrightarrow-3\sqrt{x}=-11\)
\(\Leftrightarrow\sqrt{x}=\dfrac{11}{3}\)
\(\Leftrightarrow x=\left(\dfrac{11}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{121}{9}\)
Vậy \(x=\dfrac{121}{9}\) thì \(P=\dfrac{6}{5}\)
7 tháng 10 2021
a) \(P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=-1\)
\(\Leftrightarrow-\sqrt{x}-1=\sqrt{x}-1\Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)
c) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\)
\(\Rightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Kết hợp đk:
\(\Rightarrow x\in\left\{0\right\}\)
d) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\)
7 tháng 10 2021
\(a,P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,P=-1\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\\ \Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\\ c,P\in Z\Leftrightarrow\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}\in Z\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}\in Z\\ \Leftrightarrow2⋮\sqrt{x}+1\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}+1\in\left\{1;2\right\}\left(\sqrt{x}+1\ge1\right)\\ \Leftrightarrow\sqrt{x}\in\left\{0;1\right\}\\ \Leftrightarrow x\in\left\{0;1\right\}\)
\(d,P=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
Có \(\dfrac{2}{\sqrt{x}+1}>0\left(2>0;\sqrt{x}+1>0\right)\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}< 1\Leftrightarrow P< 1\)
\(e,P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
Có \(\sqrt{x}+1\ge1\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\le2\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}\ge1-2=-1\)
\(P_{min}=-1\Leftrightarrow x=0\)
10 tháng 11 2021
\(ĐK:x\ge0;x\ne1;x\ne4\\ P=\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\\ P=\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-1-x+4}\\ P=\dfrac{2\left(\sqrt{x}-2\right)}{3\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}-4}{3\sqrt{x}+3}\)
17 tháng 7 2021
\(D=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(D=\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{x+2\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(D=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(E=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\frac{x-\sqrt{x}}{1-\sqrt{x}}\right)=\left(1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\left(1-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(E=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
18 tháng 7 2021
ĐK : a >= 0 , a khác 1
\(C=\left[\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\div\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\frac{a+\sqrt{a}-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\times\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\frac{a}{\sqrt{a}+1}\)
23 tháng 9 2021
\(B=\dfrac{x+2\sqrt{x}}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
\(=\dfrac{x+2\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{1}{\sqrt{x}-1}\)
BC
23 tháng 9 2021
Nguyễn Lê Phước Thịnh CTV Đề bài là dấu chia mà bn, có phải dấu nhân đâu
9 tháng 9 2021
\(D=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{3\sqrt{x}+1}{x-1}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\left(x\ge0;x\ne1\right)\\ D=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\\ D=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\cdot\dfrac{1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
9 tháng 5 2021
Sửa đề: \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
Ta có: \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{2}{x-1}\)
H
28 tháng 5 2023
a.
\(B=\dfrac{\sqrt{x}+1+\sqrt{x}\left(\sqrt{x}-1\right)+2\sqrt{x}}{1-x}=\dfrac{\sqrt{x}+1+x-\sqrt{x}+2\sqrt{x}}{1-x}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b.
\(P=\dfrac{B}{A}=\dfrac{x+3}{\sqrt{x}+1}:\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\left(x+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{x+3}{\sqrt{x}-1}=\dfrac{x-1+4}{\sqrt{x}-1}\)
\(=\sqrt{x}+1+\dfrac{4}{\sqrt{x}-1}\)\(=\sqrt{x}-1+\dfrac{4}{\sqrt{x}-1}+2\)
Theo BĐT AM - GM ta có: \(\sqrt{x}-1+\dfrac{4}{\sqrt{x}-1}\ge2\sqrt{\left(\sqrt{x}-1\right)\dfrac{4}{\sqrt{x}-1}}=4\)
\(\Rightarrow\dfrac{1}{P}\ge6\Rightarrow Min_{\dfrac{1}{P}}=6\)
Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{x}-1\right)^2=4\Rightarrow x=9\) (loại trường hợp \(\sqrt{x}-1=-2\))
Vậy GTNN của biểu thức \(\dfrac{1}{P}=6\) khi x = 9.
\(A=\dfrac{\sqrt{x}-\sqrt{x-1}-\sqrt{x}-\sqrt{x-1}}{x-x+1}=-2\sqrt{x-1}\)