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3: \(C=\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{2014\cdot2016}\right)\)
\(=\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{2015^2-1}\right)\)
\(=\dfrac{2^2-1+1}{2^2-1}\cdot\dfrac{3^2-1+1}{3^2-1}\cdot...\cdot\dfrac{2015^2-1+1}{2015^2-1}\)
\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2015^2}{\left(2015-1\right)\left(2015+1\right)}\)
\(=\dfrac{2\cdot3\cdot...\cdot2015}{1\cdot2\cdot...\cdot2014}\cdot\dfrac{2\cdot3\cdot...\cdot2015}{3\cdot4\cdot...\cdot2016}\)
\(=\dfrac{2015}{1}\cdot\dfrac{2}{2016}=\dfrac{2015}{1008}\)
1: \(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\right)\)
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
...
\(\dfrac{1}{50^2}< \dfrac{1}{49\cdot50}=\dfrac{1}{49}-\dfrac{1}{50}\)
Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1-\dfrac{1}{50}\)
=>\(1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}< 2-\dfrac{1}{50}\)
=>\(A=\dfrac{1}{4}\left(1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{4}\left(2-\dfrac{1}{50}\right)\)
=>\(A< \dfrac{1}{2}-\dfrac{1}{200}\)
=>A<1/2
a: \(A=\dfrac{1}{2}:\dfrac{1}{20}-\dfrac{3}{5}\)
\(=\dfrac{1}{2}\cdot\dfrac{20}{1}-\dfrac{3}{5}\)
\(=10-\dfrac{3}{5}=\dfrac{47}{5}\)
b: \(B=1,2+2,5\cdot2-4,8\)
\(=1,2+5-4,8\)
=5-3,6=1,4
c: \(C=\dfrac{3}{13}\cdot\dfrac{9}{4}+\dfrac{3}{4}\cdot\dfrac{4}{-13}-\dfrac{3}{13}\cdot\dfrac{1}{4}+\dfrac{10}{13}\)
\(=\dfrac{3}{13}\left(\dfrac{9}{4}-\dfrac{4}{4}-\dfrac{1}{4}\right)+\dfrac{10}{13}\)
\(=\dfrac{3}{13}\cdot\dfrac{4}{4}+\dfrac{10}{13}=\dfrac{3}{13}+\dfrac{10}{13}=1\)
d: \(D=\left(3\dfrac{1}{3}+5\dfrac{2}{3}\right)\cdot\left(-2014^{12}\right)^0-30\%\cdot2,5\)
\(=\left(3+\dfrac{1}{3}+5+\dfrac{2}{3}\right)-0,3\cdot2,5\)
\(=9-0,75=8,25\)
Sửa đề: 7,15:0,5+7,15x9-0,715:0,1
=7,15x2+7,15x9-7,15
=7,15x(2+9-1)
=7,15x10=71,5
7,15 : 0,5 + 7,15 x 9- 0,715:0,1
= 7,15 x 2 + 7,15 x 9 - 7,15
= 7,15 x ( 2 + 9 - 1 )
= 7,15 x 10
= 71,5
Nhận thấy:\(\dfrac{1}{2^2}< \dfrac{1}{1.2},\dfrac{1}{3^2}< \dfrac{1}{2.3},\dfrac{1}{4^2}< \dfrac{1}{3.4},...,\dfrac{1}{99^2}< \dfrac{1}{98.99},\dfrac{1}{100^2}< \dfrac{1}{99.100}\\ \)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\\ \Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ \Rightarrow A< 1-\dfrac{1}{100}\\ \Rightarrow A< \dfrac{99}{100}\)
Bạn xem lại xem có nhầm đề không nhỉ?
\(x-\dfrac{5}{3}.x=\dfrac{22}{9}\)
\(=>x.\left(1-\dfrac{5}{3}\right)=\dfrac{22}{9}\)
\(=>x.\left(\dfrac{3}{3}-\dfrac{5}{3}\right)=\dfrac{22}{9}\)
\(=>x.\left(-\dfrac{2}{3}\right)=\dfrac{22}{9}\)
\(=>x=\dfrac{22}{9}:\left(\dfrac{-2}{3}\right)=\dfrac{22}{9}\times\left(\dfrac{-3}{2}\right)\)
\(=>x=\dfrac{-11}{3}\)
\(#NqHahh\)
mk trả lời đc mỗi câu 3thui
Câu 3: số chẵn bé nhất có 4 c/s là 1000
số chẵn lớn nhất có 4 c/s là 9998
Có tất cả: (9998 - 1000) : 2 + 1 = 5000 ( số chẵn có 4 c/s )
Đ/s : 5000 số
a: \(B=\dfrac{4x+1}{2x+3}=\dfrac{4x+6-7}{2x+3}=2-\dfrac{7}{2x+3}\)
Để B min thì \(\dfrac{7}{2x+3}\) max
=>2x+3=1
=>2x=-2
=>x=-1
b: Để B max thì 2x+3=-1
=>2x=-4
=>x=-2