\(A=\dfrac{1}{299}\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)\)

\(299A=1+\dfrac{1}{2}+...+\dfrac{1}{101}-\left(\dfrac{1}{300}+\dfrac{1}{301}+...+\dfrac{1}{400}\right)\)

Thêm bớt \(\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{299}\) ta được:

\(299A=1+\dfrac{1}{2}+...+\dfrac{1}{101}+\left(\dfrac{1}{102}+...+\dfrac{1}{299}\right)-\left(\dfrac{1}{102}+...+\dfrac{1}{299}\right)-\left(\dfrac{1}{300}+...+\dfrac{1}{400}\right)\)

\(299A=\left(1+\dfrac{1}{2}+...+\dfrac{1}{299}\right)-\left(\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{400}\right)\)

\(101B=1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+\dfrac{1}{3}-\dfrac{1}{104}+....+\dfrac{1}{299}-\dfrac{1}{400}\)

\(101B=\left(1+\dfrac{1}{2}+...+\dfrac{1}{299}\right)-\left(\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{400}\right)\)

\(\Rightarrow299A=101B\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{101}{299}\)

aiđúng mik tick trong bây giờ

\(7\cdot4^{x-1}+4^{x+1}=23\)

=>\(7\cdot4^x\cdot\dfrac{1}{4}+4^x\cdot4=23\)

=>\(4^x\left(\dfrac{7}{4}+4\right)=23\)

=>\(4^x=23:\dfrac{23}{4}=4\)

=>x=1

9 lần số ban đầu là:

332-8=324

Số ban đầu là 324:9=36

a: \(\widehat{MON}+\widehat{O_1}+45^0=180^0\)

=>\(\widehat{O_1}=180^0-90^0-45^0=45^0\)

Ta có: \(\widehat{O_1}=\widehat{MNO}\left(=45^0\right)\)

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

nên OB//AM

b: Ta có: OB//AM

MA\(\perp\)AB

Do đó: OB\(\perp\)BA

\(119\times24-53\times23-24\times66\)

\(=\left(119-66\right)\times24-53\times23\)

\(=53\times24-53\times23\)

\(=53\times\left(24-23\right)\)

\(=53\times1\)

\(=53\)

\(119\cdot24-53\cdot23-24\cdot66\)

\(=24\left(119-66\right)-53\cdot23\)

\(=53\cdot24-53\cdot23=53\)

Chọn B

B

\(n^2+n-7=\left(n^2-2n\right)+\left(3n-6\right)-1\\ =n\left(n-2\right)+3\left(n-2\right)-1\\ =\left(n-2\right)\left(n+3\right)-1\)

Để: \(\left(n^2+n-7\right)⋮\left(n-2\right)\Rightarrow\left[\left(n-2\right)\left(n+3\right)-1\right]⋮\left(n-2\right)\\ \Rightarrow1⋮\left(n-2\right)\) (Vì: \(\left(n-2\right)\left(n+3\right)⋮\left(n-2\right)\forall n\inℤ\) )

\(\Rightarrow n-2\in\left\{1;-1\right\}\Rightarrow n\in\left\{3;1\right\}\)

Ta có:

\(n^2+n-7\\ =\left(n^2-2n\right)+\left(3n-6\right)-1\\ =n\left(n-2\right)+3\left(n-2\right)-1\\ =\left(n-2\right)\left(n+3\right)-1\)

Để `n^2+n-7` chia hết cho n - 2 thì: 

1 ⋮ n - 2

=> n - 2 ∈ Ư(1) = {1; -1}

=> n ∈ {3; 1} 

\(a,14364:9+20020\)

\(=1596+20020\)

\(=21616\)

\(b,\left(10356\times5-780\right):6\)

\(=\left(51780-780\right):6\)

\(=51000:6\)

\(=8500\)

`c`

\(=51000:6\)

Vậy `9276 : 39 = 237`(dư`33`)