Đáp án + Giải thích các bước giải:
1)
a)ĐKXĐ: `x>=0;x\ne2`
b)ĐKXĐ: `x>=0;x/ne 2/3`
2)
a)`\sqrt{4.36}=\sqrt{2^2.6^2}=2.6=12`
b)`\sqrt{(25)/(81) . (16)/(49)}=\sqrt{(5/9)^2 . (4/7)^2}`
`=5/9 . 4/7=(20)/(63)`
c)`(\sqrt{8}-3\sqrt{2}).2=\sqrt{16}-3\sqrt{4}`
`=\sqrt{4^2}-3\sqrt{2^2}=4-3.2=-2`
d)`(\sqrt{14}-\sqrt{7})/(1-\sqrt{2})=(-(\sqrt{14}-\sqrt{7}))/(\sqrt{2}-1)`
`=(-\sqrt{7}(\sqrt{2}-1))/(\sqrt{2}-1)=-\sqrt{7}`
3)
a)`\sqrt{19+\sqrt{136}}-\sqrt{19-\sqrt{136}}`
`=\sqrt{19+2\sqrt{34}}-\sqrt{19-2\sqrt{34}}`
`=\sqrt{17+2.\sqrt{17}.\sqrt{2}+2}-\sqrt{17-2.\sqrt{17}.\sqrt{2}+2}`
`=\sqrt{(\sqrt{17}+\sqrt{2})^2}-\sqrt{(\sqrt{17}-\sqrt{2})^2}
`=|\sqrt{17}+\sqrt{2}|-|\sqrt{17}-\sqrt{2}|`
`=\sqrt{17}+\sqrt{2}-\sqrt{17}+\sqrt{2}
`=2\sqrt{2}
b)`\root{3}{27}+\root{3}{-64}+2\root{3}{125}`
`=\root{3}{3^3}+\root{3}{(-8)^3}+2\root{3}{5^3}`
`=3+-8+2.5=5`
