Câu 1: 

a:=17+5=22

b: =40-25=15

ai giúp em đi mờ

em tick choa

Câu 1 : -123 + 77 + ( - 257 ) + 23 - 43

          = ( -123 + 23 ) + ( -257 + 77 ) - 43

          = -100 + ( -180 ) - 43 

          = -280 - 43 

          = -323

Câu 2 : ( 36 + 79 ) + ( 145 - 79 - 36 )

           = 36 + 79 + 145 - 79 - 36 

           = ( 36 - 36 ) + ( 79 - 79 ) + 145

           = 0 + 0 + 145

           = 145

Câu 1 : ( -123) + 77 + (-257) + 23 - 43 =

        = [ (-123) +23 ] + [ (-257) - 43 ) + 77

        = (-100) + (-300) + 77

        = (-400) + 77

        = -323

Câu 2 : ( 36 + 79 ) + ( 145 - 79 - 36 ) =

        = 36 + 79 + 145 - 79 - 36

        = ( 36 - 36 ) + ( 79 - 79 ) + 145

        = 0 + 0 + 145 

        = 145

Học tốt nhé###

Lời giải:

$27,75+25,27=53,02$

$12+8(-8)=12-64=-52$

$(-321)-[(-321+5^2)-225]=-321+321-25+225=-25+225=200$

27,75 + 25,27

= 53,02

--------------------

12 + 8. (-8)

= 12 + (-64)

= -52

---------------------

(-321) - [(-321 + 5²) - 225]

= (-321) - [(-321 + 25) - 225]

= (-321) - [(-296) - 225]

= (-321) - (-521)

= -321 + 521

= 200

= 25 - 225

= -200

Câu 2

\((1) MnO_2 + 4HCl \to MnCl_2 + Cl_2 + 2H_2O\\ (2) Cl_2 + H_2 \xrightarrow{as} 2HCl\\ (3) 3Cl_2 + 2Fe \xrightarrow{t^o} 2FeCl_3\\ (4) 2FeCl_3 + Fe \to 3FeCl_2\\ (5) 2NaOH + Cl_2 \to NaCl + NaClO + H_2O\)

\((1) 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ (2) 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3\\ (3) C + O_2 \xrightarrow{t^o} CO_2\\ (4) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ (5) 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ (6) 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ (7) Fe + H_2SO_4 \to FeSO_4 + H_2\\ (8) Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O\\ (9) 2Fe + 6H_2SO_4 \to Fe_2(SO_4)_3 + 3SO_2 + 6H_2O\\ (10) 2Al + 6H_2SO_4 \to Al_2(SO_4)_3 + 3SO_2 + 6H_2O\)

\(77.\left(-299\right)-77\)

\(=-23023-77\)

\(=-23100\)

77.(-299)-77=-23023-77=-23100.

a: \(-177-\left\lbrack89+\left(-177\right)+75\right\rbrack-\left(-25\right)\)

=-177-[89-177+75]+25

=-177-89+177-75+25

=-89-50

=-139

b: \(2\cdot\left\lbrack16+\left(-12\right)-5\right\rbrack+\left(-32\right)-\left(-10\right)\)

\(=2\cdot\left\lbrack16-17\right\rbrack-32+10\)

=2*(-1)-32+10

=-34+10

=-24

c: \(56\cdot53-\left\lbrack57+\left(-75\right)-\left(-9\right)\cdot2\right\rbrack\)

=2968-[57-75+18]

=2968

d: \(16\cdot\left(-45\right)+16\cdot34-\left(-16\right)\cdot\left(-89\right)\)

\(=16\left(-45+34\right)-16\cdot89\)

\(=16\left(-11-89\right)=16\cdot\left(-100\right)=-1600\)

e: \(77\cdot18-\left(-77\right)\cdot60+\left(-77\right)\cdot\left(-22\right)\)

\(=77\cdot18+77\cdot60+77\cdot22\)

\(=77\left(18+60+22\right)=77\cdot100=7700\)

dấu * là sao

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 1\)(ĐPCM)