Đặt \(A=1+5^2+5^4+...+5^{40}\)

\(\Rightarrow25A=5^2+5^4+5^6+...+5^{42}\)

Lấy \(25A-A=\left(5^2+5^4+5^6+...+5^{42}\right)-\left(1+5^2+5^4+...+5^{40}\right)\)

\(\Rightarrow24A=5^{42}-1\)

\(\Rightarrow A=\dfrac{5^{42}-1}{24}\)

nguyễn thị hương giang, cúm ơn rất nhìu !!

\(a,\left(y^{54}\right)^2=y\)\(\Rightarrow y^{108}=y\)\(\Rightarrow y=\pm1\)

\(b,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)

\(\Rightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)

\(\Rightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)

\(\Rightarrow x\left(x-1\right)^{x+2}\left(x-2\right)=0\)

\(\Rightarrow x\in\left\{0;1;2\right\}\)

\(c,x\left(6-x\right)^{2019}=\left(6-x\right)^{2019}\)

\(\Rightarrow\left(6-x\right)^{2019}\left(x-1\right)=0\)

\(\Rightarrow x\in\left\{1;6\right\}\)

\(\left(y^{54}\right)^2=y\)

\(\Rightarrow y^{108}=y\)

\(\Rightarrow y^{108}-y=0\)

\(\Rightarrow y\cdot\left(y^{107}-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}y=0\\y^{107}-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}y=0\\y^{107}=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}y=0\\y=1\end{cases}}\)

a) \(\left(y^{54}\right)^2=y\)

\(\Leftrightarrow y^{108}=y\)

\(\Leftrightarrow y^{108}-y=0\)

\(\Leftrightarrow y.\left(y^{107}-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=0\\y^{107}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}y=0\\y=1\end{cases}}}\)

Vậy \(y\in\left\{0;1\right\}\)

\(1+5^2+5^4+...+5^{2x}\left(1\right)=\dfrac{25^6-1}{24}\)

Đặt \(\left(1\right)=A\)

\(\Rightarrow A=1+5^2+...+5^{2x}\)

\(\Rightarrow5^2A=5^2+5^4+...+5^{2x+2}\)

\(\Rightarrow25A=5^2+5^4+...+5^{2x+2}\)

\(\Rightarrow25A-A=5^2+5^4+...+5^{2x+2}-1-5^2-...-5^{2x}\)

\(\Rightarrow24A=5^{2x+2}-1\)

\(\Rightarrow A=\dfrac{5^{2x+2}-1}{24}\)

Mà: \(A=\dfrac{25^6-1}{24}\)

\(\Rightarrow\dfrac{5^{2x+2}-1}{24}=\dfrac{\left(5^2\right)^6-1}{24}\)

\(\Rightarrow5^{2x+2}-1=5^{12}-1\)

\(\Rightarrow5^{2x+2}=5^{12}\)

\(\Rightarrow2x+2=12\)

\(\Rightarrow2x=10\)

\(\Rightarrow x=\dfrac{10}{2}\)

\(\Rightarrow x=5\)

    57 \(\times\) 46 + 57 \(\times\) 54

= 57    \(\times\) ( 46 + 54)

= 57 \(\times\) 100

= 5700

137 + 355 + 63 + 45

= ( 137 + 63) + ( 355 + 45)

= 200 + 400

= 600

215 \(\times\) 48 - 215 \(\times\) 38

= 215 \(\times\) ( 48 - 38)

= 215 \(\times\) 10

= 2150 

4 \(\times\) 50 \(\times\) 25 

= ( 4 \(\times\) 25) \(\times\) 50

= 100 \(\times\) 50

= 5 000

57×(46+54) =57×100=5700

137+355+63+45=(137+63) +(355+45) =200+400=600

215×48-215×38=215×(48-38) =215×10=2150

4×50×25=(4×25) ×50=100×50=5000

    57 $\times$ 46 + 57 $\times$ 54

= 57    $\times$ ( 46 + 54)

= 57 $\times$ 100

= 5700

137 + 355 + 63 + 45

= ( 137 + 63) + ( 355 + 45)

= 200 + 400

= 600

215 $\times$ 48 - 215 $\times$ 38

= 215 $\times$ ( 48 - 38)

= 215 $\times$ 10

= 2150 

4 $\times$ 50 $\times$ 25 

= ( 4 $\times$ 25) $\times$ 50

= 100 $\times$ 50

= 5 000

Dễ thấy tổng trên có: 200-101+1=100 (số hạng) và 1/101 là số hạng lớn nhất  (vì\(\frac{1}{101}>\frac{1}{102}>\frac{1}{103}>....>\frac{1}{200}\))

=>\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+....+\frac{1}{200}<\frac{1}{101}.100=\frac{100.}{101}<1\)

=>\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+....+\frac{1}{200}<1\)

1.

x-3-3x-2=-15

-2x-5=-15

-2x=-15+5

-2x=-10

x=-10/(-2)

x=5

\(\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\times\left(1-\dfrac{1}{6}\right)\times\dots\times\left(1-\dfrac{1}{99}\right)\times\left(1-\dfrac{1}{100}\right)\) (sửa đề)

\(=\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\times\dfrac{5}{6}\times\dots\times\dfrac{98}{99}\times\dfrac{99}{100}\)

\(=\dfrac{2\times3\times4\times5\times\dots\times98\times99}{3\times4\times5\times6\times\dots\times99\times100}\)

\(=\dfrac{2}{100}\)

\(=\dfrac{1}{50}\)

Ta có M 2 ^    −    N 0 ^    =    35 °  (đề bài)      (1)

Lại có a // b nên M 2 ^    +    N 2 ^    =    180 °  (2) (hai góc trong cùng phía)

Từ (1) và (2)   ⇒ 2 M 2 ^   =   215 ° ⇒ M 2 ^   =   107.5 ° .

Từ (1) có  N 1 ^ = 107.5 ° − 35 ° = 72.5 ° .

Do a // b nên : N 2 ^ = M 2 ^ = 107.5 ° (hai góc so le trong).

N 1 ^ = M 2 ^ = 72.5 ° (hai góc so le trong)