\(M=\dfrac{202}{7\cdot1010}+\dfrac{202}{10\cdot1313}+\dfrac{202}{13\cdot1616}+...+\dfrac{202}{91\cdot9494}\\ =\dfrac{202}{7\cdot10\cdot101}+\dfrac{202}{10\cdot13\cdot101}+\dfrac{202}{13\cdot16\cdot101}+...+\dfrac{202}{91\cdot94\cdot101}\\ =\dfrac{202}{101}\cdot\left(\dfrac{1}{7\cdot10}+\dfrac{1}{10\cdot13}+\dfrac{1}{13\cdot16}+...+\dfrac{1}{91\cdot94}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+...+\dfrac{3}{91\cdot94}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{91}-\dfrac{1}{94}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{7}-\dfrac{1}{94}\right)\\ =\dfrac{2}{3}\cdot\dfrac{87}{658}\\ =\dfrac{29}{329}\)

Chứng minh cái gì? chắc là so sánh

\(7^{714}< 8^{714}\)

\(2^{1999}>2^{1998}=\left(2^3\right)^{666}=8^{666}\)

\(\Rightarrow2^{1999}>8^{666}>8^{714}>7^{714}\)

a: \(11^2=11\cdot11=121;111^2=12321\)

b: Dự đoán: \(1111^2=1234321\)

Tính: \(1111^2=1111\cdot1111=1234321\)

trình bày và giải thích giups em đk ạ

\(\overline{abc0cd}\\ =a\cdot100000+b\cdot10000+c\cdot1000+0\cdot100+c\cdot10+d\\ =a\cdot10^5+b\cdot10^4+c\cdot10^3+0\cdot10^2+c\cdot10+d\)

\(\overline{abc0cd}=a\cdot10^5+b\cdot10^4+c\cdot10^3+0\cdot10^2+c\cdot10+d\)

\(=a\cdot10^5+b\cdot10^4+c\left(10^3+10\right)+d\)

\(5n+12⋮n-5\)

=>\(5n-25+37⋮n-5\)

=>\(37⋮n-5\)

=>\(n-5\in\left\{1;-1;37;-37\right\}\)

=>\(n\in\left\{6;4;42;-32\right\}\)

mà n>=0

nên \(n\in\left\{6;4;42\right\}\)

`5n + 12 ⋮ n - 5`

`=> 5n - 25 + 37  ⋮ n - 5`

`=> 5(n - 5) + 37  ⋮ n - 5`

Do `n+5  ⋮ n - 5 `

`=> 5(n+5)  ⋮ n - 5`

`=> 37  ⋮ n - 5`

`=> n - 5` thuộc `Ư(37) =` {`-37;-1;1;37`}

`=> n` thuộc {`-32;4;6;42`}

Mà `n` thuộc `N`

`=> n` thuộc {`4;6;42`}

Vậy ...

4x+x-15=55

=>5x=55+15=70

=>\(x=\dfrac{70}{5}=14\)

= 4.x+x=55-15

=4.x+x=40

=x(4+1)=40

=x.5=40

=x=40:5

=x=8

Chúc bạn học tốt ^^

8 : 24 + 26 = \(\dfrac{1}{3}\) + 26 = \(\dfrac{1}{3}\) + \(\dfrac{78}{3}\) = \(\dfrac{79}{3}\)

\(C=2+2^2+2^3+...+2^{100}\\ 2C=2^2+2^3+2^4+...+2^{101}\\ 2C-C=\left(2^2+2^3+2^4+...+2^{101}\right)-\left(2+2^2+2^3+...+2^{100}\right)\\ C=2^{101}-2\)

C = \(2+2^2+...+2^{100}\)

2C =\(2^2+2^3+...+2^{101}\)

2C - C = \(\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)

C = \(2^{101}-2\)