\(RÚTGỌNBIỂUTHỨC:\frac{a-3\sqrt{a}}{\sqrt{a}-3}-\frac{a+\sqrt{a}+3}{\sqrt{a}+3}\)
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AH
Akai Haruma
Giáo viên
27 tháng 7 2020
Lời giải:
ĐKXĐ: a\geq 0; a\neq 1; a\neq 9$
Ta có:
\(G=\left[\frac{a\sqrt{a}-3}{(\sqrt{a}+1)(\sqrt{a}-3)}-\frac{2(\sqrt{a}-3)^2}{(\sqrt{a}+1)(\sqrt{a}-3)}-\frac{(\sqrt{a}+3)(\sqrt{a}+1)}{(\sqrt{a}-3)(\sqrt{a}+1)}\right].\frac{a+8}{a-1}\)
\(=\frac{a\sqrt{a}-3-2(\sqrt{a}-3)^2-(\sqrt{a}+3)(\sqrt{a}+1)}{(\sqrt{a}+1)(\sqrt{a}-3)}.\frac{a+8}{a-1}\)
\(=\frac{a\sqrt{a}-3a+8\sqrt{a}-24}{(\sqrt{a}+1)(\sqrt{a}-3)}.\frac{a+8}{a-1}=\frac{(\sqrt{a}-3)(a+8)}{(\sqrt{a}+1)(\sqrt{a}-3)}.\frac{a+8}{a-1}=\frac{(a+8)^2}{(\sqrt{a}+1)(a-1)}\)
1 tháng 2 2022
\(=\dfrac{a\sqrt{a}-3-2\left(a-6\sqrt{a}+9\right)-a-4\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{a-1}{a+8}\)
\(=\dfrac{a\sqrt{a}-a-4\sqrt{a}-6-2a+12\sqrt{a}-18}{\left(\sqrt{a}-3\right)}\cdot\dfrac{\sqrt{a}-1}{a+8}\)
\(=\dfrac{a\sqrt{a}-3a+8\sqrt{a}-24}{\left(\sqrt{a}-3\right)}\cdot\dfrac{\sqrt{a}-1}{a+8}=\sqrt{a}-1\)
HT
17 tháng 6 2019
a/ \(A=\frac{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}{2-\sqrt{3}}+\frac{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}{2+\sqrt{3}}\)
\(A=\frac{2+\sqrt{3}+2-\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{4}{1}=4\)
b/\(A=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\)
\(A=\frac{\sqrt{2}-1}{3-2\sqrt{2}}-\frac{\sqrt{2}+1}{3+2\sqrt{2}}\)
\(A=\frac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{2}+1\right)\left(3-2\sqrt{2}\right)}{9-8}\)
\(A=3\sqrt{2}+4-3-2\sqrt{2}-3\sqrt{2}+4-3+2\sqrt{2}=8\)
c/ \(A=\frac{\left(\sqrt{5}+\sqrt{3}\right)^2+\left(\sqrt{5}-\sqrt{3}\right)^2}{5-3}\)
\(A=\frac{5+2\sqrt{15}+3+5-2\sqrt{15}+3}{2}=8\)
d/ theo câu c có \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=8\)
\(\Rightarrow A=8-\frac{\left(\sqrt{5}+1\right)^2}{5-1}=\frac{32-5-2\sqrt{5}-1}{4}=\frac{2\left(13-\sqrt{5}\right)}{4}=\frac{13-\sqrt{5}}{2}\)
4 tháng 8 2019
\(\frac{5\sqrt{3}}{\sqrt{3-\sqrt{5}}-\sqrt{3}}-\frac{5\sqrt{3}}{\sqrt{3-\sqrt{5}}+\sqrt{3}}\)
\(=\frac{5\sqrt{3}\left(\sqrt{3-\sqrt{5}}+\sqrt{3}\right)-5\sqrt{3}\left(\sqrt{3-\sqrt{5}}-\sqrt{3}\right)}{\left(\sqrt{3-\sqrt{5}}-\sqrt{3}\right)\left(\sqrt{3-\sqrt{5}}+\sqrt{3}\right)}\)
\(=\frac{5\sqrt{3}.2\sqrt{3}}{3-\sqrt{5}-3}=-\frac{30}{\sqrt{5}}=-\frac{\sqrt{5}.\sqrt{180}}{\sqrt{5}}=-\sqrt{180}\)
KT
11 tháng 7 2018
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
Bang am 1