`2^(x+2)-2^x=96`

`=>2^x*2^2-2^x=96`

`=>2^x*4-2^x=96`

`=>2^x*(4-1)=96`

`=>2^x*3=96`

`=>2^x=96:3`

`=>2^x=32`

`=>2^x=2^5`

`=>x=5`

Vậy: ...

`a)6/(x-5)+2/(x-8)=18/((x-5)(8-x))-1` `(ĐK:x<>5;x<>8)`

`<=>(6(x-8))/((x-5)(x-8))+(2(x-5))/((x-5)(x-8))=-18/((x-5)(x-8))-((x-5)(x-8))/((x-5)(x-8))`

`<=>6(x-8)+2(x-5)=-18-((x-5)(x-8))`

`<=>6x-48+2x-10=-18-x^2+13x-40`

`<=>8x-58=-x^2+13x-58`

`<=>x^2-13x+8x=0`

`<=>x^2-5x=0` 

`<=>x(x-5)=0`

`<=>x=0(tm)` hoặc `x=5(ktm)`

`b)(x^2-x)/(x+3)-x^2/(x-3)=(7x^2-3x)/(9-x^2)` `(ĐK:x<>+-3)` 

`<=>(x^2-x)/(x+3)-x^2/(x-3)=(3x-7x^2)/(x^2-9)`

`<=>(x-3)(x^2-x)-x^2(x+3)=3x-7x^2`

`<=>x^3-x^2-3x^2+3x-x^3-3x^2=3x-7x^2`

`<=>-7x^2=3x-7x^2`

`<=>3x=0`

`<=>x=0(tm)` 

`3/((x-1)(x-2))-2/((x-1)(x-3))=1/((x-2)(x-3))` (ĐK: `x<>1;x<>2;x<>3`)

`<=>(3(x-3))/((x-1)(x-2)(x-3))-(2(x-2))/((x-1)(x-2)(x-3))=(x-1)/((x-1)(x-2)(x-3))`

`<=>3(x-3)-2(x-2)=x-1`

`<=>3x-9-2x+4=x-1`

`<=>x-5=x-1`

`<=>-5=-1` 

Điều này vô lý 

`=>` Không có x thỏa mãn

a) Đặt: `B=3+3^2+...+3^99`

`3B=3^2+3^3+...+3^100`

`3B-B=(3^2+3^3+...+3^100)-(3+3^2+..+3^99)`

`2B=3^100-3`

`B=(3^100-3)/2`

`=>A=2*(3^100-3)/2:(3^99-1)`

`=(3^100-3):(3^99-1)`

`=3(3^99-1):(3^99-1)`

`=3` 

b) Đặt: `C=1+5+5^2+...+5^99`

`5C=5+5^2+..+5^100`

`5C-C=(5+5^2+..+5^100)-(1+5+5^2+...+5^99)`

`4C=5^100-1`

`C=(5^100-1)/4`

`=>B=(5^100-1):(4*(5^100-1)/4)`

`=(5^100-1):(5^100-1)`

`=1`

`9)(4/7-1/2x)^2=4/9`

`=>(4/7-1/2x)^2=(2/3)^2`

TH1: `4/7-1/2x=2/3`

`=>1/2x=4/7-2/3`

`=>1/2x=-2/21`

`=>x=-2/21:1/2`

`=>x=-4/21`

TH2:`4/7-1/2x=-2/3`

`=>1/2x=4/7+2/3`

`=>1/2x=26/21`

`=>x=26/21:1/2`

`=>x=52/21`

`10)3(x-1/2)^3=81`

`=>(x-1/2)^3=81/3`

`=>(x-1/2)^3=27`

`=>(x-1/2)^3=3^3`

`=>x-1/2=3`

`=>x=3+1/2`

`=>x=7/2`

anh xem lại bài nhé `a>=4` thì chẳng phải `1/a<=1/4` chứ sao `>=` được ạ 

TH1: `n=2k`

`=>n(n+15)`

`=2k*(2k+15)` chia hết cho 2 

`=>` ĐPCM

TH2: `n=2k+1` 

`=>(2k+1)(2k+1+15)`

`=>(2k+1)(2k+16)`

`=>2(2k+1)(k+8)` chia hết cho 2 

`=>` ĐPCM 

`2+2^2+...+2^20`

`=(2+2^3)+(2^2+2^4)+...+(2^18+2^20)`

`=(2+8)+2^2*(2+8)+...+2^18*(2+8)`

`=10+2^2*10+...+2^18*10`

`=10*(1+2^2+...+2^18)` chia hết cho 10 

`=>2+2^2+...+2^20` chia hết cho 10 

`1)P:(4x^2-16)/(2x+1)=(4x^2+4x+1)/(x-2)`

`=>P=(4x^2+4x+1)/(x-2)*(4x^2-16)/(2x+1)`

`=>P=(2x+1)^2/(x-2)*(4(x^2-4))/(2x+1)`

`=>P=(2x+1)^2/(x-2)*(4(x-2)(x+2))/(2x+1)` 

`=>P=(2x+1)*4(x+2)`

`=>P=4(2x^2+5x+2)`

`=>P=8x^2+20x+8` 

`2)(2x^2+4x+8)/(x^3-3x^2-x+3):P=(x^3-8)/((x+1)(x-3))`

`=>P=(2x^2+4x+8)/(x^3-3x^2-x+3):(x^3-8)/((x+1)(x-3))`

`=>P=(2(x^2+2x+4))/((x^2-1)(x-3))*((x+1)(x-3))/((x-2)(x^2+2x+4))`

`=>P=(2(x^2+2x+4))/((x-1)(x+1)(x-3))*((x+1)(x-3))/((x-2)(x^2+2x+4))`

`=>P=2/((x-1)(x-2))`

`=>P=2/(x^2-3x+2)` 

Một người ăn hết số lương thực đó trong:

`120 xx 20 = 2400` (ngày)

Số lương thực đó đủ cho 150 người ăn trong;

`2400:150=16 ` (ngày)

ĐS: ...