Rút gọn biểu thức
a,\(A=\frac{2}{x^2-y^2}\sqrt{\frac{3x^2+6xy+3y^2}{4}}\)
b, \(B=\frac{1}{2a-1}\sqrt{5a^4\left(1-4a+4a^2\right)}\)
K
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Những câu hỏi liên quan
8 tháng 7 2018
\(a.A=\dfrac{2}{x^2-y^2}.\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}=\dfrac{2}{\left(x-y\right)\left(x+y\right)}.\dfrac{\left(x+y\right)\sqrt{3}}{2}=\dfrac{\sqrt{3}}{x-y}\) ( x # y )
\(b.\dfrac{1}{2x-1}.\sqrt{5a^4\left(1-4x+4a^2\right)}=\dfrac{1}{2a-1}.\left(2a-1\right)a^2\sqrt{5}=a^2\sqrt{5}\) ( a # \(\dfrac{1}{2}\) )
TH
1
2 tháng 10 2016
\(B=\frac{1}{2a-1}.\sqrt{5a^4\left(2a-1\right)^2}=\sqrt{5}a^2.\frac{\left|2a-1\right|}{2a-1}\)
Nếu \(a>\frac{1}{2}\) thì \(B=\sqrt{5}a^2\)
Nếu \(a< \frac{1}{2}\) thì \(B=-\sqrt{5}a^2\)
PK
2
HH
27 tháng 9 2020
\(B=\frac{1}{2a-1}\sqrt{5a^4\left(1-4a+4a^2\right)}\)
\(B=\frac{2\left|a\right|}{2a-1}\sqrt{5\left[1-2.2a+\left(2a\right)^2\right]}\)
\(B=\frac{2a}{2a-1}\sqrt{5\left(1-2a\right)^2}\)
\(B=\frac{2a\left|1-2a\right|}{2a-1}\sqrt{5}\)
\(=\frac{2a\left(2a-1\right)}{2a-1}\sqrt{5}=2a\sqrt{5}\)
NN
27 tháng 9 2020
\(ĐKXĐ:a\ne\frac{1}{2}\)
\(B=\frac{1}{2a-1}.\sqrt{5a^4.\left(1-4a+4a^2\right)}\)
\(=\frac{1}{2a-1}.\sqrt{5a^4.\left(1-2a\right)^2}\)
\(=\frac{1}{2a-1}.\sqrt{5}.\sqrt{a^4}.\sqrt{\left(1-2a\right)^2}\)
\(=\frac{1}{2a-1}.\sqrt{5}.a^2.\left|1-2a\right|=\frac{\sqrt{5}.a^2.\left|1-2a\right|}{2a-1}\)
+) Nếu \(a< \frac{1}{2}\)\(\Rightarrow\left|1-2a\right|=1-2a=-\left(2a-1\right)\)
\(\Rightarrow B=\frac{-\sqrt{5}.a^2.\left(2a-1\right)}{2a-1}=-\sqrt{5}.a^2\)
+) Nếu \(a>\frac{1}{2}\)\(\Rightarrow\left|1-2a\right|=-\left(1-2a\right)=-1+2a=2a-1\)
\(\Rightarrow B=\frac{\sqrt{5}.a^2.\left(2a-1\right)}{2a-1}=\sqrt{5}.a^2\)
LT
1
20 tháng 7 2019
\(\frac{2}{2a-1}.\sqrt{5x^4\left(1-4a+4a^2\right)}\)
\(=\frac{2}{2a-1}.\sqrt{5x^4\left(2a-1\right)^2}\)
\(=\frac{2}{2a-1}.x^2.\left(2a-1\right).\sqrt{5}\)
\(=2\sqrt{5}x^2\)
TN
0
5 tháng 9 2020
a) \(2\sqrt{3x}-4\sqrt{3x}+27-2\sqrt{3x}=27-4\sqrt{3x}\)
b) \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{8x}+28=3\sqrt{2x}+2\sqrt{8x}+28=3\sqrt{2x}+4\sqrt{2x}+28=7\sqrt{2x}+28\)
c) \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{\left(x-y\right)\left(x+y\right)}.\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{\sqrt{6}}{x-y}\)
d) \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4x+4a^2\right)}=\frac{2}{2a-1}\sqrt{5a^2\left(2a-1\right)^2}=\frac{2}{2a-1}.\sqrt{5}\left|a\left(2a-1\right)\right|=2a\sqrt{5}\)
Thiếu ĐKXĐ : ..............
5 tháng 9 2020
a) Ta có: \(2\sqrt{3x}-4\sqrt{3x}+27-2\sqrt{3x}\)
\(=27-4\sqrt{3x}\)
b) Ta có: \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{8x}+28\)
\(=3\sqrt{2x}-5.2\sqrt{2x}+7.2\sqrt{2x}+28\)
\(=3\sqrt{2x}-10\sqrt{2x}+14\sqrt{2x}+28\)
\(=7\sqrt{2x}+28\)
c) Ta có: \(\frac{2}{x^2-y^2}.\sqrt{\frac{3\left(x+y\right)^2}{2}}\)
\(=\sqrt{\frac{4}{\left(x-y\right)^2.\left(x+y\right)^2}.\frac{3\left(x+y\right)^2}{2}}\)
\(=\sqrt{\frac{2.3}{\left(x-y\right)^2}}\)
\(=\frac{1}{x-y}.\sqrt{6}\)
d) Ta có: \(\frac{2}{2a-1}.\sqrt{5a^2.\left(1-4a+4a^2\right)}\)
\(=\sqrt{\frac{4}{\left(2a-1\right)^2}.5a^2.\left(2a-1\right)^2}\)
\(=2a.\sqrt{5}\)
\(\frac{\sqrt{3x^2+6xy+3y^2}}{x^2-y^2}\)
<=>\(\frac{\sqrt{3.\left(x+y\right)^2}}{\left(x-y\right).\left(x+y\right)}\)
<=>\(\frac{\sqrt{3}\left|x+y\right|}{\left(x-y\right).\left(x+y\right)}.\)
<=>\(\frac{\sqrt{3}}{x-y}\)