Bài 2. (2 điểm) Tìm x, biết: a) $\frac{3}{4}$ + $\frac{1}{4}$.x = $\frac{-1}{3}$ b) (4$\sqrt{x}$ - 9)(3x + $\frac{1}{2}$) = 0 c) 3.$4^{x + 1}$ - 112 = 5.$2^{2x}$

Đáp án+Giải thích các bước giải:

 `a)`

`3/4+1/4. x=\frac{-1}{3}`

`=>1/4. x=-1/3-3/4`

`=>1/4.x=-4/12-9/12`

`=>1/4.x=-13/12`

`=>x=(-13/12):1/4`

`=>x=-13/3`

Vậy `x=-13/3`

`b)`

`(4\sqrt{x}-9).(3x+1/2)=0` `(đk:x\ge0)`

`+)TH1:4\sqrt{x}-9=0`

`=>4\sqrt{x}=9`

`=>\sqrt{x}=9/4`

`=>x=(9/4)^{2}`

`=>x=81/16(tmđk)`

`+)TH2: 3x+1/2=0`

`=>3x=-1/2`

`=>x=-1/6`

Vậy `x\in{81/16;-1/6}`

`c)`

`3.4^{x+1}-112=5.2^{2x}`

`=>3.(2^{2})^{x+1}-112=5.2^{2x}`

`=>3.2^{2.(x+1)}-5.2^{2x}=112`

`=>3.2^{2x+2}-5.2^{2x}=112`

`=>3.2^{2x}.2^{2}-5.2^{2x}=112`

`=>2^{2x}.(3.4-5)=112`

`=>2^{2x}.7=112`

`=>2^{2x}=112:7`

`=>2^{2x}=16`

`=>2^{2x}=2^{4}`

`=>2x=4`

`=>x=2`

Vậy `x=2`