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Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2016}{2017}\)
\(=\frac{1.2.3......2016}{2.3.4.......2017}\)
\(=\frac{1}{2017}\)
Bài 2:
b: x+25%x=-1,25
=>1,25x=-1,25
hay x=-1
c: x-75%x=1/4
=>1/4x=1/4
hay x=1
Bài 2:
a: =3/2-11/4=6/4-11/4=-5/4
b: =-49/6-17/2=-49/6-51/6=-100/6=-50/3
P \(=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
P\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{50^2-1}{50^2}\)
P \(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)
P\(=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)
P\(=\frac{1.51}{50.2}=\frac{51}{100}\)
Câu 1 :
Ta có :
\(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
\(A=\frac{4-1}{4}+\frac{9-1}{9}+\frac{16-1}{16}+...+\frac{10000-1}{10000}\)
\(A=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+...+\frac{100^2-1}{100^2}\)
\(A=\frac{2^2}{2^2}-\frac{1}{2^2}+\frac{3^2}{3^2}-\frac{1}{3^2}+\frac{4^2}{4^2}-\frac{1}{4^2}+...+\frac{100^2}{100^2}-\frac{1}{100^2}\)
\(A=1-\frac{1}{2^2}+1-\frac{1}{3^2}+1-\frac{1}{4^2}+...+1-\frac{1}{100^2}\)
\(A=\left(1+1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
Do từ \(2\) đến \(100\) có \(100-2+1=99\) số \(1\) nên :
\(A=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)< 99\) \(\left(1\right)\)
Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) lại có :
\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(B< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B< 1-\frac{1}{100}< 1\)
\(\Rightarrow\)\(A=99-B>99-1=98\)
\(\Rightarrow\)\(A>98\) \(\left(2\right)\)
Từ (1) và (2) suy ra :
\(98< A< 99\)
Vậy A không phải là số nguyên
Chúc bạn học tốt ~
\(\frac{1}{2}A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+......+\frac{1}{2^{11}}\)
\(\frac{1}{2}A-A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+......+\frac{1}{2^{11}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(-\frac{1}{2}A=\frac{1}{2^{11}}-\frac{1}{2}\)
BAN TU LAM NOT NHE
Ta có:
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)
\(\Leftrightarrow\)\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(\Leftrightarrow\)\(2A-A=A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(\Leftrightarrow\)\(=\)\(1+\frac{1}{2}+...+\frac{1}{2^9}-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)
\(\Leftrightarrow\)\(=\)\(\left(1-\frac{1}{2^{10}}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{2^9}-\frac{1}{2^9}\right)\)
\(\Leftrightarrow\)\(=\)\(1-\frac{1}{2^{10}}\)
\(\Leftrightarrow\)\(=\)\(1-\frac{1}{1024}\)
\(\Leftrightarrow\)\(=\)\(\frac{1023}{1024}\)
\(\Leftrightarrow\)Vậy \(A=\frac{1023}{1024}\)