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Ta xét tử số 3+5+...+99+101
=[(101-3):2+1].[101+3):2]
=50.52
=2600 (1)
Ta xét mẫu số13.47+13.53
=13.(47+53)
=13.100
=1300 (2)
Thay (1) vả (2) vào biểu thức A ta được ;2600/1300=2
\(A=\frac{3+5+...+99+101}{13.47+13.53}\)
\(A=\left(3+5+...+99+101\right)\div\left(13.47+13.53\right)\)
\(A=\left\{101+3\cdot\left[\left(101-3\right)\cdot2+1\right]\div2\right\}\div\left[13.\left(47+53\right)\right]\)
\(A=2548\div1300\)
\(A=\frac{49}{25}\)
\(A=\dfrac{101\cdot\dfrac{102}{2}}{\left(101-100\right)+99-98+...+3-2+1}\)
\(=\dfrac{101\cdot51}{1+1+...+1}=\dfrac{101\cdot51}{51}=101\)
\(B=\dfrac{37\cdot43\left(101-101\right)}{2+4+...+100}=0\)
a, \(A=\dfrac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)
Ta có: \(T=101+100+99+98+...+3+2+1\) \(=\dfrac{\left(101+1\right).101}{2}\)
\(=\dfrac{102.101}{2}\Leftrightarrow51.101\)
\(M=101-100+99-98+...+3-2+1\)
Ta có: \(101:2=50\) (dư \(1\))
\(\Rightarrow M=\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1\)
Có \(50\) dấu ngoặc tròn "\(\left(\right)\)"
\(\Rightarrow M=1+1+...+1+1=51.1=51\)
\(M\) có \(51\) số \(1\)
\(\Rightarrow A=\dfrac{T}{M}=\dfrac{51.101}{51}=101\)
Vậy \(A=101\)
b, \(B=\dfrac{3737.43-4343.37}{2+4+6+...100}\)
Ta có: \(T=3737.43-4343.37\)
\(T=37.101.43-43.101.37\)
\(T=0\)
\(\Rightarrow\) \(B=\dfrac{T}{2+4+6+...+100}=\dfrac{0}{2+4+6+...+100}\) \(=0\)
Vậy \(B=0\)
\(A=\dfrac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}.\)
\(A=\dfrac{\left[\dfrac{\left(101-1\right)}{1}+1\right]\left[\dfrac{101+1}{2}\right]}{\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1}.\)
\(A=\dfrac{101.51}{1+1+1+...+1+1}\) (có 51 số 1).
\(A=\dfrac{5151}{51}=101.\)
Vậy \(A=101.\)
Ta có:
A = \(\dfrac{101+100+99+98+...+1}{101-100+99-98+...+3-2+1}\)
= \(\dfrac{101+\left(100+1\right)+\left(99+2\right)+...+\left(51+50\right)}{\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1}\)
= \(\dfrac{101+101+101+...+101}{1+1+1+...+1}\) (51 số 101 và 51 số 1)
= \(\dfrac{101.51}{51}\)
= 101
Vậy A = 101
Gọi \(101+100+99+98+...+3+2+1\) là \(A\)
Gọi \(101-100+99-98+...+3-2+1\) là \(B\)
Ta có:
\(A=1+2+3+...+98+99+100+101\\ =\dfrac{101\cdot\left(101+1\right)}{2}\\ =\dfrac{101\cdot102}{2}\\ =5151\)
\(B=101-100+99-98+...+3-2+1\\ =\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1\\ =1+1+...+1+1\)
(có 51 số hạng 1) \(=51\cdot1\\ =51\) \(\dfrac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}=\dfrac{A}{B}=\dfrac{5151}{51}=101\)Ta có: A=1.2.3.....99.100.(\(1+\dfrac{1}{2}+\dfrac{1}{3}+......+\dfrac{1}{99}+\dfrac{1}{100}\))
\(=1.2.3...100\left[\left(1+\dfrac{1}{100}\right)+\left(\dfrac{1}{2}+\dfrac{1}{99}\right)+......+\left(\dfrac{1}{50}+\dfrac{1}{51}\right)\right]\)
=>A= 1.2...100.\(\left[\dfrac{101}{100}+\dfrac{101}{2.99}+......+\dfrac{101}{50.51}\right]\)
=1.2.....100.101\(\left[\dfrac{1}{100}+\dfrac{1}{2.99}+.....+\dfrac{1}{50.51}\right]⋮101\)
Vậy A chia hết cho 101
a) \(A=\dfrac{3737.43-4343.37}{2+4+6+...+2018}\)
\(\Leftrightarrow A=\dfrac{3737.43-43.101.37}{2+4+6+...+2018}\)
\(\Leftrightarrow A=\dfrac{3737.43-43.3737}{2+4+6+...+2018}\)
\(\Leftrightarrow A=\dfrac{0}{2+4+6+...+2018}\)
\(\Leftrightarrow A=0\)
b) \(B=\dfrac{101+100+99+...+3+2+1}{101-100+99-98+...+3-2+1}\)
\(\Leftrightarrow B=\dfrac{\left(101+1\right).101:2}{\left(101+99+97+...+1\right)-\left(100+98+96+...+2\right)}\)
\(\Leftrightarrow B=\dfrac{5151}{\left[\left(101+1\right).51:2\right]-\left[\left(100+2\right).50:2\right]}\)
\(\Leftrightarrow B=\dfrac{5151}{2601-2550}\)
\(\Leftrightarrow B=\dfrac{5151}{51}\)
\(\Leftrightarrow B=101\)
cái này bấm máy vina là ra em à
shift log
=>381.3