![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\sqrt{4x-1-2\sqrt{4x-1}+1}+\sqrt{4x-1+2\sqrt{4x-1}+1}\)
\(=\sqrt{\left(\sqrt{4x-1}-1\right)^2}+\sqrt{\left(\sqrt{4x-1}+1\right)^2}\)
\(=\left|\sqrt{4x-1}-1\right|+\sqrt{4x-1}+1\)
\(=\left[{}\begin{matrix}2\sqrt{4x-1}\text{ nếu }x\ge\dfrac{1}{2}\\2\text{ nếu }\dfrac{1}{4}\le x< \dfrac{1}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\frac{4x-x^3-x+4x^3}{1-4x^2}:\frac{4x^2-x^4+1-4x^2}{1-4x^2}\)
\(=\frac{3x^3+3x}{1-4x^2}:\frac{1-x^4}{1-4x^2}\)
\(=\frac{3x\left(x^2+1\right)}{\left(1-x^2\right)\left(1+x^2\right)}\)
\(=\frac{3x}{1-x^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1-\left(\dfrac{2}{1+2\sqrt{x}}-\dfrac{5\sqrt{x}}{4x-1}-\dfrac{1}{1-2\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{4x+4\sqrt{x}+1}\)
\(A=1-\dfrac{2\left(2\sqrt{x}-1\right)-5\sqrt{x}+\left(2\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{\left(2\sqrt{x}+1\right)^2}\)
\(A=1-\dfrac{4\sqrt{x}-2-5\sqrt{x}+2\sqrt{x}+1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{\left(2\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(A=1-\dfrac{\sqrt{x}-1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{\left(2\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(A=1-\dfrac{2\sqrt{x}+1}{2\sqrt{x}-1}=\dfrac{2\sqrt{x}-1-2\sqrt{x}-1}{2\sqrt{x}-1}=\dfrac{-2}{2\sqrt{x}-1}\)
Tick hộ nha
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì em ghi không rõ nên cô sẽ hiểu là:
Rút gọn \(H=2x-3+\sqrt{4x^2-4x+1}\)
Ta có \(H=2x-3+\sqrt{\left(2x-1\right)^2}\)
Với \(2x-1\ge0\Leftrightarrow x\ge\frac{1}{2}\) , \(H=2x-3+2x-1=4x-4\)
Với \(x< \frac{1}{2},H=2x-3-2x+1=-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=5x-\sqrt{4x^2-4x+1}\)
\(=5x-\sqrt{\left(2x-1\right)^2}\)
\(=5x-\left|2x-1\right|\)
+) Với x < 1/2
A = 5x - [ -( 2x - 1 ) ] = 5x - ( 1 - 2x ) = 5x - 1 + 2x = 7x - 1
+) Với x ≥ 1/2
A = 5x - ( 2x - 1 ) = 5x - 2x + 1 = 3x + 1
b) Với x = -2 < 1/2
=> A = 7.(-2) - 1 = -14 - 1 = -15
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{4x-2\sqrt{4x-1}}+\sqrt{4x+2\sqrt{4x-1}}\)(với \(x\ge\dfrac{1}{4}\))
\(=\sqrt{\left(\sqrt{4x-1}-1\right)^2}+\sqrt{\left(\sqrt{4x-1}+1\right)^2}\)
\(=\left(\sqrt{4x-1}-1\right)+\left(\sqrt{4x-1}+1\right)\)
\(=2\sqrt{4x-1}\) (với \(x\ge\dfrac{1}{4}\))
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}=\left|3a^2\right|=3a^2\)
b) \(2\sqrt{a^2}-5a=2\left|a\right|-5a=-2a-5a=-7a\)
c) \(\sqrt{16\left(1+4x+4x^2\right)}=\sqrt{\left[4\left(1+2x\right)\right]^2}=\left|4\left(1+2x\right)\right|=4\left(1+2x\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
12b:
\(\left(\dfrac{1}{1-2\sqrt{x}}-1\right):\left(\dfrac{1}{4x-1}+\dfrac{1}{2\sqrt{x}+1}\right)\)
\(=\left(\dfrac{-1}{2\sqrt{x}-1}-1\right):\dfrac{1+2\sqrt{x}-1}{4x-1}\)
\(=\dfrac{-1-2\sqrt{x}+1}{2\sqrt{x}-1}\cdot\dfrac{4x-1}{2\sqrt{x}}\)
\(=\dfrac{-\left(4x-1\right)}{2\sqrt{x}-1}=-2\sqrt{x}-1\)