`H=1+2.6+3.6^2+4.6^3+...+100.6^99`

`6H = 6+2.6^2+3.6^3+4.6^4+...+100.6^100`

`6H - H = (6+2.6^2+3.6^3+4.6^4+...+100.6^100)-(1+2.6+3.6^2+4.6^3+...+100.6^99)`

`5H = (6 - 2.6) + (2.6^2 - 3.6^2) + (3.6^3  - 4.6^3) + ... + (99. 6^99 - 100.6^99) + 100.6^100 - 1`

`5H = 100.6^100 - 1 + (-6) + (-6^2) + (-6^3) + ... + (-6^99)`

`5H = 100.6^100 - 1 - (6+6^2+6^3 + ... + 6^99)`

Đặt `S = 6+6^2+6^3 + ... + 6^99`

`6S = 6^2+6^3+6^4 + ... + 6^100`

`6S - S = (6^2+6^3+6^4 + ... + 6^100) - ( 6+6^2+6^3 + ... + 6^99)`

`5S = 6^100 - 6`

`S = ( 6^100 - 6)/5`

Khi đó: `5H = 100.6^100 - 1 - S`

`5H = 100.6^100 - 1 - ( 6^100 - 6)/5`

`5H = (500.6^100)/5 - 5/5 - ( 6^100 - 6)/5`

`5H =  (500.6^100 - 5 - 6^100 + 6)/5`

`H = (499 . 6^100 + 1)/5`

Vậy ...

Ta có: \(\widehat{xOt}+\widehat{xOy}=180^0\)(hai góc kề bù)

=>\(\widehat{xOt}=180^0-70^0=110^0\)

Ta có: \(\widehat{xOt}=\widehat{yOz}\)(hai góc đối đỉnh)

mà \(\widehat{xOt}=110^0\)

nên \(\widehat{yOz}=110^0\)

Ta có: \(\widehat{yOz}+\widehat{zOt}=180^0\)(hai góc kề bù)

=>\(\widehat{zOt}=180^0-110^0=70^0\)

Do `Oz` đối tia `Ox`

=> \(\widehat{xOz}=180^o\)

Mà \(\widehat{xOy}+\widehat{yOz}=\widehat{xOz}\) (Vì `Oy` nằm giữa `Oz` và `Ox`)

=> \(70^o+\widehat{yOz}=180^o\)

=> \(\widehat{yOz}=180^o-70^o\)

=> \(\widehat{yOz}=110^o\)

Lại có: \(\left\{{}\begin{matrix}\widehat{xOy}=\widehat{tOz}\\\widehat{yOz}=\widehat{xOt}\end{matrix}\right.\) (Các cặp góc đối đỉnh)

=> \(\left\{{}\begin{matrix}\widehat{zOt}=70^o\\\widehat{xOt}=110^o\end{matrix}\right.\)

Vậy ...

\(\left(-\dfrac{3}{5}\right)^2.\dfrac{5}{11}+\dfrac{9}{25}.\left(-\dfrac{16}{11}\right)\)

\(=\dfrac{9}{25}.\dfrac{5}{11}+\dfrac{9}{25}.\left(-\dfrac{16}{11}\right)\)

\(=\dfrac{9}{25}.\left[\dfrac{5}{11}+\left(-\dfrac{16}{11}\right)\right]\)

\(=\dfrac{9}{25}.\left(-1\right)\)

\(=-\dfrac{9}{25}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có: 

\(\dfrac{x+1}{3}=\dfrac{y-2}{4}=\dfrac{\left(x+1\right)-\left(y-2\right)}{3-4}=\dfrac{x+1-y+2}{-1}=\dfrac{x-y+3}{-1}=\dfrac{18}{-1}\)

`= -18`

Suy ra: \(\left\{{}\begin{matrix}\dfrac{x+1}{3}=-18\\\dfrac{y-2}{4}=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1=-54\\y-2=-72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-55\\y=-70\end{matrix}\right.\)

Vậy ....

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x+1}{3}=\dfrac{y-2}{4}=\dfrac{x-y+1+2}{3-4}=\dfrac{15+3}{-1}=-18\)

=>\(\left\{{}\begin{matrix}x+1=-18\cdot3=-54\\y-2=4\cdot\left(-18\right)=-72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-54-1=-55\\y=-72+2=-70\end{matrix}\right.\)

giúp tui pls

Sửa đề; DE//BC

Xét ΔABC có \(\dfrac{AD}{AB}=\dfrac{AE}{AC}\)

nên DE//BC

\(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\dfrac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot\left(-1\right)\cdot3^{14}}=-\dfrac{3}{5}\)

\(\dfrac{\left(-3\right)^{10}.15^5}{25^3.\left(-9\right)^7}\)

\(=\dfrac{\left(-3\right)^{10}.\left(3.5\right)^5}{\left(5^2\right)^3.\left(-3^2\right)^7}\)

\(=\dfrac{\left(-3\right)^{10}.3^5.5^5}{5^6.\left(-3\right)^{14}}\)

\(=\dfrac{1.3^5.1}{5.3^4}\)

\(=\dfrac{3}{5.1}\)

\(=\dfrac{3}{5}\)

\(\dfrac{\left(\dfrac{2}{3}\right)^3\cdot\left(\dfrac{3}{4}\right)^2\cdot\left(-1\right)^5}{\left(\dfrac{2}{5}\right)\cdot\left(-\dfrac{5}{12}\right)^2}=\dfrac{\dfrac{2^3}{3^3}\cdot\dfrac{3^2}{4^2}\cdot\left(-1\right)}{\dfrac{2}{5}\cdot\dfrac{25}{144}}\)

\(=\dfrac{\dfrac{1}{2\cdot3}\cdot\left(-1\right)}{\dfrac{5}{72}}=-\dfrac{1}{6}\cdot\dfrac{72}{5}=-\dfrac{12}{5}\)

a: ta có: \(\widehat{MAB}=\widehat{ABC}\)

mà hai góc này là hai góc ở vị trí so le trong

nên AM//BC

ta có: \(\widehat{CAN}=\widehat{ACB}\)

mà hai góc này là hai góc ở vị trí so le trong

nên AN//BC

Ta có: AM//BC

NA//BC

mà AM,AN có điểm chung là A

nên M,A,N thẳng hàng

b: Vì M,A,N thẳng hàng nên \(\widehat{MAB}+\widehat{BAC}+\widehat{CAN}=180^0\)

=>\(\widehat{ABC}+\widehat{BAC}+\widehat{ACB}=180^0\)