`C=(\sqrt{a}+\sqrt{b})/(2\sqrt{a}-2\sqrt{b})-(\sqrt{a}-\sqrt{b})/(2\sqrt{a}+2\sqrt{b})-(2b)/(b-a)`
ĐKXĐ: `a>0; b>0; a \ne b`
Với `a>0; b>0; a \ne b` ta có:
`C=(\sqrt{a}+\sqrt{b})/(2(\sqrt{a}-\sqrt{b}))-(\sqrt{a}-\sqrt{b})/(2(\sqrt{a}+\sqrt{b}))+(2b)/((\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})`
`C=((\sqrt{a}+\sqrt{b})^2-(\sqrt{a}-\sqrt{b})^2+2b.2)/(2(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}))`
`C=(a+b+2\sqrt{ab}-a-b+2\sqrt{ab}+4b)/(2(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}))`
`C=(4b+4\sqrt{ab})/(2(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}))`
`C=(4\sqrt{b}(\sqrt{b}+\sqrt{a}))/(2(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}))`
`C=(2\sqrt{b})/(\sqrt{a}-\sqrt{b})`
Vậy `C=(2\sqrt{b})/(\sqrt{a}-\sqrt{b})` với `a>0; b>0; a\neb`
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`D=(1/(a-\sqrt{a})+1/(\sqrt{a}-1)):(\sqrt{a}+1)/(a-2\sqrt{a}+1)`
ĐKXĐ: `a>0; a \ne 1`
Với `a>0; a\ne1` thì:
`D=(1/(\sqrt{a}(\sqrt{a}-1))+1/(\sqrt{a}-1)). (\sqrt{a}-1)^2/(\sqrt{a}+1)`
`D=(1+\sqrt{a})/(\sqrt{a}(\sqrt{a}-1)). (\sqrt{a}-1)^2/(\sqrt{a}+1)`
`D= (\sqrt{a}-1)/\sqrt{a}`
Vậy `D=(\sqrt{a}-1)/\sqrt{a}` với `a>0; a\ne1`
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`E=((\sqrt{a}+2)/(a+2\sqrt{a}+1)+(\sqrt{a}-2)/(1-a)). (\sqrt{a}+1)/\sqrt{a}`
ĐKXĐ: `a>0; a\ne1`
Với `a>0; a\ne1` thì:
`E=((\sqrt{a}+2)/(\sqrt{a}+1)^2+(\sqrt{a}-2)/((1-\sqrt{a})(1+\sqrt{a}))). (\sqrt{a}+1)/\sqrt{a}`
`E=((\sqrt{a}+2)(1-\sqrt{a})+(\sqrt{a}-2)(\sqrt{a}+1))/((1-\sqrt{a})(\sqrt{a}+1)^2). (\sqrt{a}+1)/\sqrt{a}`
`E=(\sqrt{a}-a+2-2\sqrt{a}+a+\sqrt{a}-2\sqrt{a}-2)/((1-\sqrt{a})(\sqrt{a}+1)^2).(\sqrt{a}+1)/\sqrt{a}`
`E=(-2\sqrt{a})/((1-\sqrt{a})(\sqrt{a}+1)^2).(\sqrt{a}+1)/\sqrt{a}`
`E=-2/((1-\sqrt{a})(\sqrt{a}+1))`
`E=2/(a-1)`
Vậy `E=2/(a-1)` với `a>0; a\ne1`
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`M=((\sqrt{a}+\sqrt{b})^2-4\sqrt{ab})/(\sqrt{a}-\sqrt{b})-(a\sqrt{b}+b\sqrtt{a})/\sqrt{ab}`
ĐKXĐ: `a>0;b>0; a\neb`
Với `a>0;b>0;a\neb` thì
`M=(a+2\sqrt{ab}+b-4\sqrt{ab})/(\sqrt{a}-\sqrt{b})-(\sqrt{ab}(\sqrt{a}+\sqrt{b}))/\sqrt{ab}`
`M=(a-2\sqrt{ab}+b)/(\sqrt{a}-\sqrt{b})-(\sqrt{a}-\sqrt{b})`
`M=(\sqrt{a}-\sqrt{b})^2/(\sqrt{a}-\sqrt{b})-\sqrt{a}+\sqrt{b}`
`M=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}`
`M=0`
Vậy `M=0` với `a>0;b>0;a\neb`
