Rút gọn biểu thức sau: C= \(\dfrac{1}{2a-1}.\sqrt{5a^4.\left(1-4a+4a^2\right)}\)
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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 \(E=\dfrac{-a^2.\left(2a-1\right).\sqrt{5}}{2a-1}=-a^2\sqrt{5}\)
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
\(\frac{2}{2a-1}.\sqrt{5x^4\left(1-4a+4a^2\right)}\)
\(=\frac{2}{2a-1}.\sqrt{5x^4\left(2a-1\right)^2}\)
\(=\frac{2}{2a-1}.x^2.\left(2a-1\right).\sqrt{5}\)
\(=2\sqrt{5}x^2\)
2 tháng 7 2021
a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)
\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)
\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)
=-a-1
b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)
\(=\left|3a-5\right|-2a+4\)
\(=5-3a-2a+4\)
=9-5a
c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)
\(=4a-3-\left|2a-1\right|\)
\(=4a-3-2a+1\)
\(=2a-2\)
d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
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
\(=\dfrac{2\sqrt{5}\cdot a\left(2a-1\right)}{2a-1}=2a\sqrt{5}\)
TN
0
ZR
13 tháng 12 2023
Q = (1 - \(\dfrac{\sqrt{a}-4a}{1-4a}\)) : \(\left[1-\dfrac{1+2a-2\sqrt{a}\left(2\sqrt{a}+1\right)}{1-4a}\right]\)
= \(\left(\dfrac{1-4a-\sqrt{a}+4a}{1-4a}\right):\left[\dfrac{1-4a-1-2a+4a+2\sqrt{a}}{1-4a}\right]\)
= \(\dfrac{1-\sqrt{a}}{1-4a}:\left(\dfrac{-2a+2\sqrt{a}}{1-4a}\right)\)
= \(\dfrac{1-\sqrt{a}}{1-4a}.\dfrac{1-4a}{2\sqrt{a}\left(1-\sqrt{a}\right)}\)
= \(\dfrac{1}{2\sqrt{a}}\) = \(\dfrac{\sqrt{a}}{2a}\)
17 tháng 6 2017
\(\frac{\sqrt{3x^2+6xy+3y^2}}{x^2-y^2}\)
<=>\(\frac{\sqrt{3.\left(x+y\right)^2}}{\left(x-y\right).\left(x+y\right)}\)
<=>\(\frac{\sqrt{3}\left|x+y\right|}{\left(x-y\right).\left(x+y\right)}.\)
<=>\(\frac{\sqrt{3}}{x-y}\)
Ha Hoang CTV, sao bạn bỏ được dấu giá trị tuyệt đối của 1-2a vậy??