Đáp án:
Giải thích các bước giải:
a) `(4/5)^(2x+7)=625/256`
`=>(4/5)^(2x+7)=(5/4)^4`
`=>(4/5)^(2x+7)=(4/5)^(-4)`
`=>`$2x+7=-4$
`=>`$2x=-11$
`=>x=(-11)/2`
Vậy `x=(-11)/2`.
b) `(7^(x+2)+7^(x+1)+7^x)/57=(5^2x+5^(2x+1)+5^(2x+3))/131`
`=>(7^x.(7^2+7+1))/57=(5^(2x).(1+5+5^3))/131`
`=>(7^x. 57)/57=(5^(2x). 131)/131`
`=>7^x=5^(2x)`
`=>7^x=25^x`
`=>x=0`
Vậy `x=0`.
`2`.
a) `A=1/101^2+1/102^2+1/103^2+1/104^2+1/105^2`
`=>A=1/101.101+1/102.102+1/103.103+1/104.104+1/105.105`
`=>A<1/100.101+1/101.102+1/102.103+1/103.104+1/104.105`
`=>A<1/100-1/101+1/101-1/102+1/102-1/103+1/103-1/104+1/104-1/105`
`=>A<1/100-1/105`
`=>A<1/2100=1/(2^2 .3.5^2 .7)`
`=>A<1/(2^2. 3.5^2. 7`
Vậy biểu thức có giá trị bé hơn `1/(2^2. 3.5^2. 7`.
b) `3^(x+1)+3^(x+2)+3^(x+3)+...+3^(x+100)`
`=3^x.(3+3^2+3^3+...+3^100)`
`=3^x.[(3+3^2+3^3+3^4)+...+(3^97+3^98+3^99+3^100)]`
`=3^x.[(3+3^2+3^3+3^4)+...+3^96.(3+3^2+3^3+3^4)]`
`=3^x.(3+3^2+3^3+3^4).(1+...+3^96)`
`=3^x. 120.(1+...+3^96)vdots20`
`=>`Biểu thức chia hết cho `20`.