a)\(\dfrac{726486}{1333}\)
b)=(21+29)+(28+22)+(27+23)+(26+24)+25
=50.4+25
=200+25
=225
c)
= 23(39+68-7)
=23.100=2300
d)
={2011+[45-(-25)]:4}:2022
=(2011+70:4):2022
=(2011+17,5):2022
=2028,5:2022
=\(\dfrac{4057}{4044}\)

\(27\equiv0\left(mod3\right)\Rightarrow27^{2021}\equiv0\left(mod3\right)\)

\(34\equiv1\left(mod3\right)\Rightarrow34^{2020}\equiv1\left(mod3\right)\)

\(\Rightarrow27^{2021}+34^{2022}+1\equiv2\left(mod3\right)\)

Mà các số chính phương chia 3 chỉ có số dư 0 hoặc 1

\(\Rightarrow27^{2021}+34^{2022}+1\) không thể là số chính phương (do nó chia 3 dư 2)

a)\(27.65+27.35+300=27.\left(65+35\right)+300\)

\(=27.100+300=2700+300=3000\)

b)\(3838:\left[\left(190-6.5^2\right):4+3\right]\)

\(=3838:\left[\left(190-6.25\right):4+3\right]\)

\(=3838:\left[\left(190-150\right):4+3\right]\)

\(=3838:\left[40:4+3\right]=3838:\left[10+3\right]\)

\(=3838:13=\dfrac{3838}{13}\)

c)\(2022-x=2021\)

\(x=2022-2021=1\)

d)\(26+14:\left(x-5\right)=33\)

\(14:\left(x-5\right)=33-26=7\)

\(x-5=14+7=2\)

\(x=2+5=7\)

e)đề hỏi làm j thế bạn

23-(-27)+(-50) = 0

234+(-117)-100+(-234) = -217

-927-(-1421)-(930)+(-1421) = -1020

2021-[-2022-(-2021)] = 2021

1-2+3-4+...+2021-2022= 

A=1-2+3-4+...+2021-2022

=(1-2)+(3-4)+...+(2021-2022)  (có tất cả 1011 cặp)

=(-1).(-1)...(-1)

=(-1).1011=-1011

HOk tốt!!!!!!!!!!!!!

23 - ( - 27 ) + ( - 50 ) = 0

234 + ( - 117 ) - 100 + ( - 234 ) = - 217

- 927 - ( - 1421 ) - ( 930 ) + ( - 1421 ) = 985

2021 - [ - 2022 - ( - 2021 )] = 2022

1 - 2 + 3 - 4 + ...+ 2021 - 2022 

=(1-2)+(3-4)+...+(2021-2022)

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

=(-1)x(2022:2)

= -1011

\(a)23-\left(-27\right)+\left(-50\right).\)

\(=23+27-50\)

\(=50-50=0\)

\(b)234+\left(-117\right)-100+\left(-234\right)\)

\(=\left[234+\left(-234\right)\right]+\left[\left(-117\right)-100\right]\)

\(=0+\left(-217\right)\)

\(=-217\)

\(c)-927-\left(-1421\right)-\left(-930\right)+\left(-1421\right)\)

\(=-927+1421+930-1421\)

\(=\left(-927+930\right)+\left(1421-1421\right)\)

\(=3+0=3\)

\(d)2021-\left[-2022-\left(-2021\right)\right]\)

\(=2021-\left(-2022+2021\right)\)

\(=2022\)

\(e)1-2+3-4+...+2021-2022\)

\(=\left(1-2\right)+\left(3-4\right)+...+\left(2021-2022\right)\)

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

\(=-1\times1011\)

\(=-1011\)

\(#Wee\)

\(a^2-2a+1+b^2-4b+4=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-2\right)^2=0\)

=>a=1 và b=2

\(a^{27}+b^2+2022=1^{27}+2^2+2022=2022+4+1=2027\)

= 2027

a) (-365) + 75 + 365

= (-365 + 365) + 75

= 0 + 75

= 75

b) (-52).76 + (-52).24

= -52.(76 + 24)

= -52.100

= -5200

c) [15 + (2022⁰.37 - 2³] : (-17)

= [15 + (1.37 - 8)] : (-17)

= (15 + 19) : (-17)

= 34 : (-17)

= -2

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

Lời giải:
$ab+bc+ac=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}=\frac{9^2-27}{2}=27$

$\Rightarrow a^2+b^2+c^2=ab+bc+ac$

$\Leftrightarrow 2(a^2+b^2+c^2)=2(ab+bc+ac)$

$\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)=0$

$\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0$
Vì $(a-b)^2; (b-c)^2; (c-a)^2\geq 0$ với mọi $a,b,c$ nên để tổng của chúng bằng $0$ thì $(a-b)^2=(b-c)^2=(c-a)^2=0$
$\Rightarrow a=b=c$

Mà $a+b+c=9$ nên $a=b=c=3$. 

Khi đó:

$(a-4)^{2021}+(b-4)^{2022}+(c-4)^{2023}=(-1)^{2021}+(-1)^{2022}+(-1)^{2023}$

$=(-1)+1+(-1)=-1$