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\(A=1+3+3^2+...+3^{2016}\)

\(3A=3.\left(1+3+3^2+...+3^{2016}\right)\)

\(3A=3+3^2+3^3+...+3^{2017}\)

\(3A-A=\left(3+3^2+3^3+...+3^{2017}\right)-\left(1+3+3^2+...+3^{2016}\right)\)

\(2A=3^{2017}-1\)

\(A=\left(3^{2017}-1\right):2\)

\(B=1+6+6^2+...+6^{200}\)

\(6B=6.\left(1+6+6^2+...+6^{200}\right)\)

\(6B=6+6^2+6^3+...+6^{201}\)

\(6B-B=\left(6+6^2+6^3+...+3^{201}\right)-\left(1+6+6^2+...+6^{200}\right)\)

\(5B=6^{201}-1\)

\(B=\left(6^{201}-1\right):5\)

\(3^{x-2}.4=324\)

\(3^{x-2}=324:4\)

\(3^{x-2}=81\)

\(3^{x-2}=3^4\)

\(x-2=4\)

\(x=4+2\)

\(x=6\)

\(2x< 20\)

\(\Rightarrow x=\left\{0;1;2;3;4;5;6;7;8;9\right\}\)

Ta có :

\(10A=\dfrac{10\left(10^{1990}+1\right)}{10^{1991}+1}=\dfrac{10^{1991}+10}{10^{1991}+1}=\dfrac{10^{1991}+1+9}{10^{1991}+1}=1+\dfrac{9}{10^{1991}+1}\left(1\right)\)

\(10B=\dfrac{10\left(10^{1991}+1\right)}{10^{1992}+1}=\dfrac{10^{1992}+10}{10^{1992}+1}=\dfrac{10^{1992}+1+9}{10^{1992}+1}=1+\dfrac{9}{10^{1992}+1}\left(2\right)\)

Lại có : \(1+\dfrac{9}{10^{1991}+1}>1+\dfrac{9}{10^{1992}+1}\)

\(\Leftrightarrow10A>10B\Leftrightarrow A>B\)

Vậy...

Sửa đề:

So sánh:

\(A=\dfrac{10^{2015}+1}{10^{2016}+1}\) và \(B=\dfrac{10^{2016}+1}{10^{2017}+1}\)

Giải:

Ta thấy: \(\left\{{}\begin{matrix}A=\dfrac{10^{2015}+1}{10^{2016}+1}< 1\\B=\dfrac{10^{2016}+1}{10^{2017}+1}< 1\end{matrix}\right.\)

\(\Rightarrow\) Áp dụng tính chất \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\) ta có:

\(B=\dfrac{10^{2016}+1}{10^{2017}+1}< \dfrac{10^{2016}+1+9}{10^{2017}+1+9}=\dfrac{10^{2016}+10}{10^{2017}+10}\)

\(=\dfrac{10\left(10^{2015}+1\right)}{10\left(10^{2016}+1\right)}=\dfrac{10^{2015}+1}{10^{2016}+1}\)

\(\Rightarrow\dfrac{10^{2016}+1}{10^{2017}+1}< \dfrac{10^{2015}+1}{10^{2016}+1}\)

Vậy \(B< A\)

Hay \(A>B\)

0.01956364

số 73 mũ 75 lớn hơn nha kết bạn nhé

\(16750< 73^{75}\)

Theo mk là như vậy còn cách làm thì mk ko biết

thông cảm

#sakurasyaoran#