với a > 1, cho biểu thức \(P=\left(\dfrac{2}{\sqrt{a+1}}+\sqrt{a-1}\right):\left(\dfrac{2}{\sqrt{a^2-1}}+1\right)\)
rút gọn biểu thức P và tim giá trị của a để P=2
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22 tháng 12 2020
Bài 1:
a) Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b) Để Q dương thì \(\dfrac{\sqrt{a}-2}{3\sqrt{a}}>0\)
mà \(3\sqrt{a}>0\forall a\) thỏa mãn ĐKXĐ
nên \(\sqrt{a}-2>0\)
\(\Leftrightarrow\sqrt{a}>2\)
hay a>4
Kết hợp ĐKXĐ,ta được: a>4
Vậy: Để Q dương thì a>4
H9
HT.Phong (9A5)
CTVHS
28 tháng 7 2023
a) \(M=3\sqrt{3}-\sqrt{12}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(M=3\sqrt{3}-2\sqrt{3}-\left|\sqrt{3}-1\right|\)
\(M=\sqrt{3}-\sqrt{3}+1\)
\(M=1\)
b) Ta có:
\(N=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(N=\left(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(N=\left(\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(N=\dfrac{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)\cdot\left(\sqrt{a}+1\right)}\)
\(N=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Theo đề ta có: \(M=2N\)
Khi: \(1=2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\)
\(\Leftrightarrow1=\dfrac{2\sqrt{a}-2}{\sqrt{a}}\)
\(\Leftrightarrow\sqrt{a}=2\sqrt{a}-2\)
\(\Leftrightarrow2\sqrt{a}-\sqrt{a}=2\)
\(\Leftrightarrow\sqrt{a}=2\)
\(\Leftrightarrow a=4\left(tm\right)\)
10 tháng 7 2021
Sửa đề: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}\right)\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\notin\left\{1;4\right\}\end{matrix}\right.\)
Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a+3\sqrt{a}+2-a+3\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{6\sqrt{a}}\)
\(=\dfrac{a-4}{6a\left(\sqrt{a}-1\right)}\)
c) Thay \(a=9-4\sqrt{5}\) vào Q, ta được:
\(Q=\dfrac{5-4\sqrt{5}}{6\left(9-4\sqrt{5}\right)\left(\sqrt{5}-3\right)}\)
\(=\dfrac{5-4\sqrt{5}}{6\left(9\sqrt{5}-27-20+12\sqrt{5}\right)}\)
\(=\dfrac{5-4\sqrt{5}}{6\left(21\sqrt{5}-47\right)}\)
\(=\dfrac{\left(5-4\sqrt{5}\right)\left(21\sqrt{5}+47\right)}{-24}\)
\(=\dfrac{105\sqrt{5}+235-420-188\sqrt{5}}{-24}\)
\(=\dfrac{-83\sqrt{5}-185}{-24}=\dfrac{83\sqrt{5}+185}{24}\)
22 tháng 9 2021
a: Ta có: \(P=\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\left(\dfrac{1}{\sqrt{a}}+1\right)\)
\(=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\dfrac{1+\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2}{1-\sqrt{a}}\)
16 tháng 2 2021
ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
1) Ta có: \(N=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\cdot\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\cdot\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
\(=1-a\)
2) Để N=-2016 thì 1-a=-2016
\(\Leftrightarrow1-a+2016=0\)
\(\Leftrightarrow2017-a=0\)
hay a=2017(thỏa ĐK)
Vậy: Để N=-2016 thì a=2017
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 \(S=\varnothing\)
26 tháng 12 2021
a: \(A=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
27 tháng 12 2021
\(ĐK:a>0;a\ne1;a\ne4\\ a,A=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,A>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow a>4\)
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: \(0< x< 4\)
\(\text{#}\mathit{Toru}\)
\(P=\left(\dfrac{2+\sqrt{a+1}\sqrt{a-1}}{\sqrt{a+1}}\right):\left(\dfrac{2+\sqrt{a+1}\sqrt{a-1}}{\sqrt{a+1}\sqrt{a-1}}\right)\)
\(=\dfrac{2+\sqrt{a+1}\sqrt{a-1}}{\sqrt{a+1}}.\dfrac{\sqrt{a+1}\sqrt{a-1}}{2+\sqrt{a+1}\sqrt{a-1}}\)
\(=\sqrt{a-1}\)
P=2\(\Rightarrow\sqrt{a-1}=2\Leftrightarrow a-1=4\Leftrightarrow a=5\left(tm\right)\)