Đề rõ hơn đi ạ?

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

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

\(\left(\dfrac{x+\sqrt{x}+1}{x\sqrt{x}-1}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{6}{x-1}\\ =\left(\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}\right)^3-1}+\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{x-1}{6}\\ =\left(\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{x-1}{6}\\ =\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{x-1}{6}\\ =\dfrac{2}{\sqrt{x}-1}\cdot\dfrac{x-1}{6}\\ =\dfrac{\sqrt{x}+1}{3}\)

`4/6 xx 1/5 + 4/6 xx 3/5 + 2/6 xx 4/5`

`= 4/6 xx (1/5 + 3/5) + 2/6 xx 4/5`

`= 4/6 xx 4/5 + 2/6 xx 4/5`

`= (4/6 +2/6) xx 4/5`

`= 6/6 xx 4/5`

`=1 xx 4/5`

`=4/5`

` 2 1/3 + 1 3/5`

`= (2xx3+1)/3 + (1xx5+3)/5`

`= 7/3 + 8/5`

`=59/15`

__

`4 5/9 -2 5/6`

`= (4xx9+5)/9 - (2xx6+5)/6`

`= 41/9 - 17/6`

`=31/18`

__

`3 2/4 xx 2 2/7`

`= (3xx4+2)/4 xx (2xx7+2)/7`

`= 14/4 xx 16/7`

`=8`

__

`4 2/5 : 4 1/8`

`= (4xx5+2)/5 : (4xx8+1)/8`

`= 22/5 xx 8/33`

`=16/15`

thường thường trên lớp k cần đkxđ nên không biết, camon.