Tính nhanh : P= -7/4 x (33/12 + 3333/2020 + 333333/303030 + 33333333/42424242)
7/4.( 33/12 + 3333/2020 + 333333/303030 + 33333333/42424242) = ?
\(\frac{7}{4}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
=\(\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
=\(\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
=\(\frac{231}{4}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
=\(\frac{231}{4}\left(\frac{1}{3}-\frac{1}{7}\right)\)
=\(\frac{231}{4}.\frac{4}{21}\)
= 11
-7/4 .(33/12+3333/2020+333333/303030+33333333/42424242)
\(\dfrac{-7}{4}.\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)
\(=\dfrac{-7}{4}.\left(\dfrac{33}{12}.\dfrac{33.101}{20.101}.\dfrac{33.10101}{30.10101}.\dfrac{33.1010101}{42.1010101}\right)\)
\(=\dfrac{-7}{4}.\left(\dfrac{33}{12}.\dfrac{33}{20}.\dfrac{33}{30}.\dfrac{33}{42}\right)\)
\(=\dfrac{-7}{4}.33.\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)
\(=\dfrac{-7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(=\dfrac{-7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(=\dfrac{-7}{4}.33.\dfrac{4}{21}\)
\(=33.\dfrac{-1}{3}\)
\(=11\)
xl nhé kết quả là -11 mk viết thiếu dấu -
tìm x biết
-7/4 nhân x nhân (33/12+3333/2020+333333/303030+33333333/42424242)=22
Tìm x biết: -7/4x.(33/12+3333/2020+333333/303030+33333333/42424242)=22
Kết quả : Viết lại biểu thức đã cho
=> -7/4x . ( 33/12 + 33/20 + 33/30 + 33/42 ) = 22
-7/4x . 33 . ( 1/12 + 1/20 + 1/30 + 1/42 ) = 22
-231/4x . ( 1/3 . 4 + 1/ 4. 5 + 1/5 . 6 + 1/ 6. 7 ) = 22
-231/4x . ( 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 ) = 22
-231/4x . ( 1/3 - 1/7 ) = 22
-231/4x . 4/21 = 22
-11x = 22
x = 22 : -11
x = -2
Vậy x = -2
-7/4x.(33/12+3333/2020+333333/303030+33333333/42424242)=22
\(D=\dfrac{7}{4}.\dfrac{33}{12}+\dfrac{3333}{2020}\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\)
Cái ở giữa 3333/2020 và 333333/3030303 là j vậy ạ
Tính
\(\dfrac{7}{4}.(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242})\)
\(\dfrac{7}{4}.\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)
\(=\dfrac{7}{4}.\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(=\dfrac{7}{4}.33\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)
\(=\dfrac{7}{4}.33\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\right)\)
\(=\dfrac{7}{4}.33\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(=\dfrac{7}{4}.33\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(=\dfrac{7}{4}.33\left(\dfrac{7}{21}-\dfrac{3}{21}\right)\)
\(=\dfrac{7}{4}.33.\dfrac{4}{21}\)
\(=\dfrac{231}{4}.\dfrac{4}{21}\)
\(=\dfrac{231}{21}=11\)
tìm x:
a) -7/4 . x ( 33/ 12 + 3333/2020 + 333333/303030 + 33333333/42424242)
b) 1+5 +9 + 13 + 15 +...+x = 501501
Tính nhanh
1/33 x ( 33/12+33/2020+333333/303030+33333333/42424242)
Sửa đề:
\(\frac{1}{33}\times\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
\(=\frac{1}{33}\times\frac{33}{12}+\frac{1}{33}\times\frac{3333}{2020}+\frac{1}{33}\times\frac{333333}{303030}+\frac{1}{33}\times\frac{33333333}{42424242}\)
\(=\frac{1}{12}+\frac{33\times101}{33\times101\times20}+\frac{33\times10101}{33\times10101\times30}+\frac{33\times1010101}{33\times1010101\times42}\)
\(=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=\frac{1}{3}-\frac{1}{7}\)
\(=\frac{4}{21}\)