a, 

= 51 - 2022 + 2022

= 51

b, = -2019 - 11 + 2019

    = -11

c, = 27 - 52 - 37 + 52

= -10

d, = 19 - 32 + 41 - 19 + 25 - 41

= -7

a)    (51 - 2022) + 2022

=51+(-2022+2022)

=51+0

=51

b)    (-2019) - (11 - 2019)

=(-2019)-11+2019

=(-2019+2019)-11

=0-11

=-11

c)    (27 - 52) - (37 - 52)

=27-52-37+52

=[27+(-37)]+(-52+52)

=-10+0

=-10

d) (19 - 32 + 41) - (19 - 25 + 41)

=19-32+41-19+25-41

=(19-19)+(-32-25)+(41-41)

=0+(-57)+0

=-57

sách Nâng cao phát triển lớp 6 tập 1

(3965-2378)-(437-1378)-528

=3965-2378-437-1378-528

=(3965-(437+528))-(2378-1378)

=3000-1000

=2000

Bài 1 :

\((3965-2378)-(437-1378)-528\)

\(=3965-2378-437+1378-528\)

\(=3965-(2378-1378)-(437+528)\)

\(=3965-1000-965\)

\(=(3965-965)-1000\)

\(=3000-1000\)

\(=2000\)

Bài 2 :

\((-x+31)-39=-69+11\)

\((-x+31)-39=-58\)

\(-x+31=-58+39\)

\(-x+31=-19\)

\(-x=-19-31\)

\(-x=-50\)

\(=>x=50\)

a: \(\frac{47}{15}>\frac{15}{15}=1;1=\frac{35}{35}>\frac{29}{35}\)

Do đó: \(\frac{47}{15}>\frac{29}{35}\)

b: \(\frac{1999}{2001}<\frac{2001}{2001}=1;1=\frac{11}{11}<\frac{12}{11}\)

Do đó: \(\frac{1999}{2001}<\frac{12}{11}\)

c: \(\frac{2021}{2020}>\frac{2020}{2020}=1;1=\frac{2022}{2022}>\frac{2019}{2022}\)

Do đó: \(\frac{2021}{2020}>\frac{2019}{2022}\)

=2021/2022(-9/11-2/11)

=-2021/2022

\(35:5+3^7:3^4-2022\)

\(=35:5+3^{7-4}-2022\)

\(=35:5+3^3-2022\)

\(=35:5+27-2022\)

\(=7+27-2022\)

\(=34-2022\)

\(=-1988\)

35:5+3⁷:3⁴-2022

=7+3³-2022

=7+27-2022

=34-2022

=-1988

A = \(\dfrac{2020}{2021}\) + \(\dfrac{2021}{2022}\) ;  B = \(\dfrac{2020+2021}{2021+2022}\)

B = \(\dfrac{2020+2021}{2021+2022}\)   = \(\dfrac{2020}{2021+2022}\) + \(\dfrac{2021}{2021+2022}\)

\(\dfrac{2020}{2021}\)   > \(\dfrac{2020}{2021+2022}\)

\(\dfrac{2021}{2022}\)     > \(\dfrac{2021}{2021+2022}\)

Cộng vế với vế ta có:

A = \(\dfrac{2020}{2021}\) + \(\dfrac{2021}{2022}\) > \(\dfrac{2020}{2021+2022}\) + \(\dfrac{2021}{2021+2022}\) = B

Vậy A > B

A =  \(\dfrac{10^{10}-1}{10^{11}-1}\) 

A \(\times\) 10 = \(\dfrac{(10^{10}-1)\times10}{10^{11}-1}\) = \(\dfrac{10^{11}-10}{10^{11}-1}\) = 1 - \(\dfrac{9}{10^{11}-1}\) < 1

B = \(\dfrac{10^{10}+1}{10^{11}+1}\)

B \(\times\) 10 = \(\dfrac{(10^{10}+1)\times10}{10^{11}+1}\)  = \(\dfrac{10^{11}+10}{10^{11}+1}\) = 1 + \(\dfrac{9}{10^{11}+1}\) > 1

Vì 10 A< 1< 10B

Vậy A < B

;-;;?

\(\frac{5^2\cdot5^3-\left\lbrack841\cdot2022^0-\left(37-6^2\right)^3\right\rbrack}{6+3^5+3^3}\)

\(=\frac{5^5-\left\lbrack841-1^3\right\rbrack}{6+3^3\left(3^2+1\right)}\)

\(=\frac{5^5-840}{6+27\cdot10}=\frac{2285}{276}\)