a: \(=\dfrac{-36}{35}\cdot\dfrac{14}{9}\cdot\dfrac{-1}{8}=\dfrac{36}{9}\cdot\dfrac{14}{35}\cdot\dfrac{1}{8}=4\cdot\dfrac{1}{8}\cdot\dfrac{2}{5}=\dfrac{1}{5}\)

b: \(=\left(6+\dfrac{1}{8}-\dfrac{1}{2}\right)\cdot4\)

\(=\dfrac{48+1-4}{8}\cdot4=\dfrac{45}{2}\)

Giải:

a) \(-\dfrac{36}{35}:\dfrac{9}{14}.\left(-\dfrac{1}{2}\right)^3\)

\(=-\dfrac{36}{35}.\dfrac{14}{9}.\left(-\dfrac{1}{8}\right)\)

\(=-\dfrac{36.14.\left(-1\right)}{35.9.8}\)

\(=-\dfrac{9.4.2.7.\left(-1\right)}{5.7.9.2.4}\)

\(=-\dfrac{-1}{5}\)

Vậy ...

b) \(\left[6+\left(\dfrac{1}{2}\right)^3-\dfrac{1}{2}\right]:\dfrac{3}{12}\)

\(=\left[6+\dfrac{1}{8}-\dfrac{1}{2}\right].\dfrac{12}{3}\)

\(=\left[\dfrac{48}{8}+\dfrac{1}{8}-\dfrac{4}{8}\right].4\)

\(=\dfrac{48+1-4}{8}.4\)

\(=\dfrac{45}{8}.4\)

\(=\dfrac{45}{2}\)

Vậy ...

a) = 30

b) = 5

c) = 70

a)12+24+35-2-14-25=30

b)1-2+3-4+5-6+7-8+9=5

c)25-12-3+24+36=70

a) = 30

b) = 5

c) = 70

c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)

\(=\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{-1}{4}+\dfrac{-7}{4}\)

\(=1-2=-1\)

Giải:

a)-1/12+4/3=-1/12+16/12=15/12=5/4

b)(-4/14-3/15)-(1/5-20/35-(-1)).7

=-17/35-22/35.7

=-17/35-22/5

=-171/35

c)3/5+-5/20+30/75+-7/4

=3/5+-1/4+2/5+-7/4

=(3/5+2/5)+(-1/4+-7/4)

=1+-2

=-1

d)5/6.-12/14+7/13

=-5/7+7/13

=-16/91

e)2/-9-5/-36-1/4

=-1/12-1/4

=-1/3

f)2/23+-5/12+7/18+21/23+-7/12

=(2/23+21/23)+(-5/12+-7/12)+7/18

=1+-1+7/18

=7/18

xcsxccsz

a) 1+2+3+...+99+100

giải

Từ 1 đến 100  có 100 số.Như vậy,số cặp số là:

100:2=50(cặp)

Mỗi cặp số có tổng bằng:

1+100(2+99)(3+98)...=11

Kết quả của phép tính là:

101x50=5050

Đáp số:5050

1+2+...+100=(1+100).100/2=5150

a)5050

b)325

c)Nguyễn Văn Chịu

Bài 2. > , < ,  = ?

5/7 < 4/3             2/5 < 6/10            1/4 = 3/12                27/36 >

2/9           7/6 > 7/9

7/2 = 2/7        15/23 < 1                  27/9 > 2          14/15 < 1          51/17 < 4

sai rôi

1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)

\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)

\(\left|4-2x\right|=\dfrac{1}{3}.3\)

\(\left|4-2x\right|=1\)

=>\(4-2x=\pm1\)

+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)

\(2x=4-1\) \(2x=4-\left(-1\right)\)

\(2x=3\) \(2x=4+1\)

\(x=3:2\) \(2x=5\)

\(x=1,5\) \(x=5:2\)

Vậy x=1,5 \(x=2,5\)

Vậy x=2,5

2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)

\(9:\left|x+\left(-1\right)\right|=-3\)

\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)

\(\left|x+\left(-1\right)\right|=-3\)

=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )

Vậy x = \(\varnothing\)

3) \(\left|\dfrac{3}{2}-x\right|+\left(-\dfrac{4}{5}\right)=-\dfrac{14}{5}\)

\(\left|\dfrac{3}{2}-x\right|=\left(-\dfrac{14}{5}\right)-\left(-\dfrac{4}{5}\right)\)

\(\left|\dfrac{3}{2}-x\right|=\left(-\dfrac{14}{5}\right)+\dfrac{4}{5}\)

\(\left|\dfrac{3}{2}-x\right|=-\dfrac{10}{5}\)

\(\left|\dfrac{3}{2}-x\right|=-2\)

=> \(\dfrac{3}{2}-x\) không có giá trị nào thỏa mãn ( Vì giá trị tuyệt đối không thể là số nguyên âm )

Vậy x=\(\varnothing\)