\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+....+\frac{3}{40\cdot43}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{40}-\frac{1}{43}\)

\(=1-\frac{1}{43}=\frac{42}{43}\)

=42/32

1/2×(2×x-3)+105/2=-137/2

1/2×(2×x-3)=-137/2-105/2

1/2×(2×x-3)=-242/2

2×x-3=-242/2:1/2

2×x-3=-242

2.x=(-242)+3

2.x=239

x=239:2

x=239/2

A,-12×25-25×75-25×13

=[(-12)-75-13].25

=(-100).25

-2500

B,-50/7×49/10-35/2×-10/7+ -25/3× -9/5

=(-35)-(-25)+(-13)

=-23

C,-354+265-156+125

=[(-354)-156]+(265+125)

=(-510)+277

=-233

D,-74+40-50+16-35

=(-74)+40+(-50)+16+(-35)

=[(-74)+(-35)]+[40+(-50)+16]

=(-109)+26

=-83

E,-2/3×4/5+-4/5×4/3-0,125

=-2/3.4/5+4/5.(-4/3)-0,125

=4/5.[-2/3+(-4/3)]-0,125

=4/5.(-2)-0,125

=-8/5-0,125

=(-1,6)+(-0,125)

=-1,725

-2/3×4/5+ -4/5×4/3-0,125

Giúp e nốt ik ạ

a) 50 + 48 + 46 + ... + 4 - 47 - 45 - 43 - ... - 1

= (50 - 45) + (48 - 43) + (46 - 41) + ... + (6 - 1) + (4 - 47)

=72

Cứ gộp nhóm làm sao cho trong ngoặc đó bằng 5

b) 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - 11 - 12 + ... + 50 - 51 - 52 + 53 + 54

= (1 + 54) + (2 + 53) - (3 + 52) - (4 + 51) + ... + (25 + 30) + (26 + 29) - (27 + 28)

=55

Cứ gộp nhóm làm sao cho trong ngoặc đó bằng 55. Còn dấu đằng trước nhóm thì theo dấu đề bài cho

~ Học tốt ~

bạn làm đầy đủ hơn đc k

Ta có: \(\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{40.43}+\frac{3}{43.46}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(=1-\frac{1}{46}\)

Vì \(\frac{1}{46}>0\Rightarrow1-\frac{1}{46}< 1\)

Vậy \(\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{43.46}< 1\)

a: \(=\dfrac{4\cdot2+4\cdot9}{55}+\dfrac{5}{6}=\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{49}{30}\)

b: \(=\dfrac{3}{2}\cdot\dfrac{3}{5}-\left(\dfrac{3}{7}+\dfrac{3}{20}\right)\cdot\dfrac{10}{3}\)

\(=\dfrac{9}{10}-\dfrac{81}{140}\cdot\dfrac{10}{3}\)

\(=\dfrac{9}{10}-\dfrac{27}{14}=\dfrac{-36}{35}\)

c: \(=15+\dfrac{3}{13}-3-\dfrac{4}{7}-8-\dfrac{3}{13}\)

\(=4-\dfrac{4}{7}=\dfrac{24}{7}\)

d: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=5\)

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\) < 1

\(S=3\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{40.43}+\frac{1}{43.46}\right)\)

\(S=3.\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\right)\)

\(\Rightarrow S=1-\frac{1}{46}\Rightarrow S< 1\left(đpcm\right)\)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

= \(1-\frac{1}{46}< 1\)

\(\Rightarrow S< 1\left(đpcm\right)\)