tìm a
\(\left(2a^2+4a+96\right)⋮\left(a+4\right)̸\)
K
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ZR
13 tháng 12 2023
Q = (1 - \(\dfrac{\sqrt{a}-4a}{1-4a}\)) : \(\left[1-\dfrac{1+2a-2\sqrt{a}\left(2\sqrt{a}+1\right)}{1-4a}\right]\)
= \(\left(\dfrac{1-4a-\sqrt{a}+4a}{1-4a}\right):\left[\dfrac{1-4a-1-2a+4a+2\sqrt{a}}{1-4a}\right]\)
= \(\dfrac{1-\sqrt{a}}{1-4a}:\left(\dfrac{-2a+2\sqrt{a}}{1-4a}\right)\)
= \(\dfrac{1-\sqrt{a}}{1-4a}.\dfrac{1-4a}{2\sqrt{a}\left(1-\sqrt{a}\right)}\)
= \(\dfrac{1}{2\sqrt{a}}\) = \(\dfrac{\sqrt{a}}{2a}\)
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: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
26 tháng 5 2022
\(A=\left(\dfrac{-\left(2a-1\right)}{2a+1}+\dfrac{\left(2a-1\right)^2}{2a+1}\cdot\dfrac{1}{\left(2a-1\right)\left(2a+1\right)}\right)\cdot\left(\dfrac{4a\left(a+1\right)+1}{4a^2}\right)-\dfrac{1}{2a}\)
\(=\left(\dfrac{-\left(2a-1\right)}{2a+1}+\dfrac{2a-1}{\left(2a+1\right)^2}\right)\cdot\dfrac{4a^2+4a+1}{4a^2}-\dfrac{1}{2a}\)
\(=\dfrac{-\left(2a-1\right)\left(2a+1\right)}{\left(2a+1\right)^2}\cdot\dfrac{\left(2a+1\right)^2}{4a^2}-\dfrac{1}{2a}\)
\(=\dfrac{-\left(4a^2-1\right)}{4a^2}-\dfrac{2a}{4a^2}\)
\(=\dfrac{-4a^2-2a+1}{4a^2}\)
KN
27 tháng 9 2020
a) \(ĐK:a\ne1;a\ne0\)
\(A=\left[\frac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\frac{1-2a^2+4a}{a^3-1}+\frac{1}{a-1}\right]:\frac{a^3+4a}{4a^2}=\left[\frac{a^2-2a+1}{a^2+a+1}-\frac{1-2a^2+4a}{a^3-1}+\frac{a^2+a+1}{a^3-1}\right].\frac{4a^2}{a^3+4a}\)\(=\left[\frac{a^3-3a^2+3a-1}{a^3-1}-\frac{1-2a^2+4a}{a^3-1}+\frac{a^2+a+1}{a^3-1}\right].\frac{4a^2}{a^3+4a}=\frac{a^3-1}{a^3-1}.\frac{4a}{a^2+4}=\frac{4a}{a^2+4}\)
b) Ta có: \(a^2+4\ge4a\)(*)
Thật vậy: (*)\(\Leftrightarrow\left(a-2\right)^2\ge0\)
Khi đó \(\frac{4a}{a^2+4}\le1\)
Vậy MaxA = 1 khi x = 2