Rút Gọn
\(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\left(a>\frac{1}{2}\right)\)
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LT
1
20 tháng 7 2019
\(\frac{2}{2a-1}.\sqrt{5x^4\left(1-4a+4a^2\right)}\)
\(=\frac{2}{2a-1}.\sqrt{5x^4\left(2a-1\right)^2}\)
\(=\frac{2}{2a-1}.x^2.\left(2a-1\right).\sqrt{5}\)
\(=2\sqrt{5}x^2\)
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 \(a>\frac{1}{2}\)\(\Rightarrow\left|1-2a\right|=-\left(1-2a\right)=-1+2a=2a-1\)
\(\Rightarrow B=\frac{\sqrt{5}.a^2.\left(2a-1\right)}{2a-1}=\sqrt{5}.a^2\)
TN
0
TN
2
9 tháng 11 2021
\(=\dfrac{2\sqrt{5}\left|a\left(2a-1\right)\right|}{2a-1}=\dfrac{2a\left(2a-1\right)\sqrt{5}}{2a-1}=2a\sqrt{5}\)
9 tháng 11 2021
\(=\dfrac{2\sqrt{5}\cdot a\left(2a-1\right)}{2a-1}=2a\sqrt{5}\)
17 tháng 6 2017
\(\frac{\sqrt{3x^2+6xy+3y^2}}{x^2-y^2}\)
<=>\(\frac{\sqrt{3.\left(x+y\right)^2}}{\left(x-y\right).\left(x+y\right)}\)
<=>\(\frac{\sqrt{3}\left|x+y\right|}{\left(x-y\right).\left(x+y\right)}.\)
<=>\(\frac{\sqrt{3}}{x-y}\)
AQ
3
AT
30 tháng 7 2021
\(E=\dfrac{1}{2a-1}\sqrt{5a^4\left(1-4a+4a^2\right)}\left(a\ne\dfrac{1}{2}\right)\)
\(=\dfrac{1}{2a-1}\sqrt{5\left(a^2\right)^2\left(1-2a\right)^2}=\dfrac{1}{2a-1}\sqrt{5}.a^2.\left|1-2a\right|\)
Xét \(a>\dfrac{1}{2}\Rightarrow1-2a< 0\Rightarrow\dfrac{1}{2a-1}\sqrt{5}.a^2.\left|1-2a\right|\)
\(=\dfrac{1}{2a-1}\sqrt{5}.a^2.\left(2a-1\right)=\sqrt{5}a^2\)
Xét \(a< \dfrac{1}{2}\Rightarrow1-2a>0\Rightarrow\dfrac{1}{2a-1}\sqrt{5}.a^2.\left|1-2a\right|\)
\(=\dfrac{1}{2a-1}\sqrt{5}.a^2.\left(1-2a\right)=-\sqrt{5}a^2\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
\(E=\dfrac{1}{2a-1}\sqrt{5a^4\left(2a-1\right)^2}=\dfrac{a^2.\left|2a-1\right|.\sqrt{5}}{2a-1}\)
- Với \(2a-1>0\Rightarrow a>\dfrac{1}{2}\) thì \(E=\dfrac{a^2\left(2a-1\right).\sqrt{5}}{2a-1}=a^2\sqrt{5}\)
- Với \(a< \dfrac{1}{2}\) thì \(E=\dfrac{-a^2.\left(2a-1\right).\sqrt{5}}{2a-1}=-a^2\sqrt{5}\)
30 tháng 9 2016
\(\left(\frac{1}{2a-b}+\frac{3b}{b^2-4a^2}-\frac{2}{2a+b}\right):\left(1+\frac{4a^2+b^2}{4a^2-b^2}\right)\left(ĐK:2a\ne\pm b\right)\)
\(=\left(\frac{1}{2a-b}-\frac{3b}{\left(2b-b\right)\left(2a+b\right)}-\frac{2}{2a+b}\right):\frac{4a^2-b^2+4a^2+b^2}{\left(2a-b\right)\left(2a+b\right)}\)
\(=\frac{2a+b-3b-2\left(2a-b\right)}{\left(2a-b\right)\left(2a+b\right)}\cdot\frac{\left(2a-b\right)\left(2a+b\right)}{8a^2}\)
\(=\frac{2a+b-3b-4a+2b}{8a^2}=\frac{-2a}{8a^2}=-\frac{1}{4a}\)
Ta có: \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=\frac{2}{2a-1}.\sqrt{5a^2\left(2a-1\right)^2}=\frac{2}{2a-1}.a\sqrt{5}.\left(2a-1\right)=2a\sqrt{5}\)