Q = \(\left(1-\dfrac{\sqrt{a}-4a}{1-4a}\right)\) : \(\left[1-\dfrac{1+2a-2\sqrt{a}\left(2\sqrt{a}+1\right)}{1-4a}\right]\) với a > 0, a ≠ \(\dfrac{1}{4}\)
Rút gọn
Giúp em với ạ ! Em cảm ơn !
Q = \(\left(1-\dfrac{\sqrt{a}-4a}{1-4a}\right)\) : \(\left[1-\dfrac{1+2a-2\sqrt{a}\left(2\sqrt{a}+1\right)}{1-4a}\right]\) với a > 0, a ≠ \(\dfrac{1}{4}\)
Rút gọn
Giúp em với ạ ! Em cảm ơn !
Q = (1 - \(\dfrac{\sqrt{a}-4a}{1-4a}\)) : \(\left[1-\dfrac{1+2a-2\sqrt{a}\left(2\sqrt{a}+1\right)}{1-4a}\right]\)
= \(\left(\dfrac{1-4a-\sqrt{a}+4a}{1-4a}\right):\left[\dfrac{1-4a-1-2a+4a+2\sqrt{a}}{1-4a}\right]\)
= \(\dfrac{1-\sqrt{a}}{1-4a}:\left(\dfrac{-2a+2\sqrt{a}}{1-4a}\right)\)
= \(\dfrac{1-\sqrt{a}}{1-4a}.\dfrac{1-4a}{2\sqrt{a}\left(1-\sqrt{a}\right)}\)
= \(\dfrac{1}{2\sqrt{a}}\) = \(\dfrac{\sqrt{a}}{2a}\)
rút gọn
\(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\) với a > 0,5
\(=\dfrac{2\sqrt{5}\left|a\left(2a-1\right)\right|}{2a-1}=\dfrac{2a\left(2a-1\right)\sqrt{5}}{2a-1}=2a\sqrt{5}\)
\(=\dfrac{2\sqrt{5}\cdot a\left(2a-1\right)}{2a-1}=2a\sqrt{5}\)
Rút gọn các biểu thức sau:
\(A=\dfrac{a^2-1}{3}\sqrt{\dfrac{9}{\left(1-a\right)^2}}\) với a < 1
\(B=\sqrt{\left(3a-5\right)^2}-2a+4\) với a < \(\dfrac{1}{2}\)
\(C=4a-3-\sqrt{\left(2a-1\right)^2}\) với a < 2
\(D=\dfrac{a-2}{4}\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\) với a < 2
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: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
Rút gọn:
\(A=\left(\dfrac{1-2a}{1+2a}-\dfrac{1-4a+4a^2}{1+2a}\cdot\dfrac{1}{1-4a^2}\right)\cdot\left(\dfrac{a+1}{a}+\dfrac{1}{4a^2}\right)-\dfrac{1}{2a}\)
\(A=\left(\dfrac{-\left(2a-1\right)}{2a+1}+\dfrac{\left(2a-1\right)^2}{2a+1}\cdot\dfrac{1}{\left(2a-1\right)\left(2a+1\right)}\right)\cdot\left(\dfrac{4a\left(a+1\right)+1}{4a^2}\right)-\dfrac{1}{2a}\)
\(=\left(\dfrac{-\left(2a-1\right)}{2a+1}+\dfrac{2a-1}{\left(2a+1\right)^2}\right)\cdot\dfrac{4a^2+4a+1}{4a^2}-\dfrac{1}{2a}\)
\(=\dfrac{-\left(2a-1\right)\left(2a+1\right)}{\left(2a+1\right)^2}\cdot\dfrac{\left(2a+1\right)^2}{4a^2}-\dfrac{1}{2a}\)
\(=\dfrac{-\left(4a^2-1\right)}{4a^2}-\dfrac{2a}{4a^2}\)
\(=\dfrac{-4a^2-2a+1}{4a^2}\)
Rút gọn:
\(E=\dfrac{1}{2a-1}\sqrt{5a^4\left(1-4a+4a^2\right)}\)
\(E=\dfrac{1}{2a-1}\sqrt{5a^4\left(1-4a+4a^2\right)}\left(a\ne\dfrac{1}{2}\right)\)
\(=\dfrac{1}{2a-1}\sqrt{5\left(a^2\right)^2\left(1-2a\right)^2}=\dfrac{1}{2a-1}\sqrt{5}.a^2.\left|1-2a\right|\)
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 \(a< \dfrac{1}{2}\Rightarrow1-2a>0\Rightarrow\dfrac{1}{2a-1}\sqrt{5}.a^2.\left|1-2a\right|\)
\(=\dfrac{1}{2a-1}\sqrt{5}.a^2.\left(1-2a\right)=-\sqrt{5}a^2\)
\(E=\dfrac{1}{2a-1}\sqrt{5a^4\left(2a-1\right)^2}=\dfrac{a^2.\left|2a-1\right|.\sqrt{5}}{2a-1}\)
- 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 \(E=\dfrac{-a^2.\left(2a-1\right).\sqrt{5}}{2a-1}=-a^2\sqrt{5}\)
Ta có: \(E=\dfrac{1}{2a-1}\cdot\sqrt{5a^4\cdot\left(4a^2-4a+1\right)}\)
\(=\dfrac{1}{2a-1}\cdot\dfrac{a^2\cdot\sqrt{5}\cdot\left(2a-1\right)}{1}\)
\(=a^2\sqrt{5}\)
\(P=\left(\dfrac{1-2a}{1+2a}1-\dfrac{1-4a+4a^2}{1+2a}\cdot\dfrac{1}{1-4a^2}\right)\cdot\left(\dfrac{1}{4a^2}+\dfrac{a+1}{a}\right)-\dfrac{1}{2a}\)
Rút gọn biểu thức sau: C= \(\dfrac{1}{2a-1}.\sqrt{5a^4.\left(1-4a+4a^2\right)}\)
\(A=\left(\dfrac{1}{2a-b}-\dfrac{a^2-1}{2a^3-b+2a-a^2b}\right)\div\left(\dfrac{4a+2b}{a^3b+ab}-\dfrac{2}{a}\right)\)
a) rút gọn biểu thức A
b)tính giá trị biểu thức A biết 4a^2+b^2=5ab a>b>0
Rút gọn:
\(A=\sqrt{\left(a-3\right)^2}-3a\) với a < 3
\(B=4a+3-\sqrt{\left(2a-1\right)^2}\) với a > 1/2
\(C=\dfrac{4}{a^2-4}\sqrt{\left(a-2\right)^2}\) với a < 2
\(D=\dfrac{a^2-9}{12}:\sqrt{\dfrac{a^2+6a+9}{16}}\) với a < -3
\(A=\left|a-3\right|-3a=3-a-3a=3-4a\)
\(B=4a+3-\left|2a-1\right|=4a+3-2a+1=2a+4\)
\(C=\dfrac{4}{a^2-4}\left|a-2\right|=\dfrac{-4\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=\dfrac{-4}{a+2}\)
\(D=\dfrac{a^2-9}{12}:\sqrt{\dfrac{\left(a+3\right)^2}{16}}=\dfrac{a^2-9}{12}:\dfrac{\left|a+3\right|}{4}=\dfrac{\left(a-3\right)\left(a+3\right).4}{-12\left(a+3\right)}=\dfrac{3-a}{3}\)
\(A=\sqrt{\left(a-3\right)^2}-3a\)
=3-a-3a
=3-4a
a)\(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3\left(x+y\right)^2}{2}}\left(x,y\ge0;x\ne y\right)\)
b)\(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\left(a>0,5\right)\)