Đáp án + giải thích các bước giải:
2/
a) `sin^2 40^0+sin^2 20^0+cos^2 40^0+sin^2 70^0`
`=sin^2 40^0 + cos^2 40^0 +sin^2 20^0+ sin^2 70^0`
`=1+sin^2 20^0+cos^2 20^0`
`=1+1`
`=2`
b) `tan a=8/15 `
`->(sin a)/(cos a)=8/15`
`->cos a=15/8 sin a`
`->cos^2a=225/64 sin^2a`
mà `sin^2a+cos^2a=1`
`->225/64 sin^2a+ sin^2a=1`
`->289/64 sin^2a=1`
`->sin^2a=64/489`
`->sina=8/17`
2/
a) `A=(4\sqrt{x}-7)/(\sqrt{x}-2) (x>=0;x\ne4)`
`x=9(TM)`
`->A=(4\sqrt{9}-7)/(\sqrt{9}-2)=(4.3-7)/(3-2)=5`
b) `B=((3\sqrt{x}-4)/(x-4)-1/(\sqrt{x}+2)):2/(\sqrt{x}+2) (x>=0;x\ne4)`
`=(3\sqrt{x}-4-(\sqrt{x}-2))/((\sqrt{x}-2)(\sqrt{x}+2)) . (\sqrt{x}+2)/2`
`=(3\sqrt{x}-4-\sqrt{x}+2)/(\sqrt{x}-2) . 1/2 `
`=(2\sqrt{x}-2)/(\sqrt{x}-2) . 1/2`
`=(\sqrt{x}-1)/(\sqrt{x}-2)`
c) `A/B=(4\sqrt{x}-7)/(\sqrt{x}-2):(\sqrt{x}-1)/(\sqrt{x}-2)=(4\sqrt{x}-7)/(\sqrt{x}-1)=(4\sqrt{x}-4-3)/(\sqrt{x}-1)=4-3/(\sqrt{x}-1)`
`A/B\inZZ`
`->3/(\sqrt{x}-1)\inZZ`
`->\sqrt{x}-1\in Ư(3)`
mà `x>=0->\sqrt{x}>=0->\sqrt{x}-1>=-1`
`->\sqrt{x}-1\in{-1;1;3}`
`->\sqrt{x}\in{0;2;4}`
`->\sqrt{x}\in{0;4;16}`
Kết hợp `ĐKXD`
`->x\in{0;16}`
Vậy `x=0;16`
