a)-8/11x(33/12+3333/2020/+333333/30330)
b) 0,4+2/7-2/11/ 0.6+3/7+3/1 + 0,25-0,2+1/7/ 0,75-0,6+3/7
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: \(=\dfrac{2}{3}+\dfrac{1}{3}=1\)
0,4+2/7-2/11 / 0,6+3/7-3/11 + 0,25-0,2+1/7/0,75-0,6+3/7
Đề bài là tính phải không :)
\(\frac{0,4+\frac{2}{7}-\frac{2}{11}}{0,6+\frac{3}{7}-\frac{3}{11}}+\frac{0,25-0,2+\frac{1}{7}}{0,75-0,6+\frac{3}{7}}\)
\(=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{11}}{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}+\frac{\frac{1}{4}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}}\)
\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{3\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}+\frac{\frac{1}{4}-\frac{1}{5}+\frac{1}{7}}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}\right)}\)
\(=\frac{2}{3}+\frac{1}{3}=1\)
Thực hiện phép tính bằng cách hợp lý
a)(-0,75-1/4):(-5)+1/15-(-1/5):(-3)
b)(3/25-1,12):3/7:[(3/1/2-3/2/3):1/14
c)0,4+2/7-2/11/0,6+3/7-3/11+0,25-0,2+1/7/0,75-0,6+3/7
tính nhanh
0,4+2/7-2/11 / 0,6+3/7x3/1 + 0,25-0,2+1/7 / 0,75--0,6+3/7
bài 1 : Tính bằng cách hợp lí
a) \(\frac{0,4+\frac{2}{7}-\frac{2}{11}}{0,6+\frac{3}{7}-\frac{3}{11}}+\frac{0,25-0,2+\frac{1}{7}}{0,75-0,6+\frac{3}{7}}\)
Tìm x biết:
a, .(x-3 ) + (x-2) +(x-1) +...+ 10 +11= 11
b, 3x + 3x+1 +3x+2 =351
c, [-7/4 . x ] . [33/12 + 3333/2020 +333333/303030 +33333333 / 42424242] = 22
Bài 1 : Tính nhanh
A= ( 1/2 - 1 ) . (1/3 - 1 ) . ( 1/4 - 1) .... ( 1/99 -1)
B= -7/4 . ( 33/12 + 3333/2020 + 333333/202020 + 3333/ 4242 )
C= ( 1 + 1/2) . ( 1 + 1/3) . ( 1+ 1/4) .... (1+ 1/99)
D= ( 1- 1/2 ) . ( 1- 1/3 ) . ( 1- 1/4 ) .... ( 1- 1/99)
c: \(C=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{100}{99}=\dfrac{100}{2}=50\)
d: \(D=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot..\cdot\dfrac{98}{99}=\dfrac{1}{99}\)
Tính giá trị biểu thức
a) (1-1/2).(1-1/3).(1-1/4)...(1-1/20)
b) 1+1/2+1/2^2+1/2^3+...+1/2^2012
c)7/4.(3333/1212+3333/2020+3333/3030+33334242)
d)2/3-1/4+5/11/5/12+1-7/11
e) 3/5+3/7-3/11/4/5+4/7-4/11
a) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...........\frac{19}{20}=\frac{1}{20}\)
b) \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)
=> \(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
=> \(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
=> \(A=2-\frac{1}{2^{2012}}\)
c) \(\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(=\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(=\frac{231}{4}.\frac{4}{21}=11\)
d.e) ktra lại đề
Tính :
\(B=\frac{7}{4}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
\(C=\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(B=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(=\dfrac{7}{4}\left(\dfrac{11}{4}+\dfrac{33}{20}+\dfrac{11}{10}+\dfrac{11}{14}\right)\)
\(=\dfrac{7}{4}\cdot\dfrac{11\cdot35+33\cdot7+11\cdot14+11\cdot10}{140}\)
\(=\dfrac{880}{20\cdot4}=11\)
\(C=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right)\cdot\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\dfrac{-41}{21}}\)
\(=\dfrac{\dfrac{25}{108}\cdot\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{-100}{41}}\)
\(=\dfrac{\dfrac{5751+187\cdot27}{108}}{\dfrac{-100}{41}}=100\cdot\dfrac{-41}{100}=-41\)