tính hợp lí
-8/11.(33/22+3333/2020+333333/303030)
-7/4x.(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
tìm x biết
-7/4 nhân x nhân (33/12+3333/2020+333333/303030+33333333/42424242)=22
-8/11x(33/12+3333/2020+333333/303030)
Tính nhanh : P= -7/4 x (33/12 + 3333/2020 + 333333/303030 + 33333333/42424242)
\(P=-\frac{7}{4}.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
\(P=-\frac{7}{4}.\left(\frac{33}{12}+\frac{33.101}{20.101}+\frac{33x10101}{30x10101}+\frac{33x1010101}{42x1010101}\right)\)
\(P=-\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(P=-\frac{7}{4}.\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}\right)\)
\(P=-\frac{77}{8}.\left(\frac{1}{6}+\frac{3}{10}+\frac{1}{15}+\frac{1}{21}\right)=-\frac{77}{8}.\frac{35+63+14+10}{210}=-\frac{11x122}{8x30}\)
\(P=-\frac{671}{120}\)
P = -7/4 x (33/12 + 3333/2020 + 333333/303030 + 33333333/42424242)
= -7/4 x (33/12 + 33/20 + 33/30 + 33/42)
= -7/4 x [33 x (1/12 + 1/20 + 1/30 + 1/42)]
= -7/4 x [33 x (35/420 + 21/420 + 14/420 + 10/420)]
= -7/4 x (33 x 80/420)
= -7/4 x 33 x 4/21
= -7/4 x 4/21 x 33 (= -7x4 / 4x21 x33)
= -7/21 x 33 (= -7x33 / 21 = -1x7x3x11 / 3x7)
= -11/1
= -11
Đáp số P = -11
Các số gạch đi là do rút gọn phân số nhé!!!
TÍNH HỢP LÝ :
1) \(\dfrac{-8}{11}.\left(\dfrac{33}{22}+\dfrac{3333}{2020}+\dfrac{333333}{303030}\right)\)
2) \(\dfrac{0,4+\dfrac{2}{7}-\dfrac{2}{11}}{0,6+\dfrac{3}{7}-\dfrac{3}{11}}+\dfrac{0,25-0,2+\dfrac{1}{7}}{0,75-0,6+\dfrac{3}{7}}\)
1: \(=\dfrac{-8}{11}\left(\dfrac{3}{2}+\dfrac{33}{20}+\dfrac{11}{10}\right)\)
\(=\dfrac{-8}{11}\cdot\dfrac{30+33+22}{20}=\dfrac{-8}{11}\cdot\dfrac{85}{20}=-\dfrac{34}{11}\)
2: \(=\dfrac{2}{3}+\dfrac{1}{3}=1\)
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 -
\(D=\dfrac{7}{4}.\dfrac{33}{12}+\dfrac{3333}{2020}\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\)
Cái ở giữa 3333/2020 và 333333/3030303 là j vậy ạ