a) Ta có: \(1+\left(-2\right)+3+\left(-4\right)+...+2021\)

\(=\left(1-2\right)+\left(3-4\right)+...+\left(2019-2020\right)+2021\)

\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+2021\)

\(=-1010+2021=1011\)

1+(-2)+3+(-4)+....+2021

=(1+3+5+.....+2021)-(2+4+6+....+2020)

=\(\dfrac{\left(1+2021\right)\cdot2021}{2}-\dfrac{\left(2+2020\right)\cdot2020}{2}=1010\cdot2021-1010\cdot2020\\ =1010\cdot\left(2021-2020\right)=1010\cdot1\\ =1010\)

1 + (-2) + 3 + (-4) +......+ 2021

= ( 1 - 2 ) + ( 3 - 4 ) +....+ ( 2019 - 2020 ) + 2021

= (-1) + (-1) +.....+ (-1) + 2021

Từ 1 đến 2020 có số số hạng là : 

( 2020 - 1 ) : 1 + 1 = 2020 ( số hạng ) 

⇒ 1 + 2021 = 2022

a.5/14+1/2=6/7

b.5/3-1=2/3

c.4/3-1/3=1

a) 5/14+ 1/2= 6/7

b) 5/3 - 1= 2/3

c) 4/3- 1/3 =1

a.5/14+1/2=6/7

b.5/3-1=2/3

c.4/3-1/3=1

Bài 1: 

a: \(=\dfrac{4}{8}+\dfrac{2}{8}+\dfrac{1}{8}=\dfrac{7}{8}\)

b: \(=\dfrac{4}{12}+\dfrac{2}{12}+\dfrac{1}{12}=\dfrac{7}{12}\)

B1:

a \(\dfrac{7}{8}\)

b \(\dfrac{7}{12}\)

B2:

a \(\dfrac{3}{7}\)

b \(\dfrac{5}{9}\)

c \(\dfrac{6}{5}\)

B3:

 đội còn phải sửa phần quãng đường nữa  :\(\dfrac{8}{15}\) phần quãng đường

a: =20/60+12/60-15/60=17/60

b: =7/8-1/4-2/5=5/8-2/5=25/40-16/40=9/40

\(a,\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{4}=\dfrac{20}{60}+\dfrac{12}{60}-\dfrac{15}{60}=\dfrac{17}{60}\\ b,\dfrac{7}{8}-\left(\dfrac{1}{4}+\dfrac{2}{5}\right)=\dfrac{7}{8}-\dfrac{13}{20}=\dfrac{9}{40}\)

\(a,\dfrac{1}{2-\sqrt{3}}-3\sqrt{\dfrac{1}{3}}+\sqrt{12}\\ =\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}-\dfrac{\sqrt{3^2}}{\sqrt{3}}+\sqrt{2^2.3}\\ =\dfrac{2+\sqrt{3}}{4-3}-\sqrt{3}+2\sqrt{3}\\ =2+\sqrt{3}-\sqrt{3}+2\sqrt{3}\\ =2+2\sqrt{3}\)

\(b,\dfrac{2}{1+\sqrt{2}}-\sqrt{9-\sqrt{32}}\\ =\dfrac{2\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}-\sqrt{9-4\sqrt{2}}\\ =\dfrac{2-2\sqrt{2}}{1-2}-\sqrt{\left(2\sqrt{2}\right)^2-2.2\sqrt{2}+1}\\ =-2+2\sqrt{2}-\sqrt{\left(2\sqrt{2}-1\right)^2}\\ =-2+2\sqrt{2}-\left|2\sqrt{2}-1\right|\\ =-2+2\sqrt{2}-2\sqrt{2}+1\\ =-1\)

Bài 1: 

a: =25+75=100

b: =60-17-43+12=12

c: =-2-18=-20

d: =-3+36-17=36-20=16

Bài 2: 

a: =-102

b: =-1000

c: =12x15=180

d: =21x(-10)=-210

B1

a 100

b 12

c -20

d 16

B2

a -102

b -1000

c 180

d -210

Bài 1:Tính

 a)(-12)+25+75+12 = 100

b)60+12+(-17)+(-43)=12
c)(-2)+(-87)+(-18)+87=-20
d)(-1)+(-2)+36+(-17)=16
Bài 2:Tính
b)8 . (-125)=-100
c)(-12) . (-15)=180
d)21 . (- 3)+(- 21 ) . 7=-210

a, \(1+\dfrac{3}{4}=\dfrac{7}{4}\)

b, \(\dfrac{4}{5}-\dfrac{3}{8}=\dfrac{32-15}{40}=\dfrac{17}{40}\)

c, \(1:\dfrac{2}{3}=\dfrac{1.3}{2}=\dfrac{3}{2}\)

d, \(\dfrac{2}{5}.\dfrac{5}{2}=1\)

a)1 + 3/4=7/4

   b)4/5 -3/8=17/40

  c)1 : 2/3=3/2

  d)2/5 x 5/2=1

Đáp án:

a)1 + 3/4=7/4

b)4/5 -3/8=17/40

c)1 : 2/3=3/2

d)2/5 x 5/2=1

a \(\dfrac{53}{84}\)

b \(\dfrac{20}{63}\)

c \(\dfrac{1}{60}\)

a: \(\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{3}{4}=\dfrac{56}{84}+\dfrac{60}{84}-\dfrac{63}{84}=\dfrac{53}{84}\)

b: \(=\dfrac{3}{7}-\dfrac{1}{9}=\dfrac{27}{63}-\dfrac{7}{63}=\dfrac{20}{63}\)

c: \(=\dfrac{2}{5}\cdot\dfrac{1}{8}\cdot\dfrac{1}{3}=\dfrac{2}{8}\cdot\dfrac{1}{15}=\dfrac{1}{60}\)