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B \(=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+...+\frac{50^2-1}{50^2}\)
\(=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
mà \(0
Đặt A =\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)
\(=99-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)\)
Đặt B = \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)
>\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{2}-\frac{1}{101}=\frac{99}{202}\)
Khi đó A = \(99-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)< 99-\frac{99}{202}\approx98,5\)
=> A < 98,5 (1)
Lại có B = \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
Khi đó A =\(99-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)>99-\frac{99}{100}=98,01\)
=> A > 98,01 (2)
Từ (1)(2) => 98,01 < A < 98,5
=> A không là số nguyên
S=1-1/4+1-1/9+...+1-1/x2
S=(1+1+1+...+1)-(1/4+1/9+...+1/x2)
Có (1/4+1/9+...+1/x2)<1/(1.2)+1/(2.3)+...+1/(x-1)x=1-1/x<1
=> (1/4+1/9+...+1/x2) ko là số nguyên
=>S ko là số nguyên
a: \(=\left(\dfrac{5^4}{5^2\cdot7^2}\right)^{15}:\left(\dfrac{5^6}{2\cdot7\cdot31}\right)^7\)
\(=\dfrac{5^{30}}{7^{30}}:\dfrac{5^{42}}{2^7\cdot7^7\cdot31^7}\)
\(=\dfrac{5^{30}}{7^{30}}\cdot\dfrac{2^7\cdot7^7\cdot31^7}{5^{42}}=\dfrac{2^7\cdot31^7}{7^{23}\cdot5^{12}}\)
b: \(=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{5^{29}\cdot2^8\cdot7^{48}}\)
\(=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1-7\cdot4\right)}{5^{29}\cdot2^8\cdot7^{48}}=5\cdot\left(-27\right)=-135\)
Bạn tham khảo nhé
Ta có :
\(B=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+\frac{24}{25}+...+\frac{2499}{2500}\)
\(B=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+\frac{5^2-1}{5^2}+...+\frac{50^2-1}{50^2}\)
\(B=\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{3^2}\right)+\left(1-\frac{1}{4^2}\right)+\left(1-\frac{1}{5^2}\right)+...+\left(1-\frac{1}{50^2}\right)\)
\(B=\left(1+1+1+1+...+1\right)-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-\frac{1}{5^2}-...-\frac{1}{50^2}\)
\(B=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{50^2}\right)\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{50^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
\(A< 1-\frac{1}{50}\)
\(A< \frac{49}{50}\)\(\left(1\right)\)
Lại có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{50^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{50.51}\)
\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{50}-\frac{1}{51}\)
\(A>\frac{1}{2}-\frac{1}{51}=\frac{49}{102}\)\(\left(2\right)\)
Từ (1) và (2) suy ra \(\frac{49}{102}< A< \frac{49}{50}\)
\(\Leftrightarrow\)\(49-\frac{49}{102}< 49-A< 49-\frac{49}{50}\)
\(\Leftrightarrow\)\(\frac{4949}{102}< B< \frac{2401}{50}\)
\(\Rightarrow\)\(B\notinℤ\)
Vậy B không là số nguyên
\(S=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+......+\frac{9999}{10000}\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+.......+\left(1-\frac{1}{10000}\right)\)
\(=\left(1+1+.....+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+.....+\frac{1}{10000}\right)\)
\(=99-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+.......+\frac{1}{10000}\right)\)( số các chữ số 1 bằng căn bậc 2 của mẫu rồi trừ đi 1 )
Đặt \(A=\frac{1}{4}+\frac{1}{9}+.........+\frac{1}{10000}\)
Ta có: \(4=2.2< 2.3\)\(\Rightarrow\frac{1}{4}>\frac{1}{2.3}\)
Tương tự ta có: \(\frac{1}{9}>\frac{1}{3.4}\); ........ ; \(\frac{1}{10000}>\frac{1}{100.101}\)
\(\Rightarrow A>\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{100.101}\)\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{2}-\frac{1}{101}=\frac{99}{202}\)
Ta lại có: \(4=2.2>1.2\)\(\Rightarrow\frac{1}{4}< \frac{1}{1.2}\)
Tương tự ta được: \(\frac{1}{9}< \frac{1}{2.3}\); ......... ; \(\frac{1}{10000}< \frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{100.101}\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow\frac{99}{202}< A< \frac{99}{100}\)\(\Rightarrow\)A không phải là số nguyên
\(\Rightarrow99-A\)không là số nguyên \(\Rightarrow\)S không là số nguyên ( đpcm )