Đáp án+Giải thích các bước giải:
a)
`M=\frac{1}{2\sqrt{m}-2}-\frac{1}{2\sqrt{m}+2}+\frac{\sqrt{m}}{1-m}`
ĐKXĐ: `m≥0;mne1`
`=\frac{1}{2(\sqrt{m}-1)}-\frac{1}{2(\sqrt{m}+1)}-\frac{\sqrt{m}}{m-1}`
`=\frac{\sqrt{m}+1-(\sqrt{m}-1)-2\sqrt{m}}{2(\sqrt{m}-1)(\sqrt{m}+1)}`
`=\frac{\sqrt{m}+1-\sqrt{m}+1-2\sqrt{m}}{2(\sqrt{m}-1)(\sqrt{m}+1)}`
`=\frac{-2\sqrt{m}+2}{2(\sqrt{m}-1)(\sqrt{m}+1)}`
`=\frac{-2(\sqrt{m}-1)}{2(\sqrt{m}-1)(\sqrt{m}+1)}`
`=\frac{-1}{\sqrt{m}+1}`
Vậy `M=\frac{-1}{\sqrt{m}+1}` với `m≥0;mne1`
b)
Ta có: `m=4/9`
`=>M=\frac{-1}{\sqrt{4/9}+1}`
`=\frac{-1}{2/3+1}`
`=\frac{-1}{5/3}`
`=-3/5`
Vậy `M=-3/5` với `m=4/9`
