Rút gọn biểu thức:
D = \(\dfrac{1}{\sqrt{h+2\sqrt{h-1}}}+\dfrac{1}{\sqrt{h-2\sqrt{h-1}}}\)
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H9
HT.Phong (9A5)
CTVHS
30 tháng 10 2023
a) \(H=\left(\dfrac{a-3\sqrt{a}}{a-2\sqrt{a}-3}-\dfrac{2a}{a-1}\right):\dfrac{1-\sqrt{a}}{a-2\sqrt{a}+1}\)
\(H=\left[\dfrac{\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+1\right)}-\dfrac{2a}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]:\dfrac{1-\sqrt{a}}{\left(\sqrt{a}-1\right)^2}\)
\(H=\left[\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{2a}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]:\dfrac{-\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)^2}\)
\(H=\dfrac{a-\sqrt{a}-2a}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}:\dfrac{-1}{\sqrt{a}-1}\)
\(H=\dfrac{-a-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\cdot-\left(\sqrt{a}-1\right)\)
\(H=\dfrac{-\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot-\left(\sqrt{a}-1\right)\)
\(H=\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\)
\(H=\sqrt{a}\)
b) Thay x = 2023 vào ta có:
\(H=\sqrt{2023}\)
12 tháng 7 2021
a)ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Ta có: \(D=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}-\dfrac{5}{x+\sqrt{x}-6}+\dfrac{1}{2-\sqrt{x}}\)
\(=\dfrac{x-4\sqrt{x}+4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-5\sqrt{x}-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
31 tháng 7 2018
\(D=\dfrac{1}{\sqrt{h+2\sqrt{h-1}}}+\dfrac{1}{\sqrt{h-2\sqrt{h-1}}}\)
\(=\dfrac{1}{\sqrt{h-1+2\sqrt{h-1}+1}}+\dfrac{1}{\sqrt{h-1-2\sqrt{h-1}+1}}\)
\(=\dfrac{1}{\sqrt{\left(\sqrt{h-1}+1\right)^2}}+\dfrac{1}{\sqrt{\left(\sqrt{h-1}-1\right)^2}}\)
\(=\dfrac{1}{\sqrt{h-1}+1}+\dfrac{1}{\sqrt{h-1}-1}\)
\(=\dfrac{\sqrt{h-1}-1+\sqrt{h-1}+1}{\left(\sqrt{h-1}+1\right)\left(\sqrt{h-1}-1\right)}\)
\(=\dfrac{2\sqrt{h-1}}{\left(\sqrt{h-1}+1\right)\left(\sqrt{h-1}-1\right)}\)
Thay \(h=3\) vào biểu thức ta được :
\(\dfrac{2\sqrt{3-1}}{\left(\sqrt{3-1}+1\right)\left(\sqrt{3-1}-1\right)}=\dfrac{2\sqrt{2}}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{2\sqrt{2}}{1}=2\sqrt{2}\)
Chúc bạn học tốt
30 tháng 7 2022
a: \(H=\dfrac{\sqrt{x-1}+\sqrt{x}+\sqrt{x-1}-\sqrt{x}}{x-1-x}+x\)
\(=-2\sqrt{x-1}+x\)
b: \(x=\dfrac{53}{9-2\sqrt{7}}=9+2\sqrt{7}\)
Khi x=9+2 căn 7 thì \(H=-2\cdot\sqrt{8+2\sqrt{7}}+9+2\sqrt{7}\)
\(=-2\left(\sqrt{7}+1\right)+9+2\sqrt{7}\)
=-2+9=7
25 tháng 6 2023
g: \(=\left(-\sqrt{5}-2\right)\left(\sqrt{5}-2\right)\)
=-(căn 5+2)(căn 5-2)
=-(5-4)=-1
h: \(=\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\dfrac{\sqrt{30}}{3}\right)\left(\dfrac{\sqrt{30}}{5}+\sqrt{2}-\dfrac{4}{5}\sqrt{5}\right)\)
=4/5*căn 10+4/3*căn 6-16/15*căn 15+2/5*căn 15+2-4/5*căn 10+30/15+2/3*căn 15-4/3*căn 6
=4
8 tháng 8 2018
a) Rut gon H
\(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}+\dfrac{1}{2-\sqrt{a}}\)
\(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}-\dfrac{1}{\sqrt{a}-2}\)
DKXD : \(\left\{{}\begin{matrix}\sqrt{a}+3\ne0\\\sqrt{a}-2\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a\ne9\\a\ne4\end{matrix}\right.\)
Ta co : \(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{\sqrt{a}+3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{a-\sqrt{a}-6}{a+\sqrt{a}-6}\)
\(A=\dfrac{1}{\sqrt{h+2\sqrt{h-1}}}+\dfrac{1}{\sqrt{h-2\sqrt{h-1}}}\)
\(=\dfrac{1}{\sqrt{\left(\sqrt{h-1}+1\right)^2}}+\dfrac{1}{\sqrt{\left(\sqrt{h-1}-1\right)^2}}\)
\(=\dfrac{1}{\sqrt{h-1}+1}+\dfrac{1}{\left|\sqrt{h-1}-1\right|}\)
\(D=\dfrac{1}{\sqrt{h+2\sqrt{h-1}}}+\dfrac{1}{\sqrt{h-2\sqrt{h-1}}}\)
\(=\dfrac{1}{\sqrt{h-1}+1}+\dfrac{1}{\sqrt{h-1}-1}\)
\(=\dfrac{2\cdot\sqrt{h-1}}{h}\)