Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\frac{-9}{10}\cdot\frac{-10}{11}\cdot\frac{-11}{12}\cdot\cdot\cdot\cdot\frac{-99}{100}\)
\(=\frac{9}{-10}\cdot\frac{-10}{11}\cdot\frac{11}{-12}\cdot\cdot\cdot\cdot\frac{99}{-100}\)
\(=\frac{-9}{100}\)
cảm ơn bạn nhé ST kết bạn với mình nhé để mình có thể hỏi bài từ bạn, đi mình xin bạn đấy
(1-1/3).(1-1/5).(1-1/7).(1-1/9).(1-1/11).(1-1/13).(1-1/2).(1-1/4).(1-1/6).(1-1/8).(1-1/10)
=2/3.4/5.6/7.8/9.10/11.12/13.1/2.3/4.5/6.7/8.9/10
=8/15.48/63.120/143.3/8.35/48.9/10
=384/945.360/1144.315/480
=138240/1081080.315/480
=43545600/518918400=84/1001
\(C=(\frac{2}{3}-\frac{1}{4}+\frac{5}{11}):(\frac{5}{12}+1-\frac{7}{11})\)
\(=\left(\frac{88}{132}-\frac{33}{132}+\frac{60}{132}\right):\left(\frac{55}{132}+\frac{132}{132}-\frac{84}{132}\right)=\left(\frac{115}{132}\right):\frac{103}{132}=\frac{115}{132}.\frac{132}{103}=\frac{115}{103}\)
\(D=1\frac{1}{3}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-\frac{25}{100}.\frac{1}{2}=\frac{1}{3}+\frac{1}{8}:\frac{1}{4}-\frac{1}{8}=\frac{1}{3}+\frac{1}{2}-\frac{1}{8}=\frac{8+12-3}{24}=\frac{17}{24}\)
\(E=\left(-\frac{1}{2}\right)^2-\left(-2\right)^2-5^0=\frac{1}{4}-4-1=\frac{1-16-4}{4}=\frac{-19}{4}\)
Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
sửa đề :
\(C=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{15}\right)+...+\left(1-\frac{1}{105}\right)\)
\(C=\left(1+1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{105}\right)\)
Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{105}\)
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{210}\)
\(A=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{210}\right)\)
\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{14.15}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{14}-\frac{1}{15}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{15}\right)\)
\(A=2.\frac{13}{30}\)
\(A=\frac{13}{15}\)
Thay A vào ta được :
B = \(\left(1+1+1+1+...+1\right)-\frac{13}{15}\)
B = \(14-\frac{13}{15}\)( có 14 số 1 )
B = \(\frac{197}{15}\)
\(\left(\frac{1}{10}-1\right).\left(\frac{1}{11}-1\right).....\left(\frac{1}{100}-1\right)\)
\(=\frac{-9}{10}.\frac{-10}{11}......\frac{-99}{100}\)
\(=\frac{-9}{100}\)
-9/100