Với `x >= 0,x ne 4` có:
`M=[(\sqrt{x}+1)(\sqrt{x}+2)+2\sqrt{x}(\sqrt{x}-2)-2-5\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`M=[x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`M=[3x-6\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]=[3\sqrt{x}]/[\sqrt{x}+2]`
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`N=(1/[\sqrt{a}-1]-1/\sqrt{a}):([\sqrt{a}+1]/[\sqrt{a}-2]-[\sqrt{a}+2]/[\sqrt{a}-1])`
- Biểu thức `N` là như vầy?
Với `a > 0,a ne 1,a ne 4` có:
`N=[\sqrt{a}-\sqrt{a}+1]/[\sqrt{a}(\sqrt{a}-1)]:[(\sqrt{a}+1)(\sqrt{a}-1)-(\sqrt{a}+2)(\sqrt{a}-2)]/[(\sqrt{a}-2)(\sqrt{a}-1)]`
`N=1/[\sqrt{a}(\sqrt{a}-1)].[(\sqrt{a}-2)(\sqrt{a}-1)]/[a-1-a+4]`
`N=[\sqrt{a}-2]/[3\sqrt{a}]`
Với \(x\ge0;x\ne4\)
Khi đó:
\(M=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{x-4}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}-\dfrac{2+5\sqrt{x}}{x-4}\\ =\dfrac{x+2\sqrt{x}+\sqrt{x}+2}{x-4}+\dfrac{2x-4\sqrt{x}}{x-4}-\dfrac{2+5\sqrt{x}}{x-4}\\ =\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{x-4}\\ =\dfrac{3x-6\sqrt{x}}{x-4}\\ =\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
Với \(a>0;a\ne1;a\ne4\)
Khi đó:
\(N=(\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}):\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ =\left(\dfrac{\sqrt{a}}{a-\sqrt{a}}-\dfrac{\sqrt{a}-1}{a-\sqrt{a}}\right):\left(\dfrac{a-1}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{a-4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ =\dfrac{1}{a-\sqrt{a}}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\\ =\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right).3}\\ =\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
