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1/2^2>1/2.3;1/3^2>1/3.4;......;1/9^2>1/9.10
suy ra S > 1/2.3+1/3.4+......+1/9.10
S> 1/2-1/3+1/3-1/4 +.....+1/9-1/10
S> 1/2-1/10=2/5
Vay 2/5 < S
Vậy còn S < \(\frac{8}{9}\)thì sao, bạn quên chưa chứng minh rồi
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{x\left(x+1\right)}=1-1\frac{1991}{1993}=\frac{1991}{1993}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{1991}{1993}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{1993}:2=\frac{1991}{3986}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{1991}{3986}=\frac{1}{1993}\)
=> x + 1 = 1993
=> x = 1993 - 1
=> x = 1992
chị ra tay giải cho đứa em cùng tên
50% = 1/2
x/2+2x/3 = x+4
(3x+4x)/6 =x +4
7x/6 - x =4
x/6 = 4
x = 24
\(\frac{x-12}{3}=\frac{x+1}{4}\)
=>(x-12).4=(x+1)*3
4x-48=3x+3
4x-3x=48+3
x=51
(x-12)/3=(x+1)/4
(x-12)*4=(x+1)*3
x*4-12*4=x*3+1*3
4x-48=3x+3
4x-3x=3+48
x=51
Câu a sai đề hay sao ấy
b) Không tối giản đâu nhé, cả tử và mẫu đều chia hết cho 2
bạn ơi nhưng cô giáo cho đề mk thế. bạn giải giùm mk với mai mk phải nộp rồi.
\(\frac{1}{3^2}<\frac{1}{3.4}\)
\(\frac{1}{4^2}<\frac{1}{4.5}\)
\(\frac{1}{5^2}<\frac{1}{5.6}\)
\(...\)
\(\frac{1}{100^2}<\frac{1}{100.101}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{100.101}\)
\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{101}\)
\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3}-\frac{1}{101}\)
Mà \(\frac{1}{3}<\frac{1}{2}\) nên \(\frac{1}{3}-\frac{1}{101}<\frac{1}{2}\)
hay \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{2}\)
Đặt A=1/3^2+1/4^2+1/5^2+...+1/100^2
Suy raA<1/2*3+1/3*4+1/4*5+..+1/99*100
A<1/2-1/100<1/2
Ta có điều phải chứng minh.
\(\frac{-5}{3}-\left(\frac{4}{5}-\frac{1}{2}\right)-\left|\frac{3}{4}-\frac{5}{2}+\frac{1}{3}\right|\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\left|\frac{3}{4}+\frac{-5}{2}+\frac{1}{3}\right|\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\left(\frac{3}{4}+\frac{-5}{2}+\frac{1}{3}\right)\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\frac{3}{4}+\frac{5}{2}-\frac{1}{3}\)
\(=\left(\frac{-5}{3}-\frac{1}{3}\right)+\left(\frac{1}{2}+\frac{5}{2}\right)-\left(\frac{4}{5}+\frac{3}{4}\right)\)
\(=\frac{-6}{3}+\frac{6}{2}-\left(\frac{16}{20}+\frac{15}{20}\right)\)
\(=-2+3-\frac{31}{20}\)
\(=1-\frac{31}{20}=\frac{-11}{20}\)
\(\frac{-5}{3}-\left(\frac{4}{5}-\frac{1}{2}\right)-\left|\frac{3}{4}-\frac{5}{2}+\frac{1}{3}\right|\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\left|\frac{3}{4}+\frac{-5}{2}+\frac{1}{3}\right|\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\left(\frac{3}{4}+\frac{-5}{2}+\frac{1}{3}\right)\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\frac{3}{4}+\frac{5}{2}-\frac{1}{3}\)
\(=\left(\frac{-5}{3}-\frac{1}{3}\right)+\left(\frac{1}{2}+\frac{5}{2}\right)-\left(\frac{4}{5}+\frac{3}{4}\right)\)
\(=\frac{-6}{3}+\frac{6}{2}-\left(\frac{16}{20}+\frac{15}{20}\right)\)
\(=\frac{-6}{3}+\frac{6}{2}-\left(\frac{16}{20}+\frac{15}{20}\right)\)
\(=1-\frac{31}{20}=\frac{-11}{20}\)
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có 10 cây cau trồng thành 5 hàng mỗi hàng có 4 cây ????