2² + 3²=4 + 9 = 13

= 4 + 8

=12

$\frac{18}{117}\times\frac{12}{113}+\frac{12}{113}\times\frac{8}{117}+\frac{26}{117}+\frac{101}{113}$

$=\frac{12}{113}\times\left(\frac{18}{117}+\frac{8}{117}\right)+\frac{26}{117}+\frac{101}{113}$

$=\frac{12}{113}\times\frac{26}{117}+\frac{26}{117}+\frac{101}{113}$

$=\frac{26}{117}\times\left(\frac{12}{113}+1\right)+\frac{101}{113}$

$=\frac{26}{117}\times \frac{125}{113}+\frac{101}{113}$

$=\frac{250}{1017}+\frac{101}{113}=\frac{1159}{1017}$

\(\dfrac{18}{117}\times\dfrac{12}{113}+\dfrac{12}{113}\times\dfrac{8}{177}+\dfrac{26}{117}\times\dfrac{101}{113}\)

\(=\left(\dfrac{18}{177}+\dfrac{8}{177}\right)\times\dfrac{12}{113}+\dfrac{26}{117}\times\dfrac{101}{113}\)

\(=\dfrac{26}{177}\times\dfrac{12}{113}+\dfrac{26}{117}\times\dfrac{101}{113}\)

\(=\dfrac{26}{177}\times\left(\dfrac{12}{113}+\dfrac{101}{113}\right)\)

\(=\dfrac{26}{177}\times1\)

\(=\dfrac{26}{117}\)

Đặt $A=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}$

Ta thấy:

$\frac{1}{2!}=\frac{1}{1.2};\frac{1}{3!}=\frac{1}{\mathrm{1.2.3}}<\frac{1}{2.3};...;\frac{1}{100!}=\frac{1}{\mathrm{1.2...100}}<\frac{1}{99.100}$

Cộng vế với vế ta được:

$A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}$

$\Rightarrow A<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}$

$\Rightarrow A<1-\frac{1}{100}<1$

Vậy $\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}<1$ (Đpcm)

lỗi phân số dấu __ nha 

A = 5¹⁰⁰ - 5⁹⁹ + 5⁹⁸ - 5⁹⁷ + ... + 5² - 5

5A = 5¹⁰¹ - 5¹⁰⁰ + 5⁹⁹ - 5⁹⁸ + ... + 5³ - 5²

5A + A = 5¹⁰¹ - 5

6A = 5¹⁰¹ - 5

A = \(\dfrac{5^{101}-5}{6}\)

\(\text{Đặt }A=5^{100}-5^{99}+5^{98}-5^{97}+...+5^2-5\\5A=5^{101}-5^{100}+5^{99}-5^{98}+...+5^3-5^2\\5A+A=(5^{101}-5^{100}+5^{99}-5^{98}+...+5^3-5^2)+(5^{100}-5^{99}+5^{98}-5^{97}+...+5^2-5)\\\\6A=5^{101}-5\\\Rightarrow A=\frac{5^{101}-5}{6}\)

\(7B=7^2+7^3+...+7^{100}\)

\(7B-B=7^2+7^3+...+7^{100}-\left(7+7^2+...+7^{99}\right)=7^{100}-7\)

\(\Rightarrow B=\dfrac{7^{100}-7}{6}\)

\(B=7+7^2+7^3+...+7^{99}\\7B=7^2+7^3+7^4+...+7^{100}\\7B-B=(7^2+7^3+7^4+...+7^{100})-(7+7^2+7^3+...+7^{99})\\6B=7^{100}-7\\\Rightarrow B=\frac{7^{100}-7}{6}\)

Đề chưa được rõ bạn ơi!

​

D phải bằng một biểu thức nào nữa chứ em ha!

\(3A=3+3^2+3^3+...+3^{2025}\)

\(3A-A=3+3^2+3^3+...+3^{2025}-\left(1+3+3^2+...+3^{2024}\right)=-1+3^{2025}\)

\(A=\dfrac{-1+3^{2025}}{2}\)

\(A=1+3+3^2+...+3^{2024}\\ 3A=3+3^2+3^3+...+3^{2025}\\ 3A-A=\left(3+3^2+3^3+...+3^{2025}\right)-\left(1+3+3^2+...+3^{2024}\right)\\ 2A=3^{2025}-1\\ A=\dfrac{3^{2025}-1}{2}\)

`#3107.101107`

\(\left(3^2\right)^4\div27\\ =3^{2\cdot4}\div3^3\\ =3^8\div3^3\\ =3^{8-3}=3^5\)

\(\dfrac{\left(3^2\right)^4}{27}=\dfrac{3^{2\cdot4}}{27}=\dfrac{3^8}{3^3}=3^{8-3}=3^5\)

a, \(\dfrac{1}{2}< \dfrac{12}{a}< \dfrac{4}{3}\Leftrightarrow\dfrac{1}{24}< \dfrac{1}{a}< \dfrac{1}{9}\Leftrightarrow9< a< 24\)

b, \(\dfrac{14}{5}< \dfrac{a}{5}< 4\Leftrightarrow14< a< 20\)

a) \(\dfrac{1}{2}< \dfrac{12}{a}< \dfrac{4}{3}\)

\(6\cdot\dfrac{1}{2}< 6\cdot\dfrac{12}{a}< 6\cdot\dfrac{4}{3}\)

\(3< \dfrac{72}{a}< 8\)

\(\dfrac{72}{3}>a>\dfrac{72}{8}\)

\(24>a>9\) 

Vậy: ... 

b) \(\dfrac{14}{5}< \dfrac{a}{5}< 4\)

\(\dfrac{14}{5}\times5< a< 5\times4\)

\(14< a< 20\)