\(C=\dfrac{\sqrt{\dfrac{4x^2+4x+1}{x}}}{\sqrt{x}\cdot\left|2x^2-x-1\right|}=\dfrac{\left|2x+1\right|}{\sqrt{x}}\cdot\dfrac{1}{\sqrt{x}\cdot\left|\left(x-1\right)\left(2x+1\right)\right|}\)

\(=\dfrac{1}{x\left|x-1\right|}\)

\(\left(\frac{\sqrt{x}-4x}{1-4x}-1\right):\left(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\right)\left(ĐK:0\le x\ne\frac{1}{4}\right)\)

\(=\frac{\sqrt{x}-4x+4x-1}{1-4x}:\frac{\left(1+2x\right)+2\sqrt{x}\left(1+2\sqrt{x}\right)+4x-1}{1-4x}\)

\(=\frac{\sqrt{x}-1}{1-4x}.\frac{1-4x}{10x+2\sqrt{x}}=\frac{\sqrt{x}-1}{2\sqrt{x}\left(5\sqrt{x}+1\right)}\)

N=2(2x + 5 )^2 - 3(1 + 4x )(1 - 4x)

= 2 (4x^2 + 20x + 25) - 3(1 - 16x^2)

= 8x^2 + 40x + 50 - 3 + 48x^2

= 56x^2 + 40x - 47

(.....????!!!!!!.....)

\(=\frac{\left(x+1\right)^2}{\left(x-1\right)^2}:\frac{2\left(x+1\right)^2}{4\left(x-1\right)^2}=\frac{\left(x+1\right)^2}{\left(x-1\right)^2}.\frac{4\left(x-1\right)^2}{2\left(x+1\right)^2}=2\)

câu b)  \(\sqrt{-2}\) không xác định

a) \(A=4x-\sqrt{8}-\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)

\(=4x-\sqrt{8}-\frac{\sqrt{x^2\left(x+2\right)}}{\sqrt{x+2}}=4x-\sqrt{8}-x=3x-\sqrt{8}\)

b) \(x=\sqrt{-2}\) (không thỏa mãn)

a)

\(M=2+\sqrt{\left(2x\right)^2-2.2x.3+3^2}\)

\(\Rightarrow M=2+\sqrt{\left(2x-3\right)^2}\)

\(\Rightarrow M=2+2x-3\)

\(\Rightarrow M=2x-1\)

b)

(+) x=5/2

=> \(M=2.\frac{5}{2}-1=5-1=4\)

(+) x= - 1/5

=> \(M=2.\frac{\left(-1\right)}{5}-1=-\frac{2}{5}-1=-\frac{7}{5}\)

ê căn (2x-3)^2=|2x-3| xét 2 th ra nhé

Ta có: \(\sqrt{4x^2-4x+1}-\sqrt{4x^2+4x+1}\)

\(=\sqrt{\left(2x-1\right)^2}-\sqrt{\left(2x+1\right)^2}\)

\(=\left|2x-1\right|-\left|2x+1\right|\)