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1x2 + 3x4 + 5x6 + 7x8 + 9x10
=2 + 12 +30 +56 +90
=14 + 30 + 56 + 90
=44 +56 + 90
=100 + 90
=190
tk nhé
\(\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{2005\times2006}\) =\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2005}-\frac{1}{2006}\) =\(\frac{1}{3}-\frac{1}{2006}=\frac{2003}{6018}\)
Ta có \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
Lại có B = \(\frac{1}{51.100}+\frac{1}{52.99}+...+\frac{1}{99.52}+\frac{1}{100.51}\)
=> 151B = \(\frac{151}{51.100}+\frac{151}{52.99}+...+\frac{151}{99.52}+\frac{151}{100.51}\)
=> 151B = \(\frac{1}{51}+\frac{1}{100}+\frac{1}{52}+\frac{1}{99}+...+\frac{1}{99}+\frac{1}{52}+\frac{1}{100}+\frac{1}{51}\)
=> 151B = \(2\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)
=> B = \(\frac{2}{151}.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Khi đó \(\frac{A}{B}=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{2}{151}\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)}=\frac{1}{\frac{2}{151}}=\frac{151}{2}=75,5\)
1x2+3x4+5x6+7x8+9x10+11x12+13x14+15x16
=1x2+3x2x2+5x3x2+7x4x2+9x5x2+11x6x2+13x7x2+15x8x2
=1x2+6x2+15x2+28x2+45x2+66x2+91x2+120x2
= 2x(1+6+15+28+45+66+91+120)
=2x370
=740
toi nay mk se lay y kien cua cac ban , giup mk nha
b1
a) \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{5}-\dfrac{1}{10}\)
\(=\dfrac{2}{10}-\dfrac{1}{10}\)
\(=\dfrac{1}{10}\)
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{1}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
c) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)
\(=\dfrac{1}{3}-\dfrac{1}{11}\)
\(=\dfrac{8}{33}\)
d) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=\dfrac{1}{3}-\dfrac{1}{101}\)
\(=\dfrac{98}{303}\)
= 1x2 + 3x2x2 + 5x3x2 + 7x4x2 + 9x5x2 + 11x6x2 + 13x7x2 + 15x8x2
=1x2 + 6x2 + 15 x2 + 28x2 + 45x2 + 66x2 + 91x2 + 120x2
=2 x ( 1+6+15+28+45+66+91+120)
= 2 x 370
=740
=
A = 1 x 2 + 2 x 3 + ....... + 10 x 11
3A = 1 x 2 x 3 + 2 x 3 x 3 + ..........+ 10 x 11 x 3
3A = 1 x 2 x (3-0) + 2 x 3 x (4-1) + .......... + 10 x 11 x (12 -9)
3A = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + ........... + 10 x 11 x 12 - 9 x 10 x 11
3A = (1 x 2 x 3 - 1 x 2 x 3) + ( 2 x 3 x 4 - 2 x 3 x 4) +............ + 10 x 11 x 12
3A = 10 x 11 x 12 = 1320
A = 1320 : 3 = 440
Gọi biểu thức trên là A, ta có :
A= 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101 A = 99x100x101 : 3 A = 333300
một hình chữ nhật có chu vi 72cm chiều dài 15cm và chiều rộng 9cm thì hình chữ nhật trở thành hình vuông. Tính diện tích hình chữ nhật ban đầu
giải hộ mình nhé
mk thấy bn viết đề phải đầy đủ và chính xác
\(\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}\right):\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\)
Đặt \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}-2.\frac{1}{2}-2.\frac{1}{4}-2.\frac{1}{6}-2.\frac{1}{8}-2.\frac{1}{10}\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}-1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\)
\(A=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\)
\(\Rightarrow\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}\right):\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\)
\(=\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right):\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\)
\(=1\)
~~Hok tốt~~