ta có :

`125^7 = (5^3)^7 = 5^(3.7) = 5^21`

vì `5 = 5 ; 20<21`

Vậy `5^20 < 5^21 <=> 5^20 < 125^7`

`=>` chọn đáp án `2`

sai r `2^40 : 2^20 = 2^(40-20) = 2^20`

`C = 5^200 + 2^199 + 5^198 + ... + 5^2 + 5^1`

`=> C = 5^1 + 5^2 + ... + 5^198 + 5^199 + 5^200`

`=> 5C = 5^2 + 5^3 + ... + 5^199 + 5^200 + 5^201`

`=> 5C - C = (5^2 + 5^3 + ... + 5^199 + 5^200 + 5^201) - (5^1 + 5^2 + ... + 5^198 + 5^199 + 5^200)`

`=> 4C = 5^201 - 5`

`=> C    = (5^201 - 5)/4`

`4/3 xx x = 1/1`

`=>      x   = 1 : 4/3`

`=>     x   = 3/4`

`=1/(1xx2) - 1/(2xx3) + 1/(2xx3) - 1/(3xx4) + ... + 1/(218 xx229) - 1/(219xx220)`

`=1/(1xx2) - 1/(219 xx 220)`

`=1/2 - 1/48180`

`=48178/96360`

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áp dụng :

`a/(bxxcxxz) = 1/(bxxc) - 1/(cxxz)  khi `(z - b = a)`

`a)5xx x+3xx x=1096`

`8xx x = 1096`

`x=1096:8`

`x=137`

`__________________________`

`b)189:x-37:x=8`

`152:x=8`

`x=152:8`

`x=19`

`a) 1 1/5 + 4/5 : x = 0,75`

`=> 6/5 + 4/5 : x    = 3/4`

`=>           4/5 : x    = 3/4 - 6/5`

`=>            4/5 : x   = -9/20`

`=>                    x    = 4/5 : -9/20`

`=>                     x    = -16/9`

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`b) 3(5x + 2/3)^2 = 4/3`

`=>  (5x + 2/3)^2  = 4/3 : 3`

`=>   (5x + 2/3)^2  = 4/9`

`=>`\(\left\{{}\begin{matrix}\left(5x+\dfrac{2}{3}\right)^2=\left(\dfrac{2}{3}\right)^2\\\left(5x+\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^2\end{matrix}\right.\)

`=>`\(\left\{{}\begin{matrix}5x+\dfrac{2}{3}=\dfrac{2}{3}\\5x+\dfrac{2}{3}=-\dfrac{2}{3}\end{matrix}\right.\)

`=>`\(\left\{{}\begin{matrix}5x=\dfrac{2}{3}-\dfrac{2}{3}\\5x=-\dfrac{2}{3}-\dfrac{2}{3}\end{matrix}\right.\)

`=>`\(\left\{{}\begin{matrix}5x=0\\5x=-\dfrac{4}{3}\end{matrix}\right.\)

`=>`\(\left\{{}\begin{matrix}x=0:5\\x=-\dfrac{4}{3}:5\end{matrix}\right.\)

`=>`\(\left\{{}\begin{matrix}x=0\\x=-\dfrac{4}{15}\end{matrix}\right.\)