(-1999)-(-234-1999)

=(-1999)+234+1999

=0+234

=234

ok

[- 234] -[ - 1999]= 1765.

[- 1999] - 1765 =

- 3764

*mở ngoặc đơn e đổi dấu í* mk giải chi tiết cho bn nhé  ^^

(-1999)-(-234-1999)

=-1999+234+1999

=(-1999+1999)+234

=0+234=234
hihi <3

a.263

b.-234

c.-167

a, 1000 - ( 137 + 572 ) + ( 263 - 291 )=263

b, ( -1999 ) - ( 234 - 1999 )=-234

c, ( 435 - 167 - 89 ) - ( 435 - 89 )=-167

a ) ( 1456 + 323 ) - 1456

= 1456 + 323 - 1456

= ( 1456 - 1456 ) + 323 = 323

b ) ( - 1999 ) - ( - 234 - 1999)

= - 1999 + 234 + 1999

= ( - 1999 + 1999 ) + 234 = 0 + 234 = 234

giúp mk với

a ) ( 1456 + 323 ) - 1456

= 1456 + 323 - 1456

= ( 1456 - 1456 ) + 323 = 323

b ) ( - 1999 ) - ( - 234 - 1999)

= - 1999 + 234 + 1999

= ( - 1999 + 1999 ) + 234 = 0 + 234 = 234

P/s tham khảo nha

a.(1456+23)-1456

=1456+23-1456

=(1456-1456)+23

=23

b.(-1999)-(-234-1999)

=-1999-(-234)+1999

=(-1999+1999)-(-234)

=-(-234)

=234

c.(116+124)+(215-116-124)

=116+124+215-116-124

=(166-166)+(124-124)+215

=215

d.(435-167-89)-(435-89)

=435-167-89-435+89

=(435-435)-(89-89)-167

=-167

k for more

#hungB

a) = 1456 + 23 - 1456

    = ( 1456 - 1456 ) + 23

    = 0 + 23

    = 23

b) = ( -1999 ) + 234 + 1999

    = ( -1999 + 1999 ) + 234

    = 0 + 234

    = 234

c) = 116 + 124 + 215 - 116 - 124

    = ( 116 - 116 ) + ( 124 - 124 ) + 215

    = 0 + 0 + 215

    = 215

d) = 435 - 167 - 89 - 435 + 89

    = ( 435 - 435 ) + ( -89 + 89 ) - 167

    = 0 + 0 - 167

    = -167

a) 3/4 < 1

115/14 > 1 

=> 3/4 < 115/14

b) 2000/1999 > 1

1998/1999 < 1

=> 2000/1999>1998/1999

c)-17/234 < 0 , 1/1995 > 0

=> -17/234 < 1/1995

d) 23/-15 < 0 ; -17/-49>0

=> 23/-15 < -17/-49

\(A=\dfrac{1999^{1999}+1}{1999^{1998}+1}\)

\(\dfrac{1}{1999}A=\dfrac{1999^{1999}+1}{1999^{1999}+1999}\)

\(\dfrac{1}{1999}A=\dfrac{1999^{1999}}{1999^{1999}}-\dfrac{1998}{1999^{1999}+1999}\)

\(\dfrac{1}{1999}A=1-\dfrac{1998}{1999^{1999}+1999}\)

\(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}\)

\(\dfrac{1}{1999}B=\dfrac{1999^{2000}+1}{1999^{2000}+1999}\)

\(\dfrac{1}{1999}B=\dfrac{1999^{2000}}{1999^{2000}}-\dfrac{1998}{1999^{2000}+1999}\)

\(\dfrac{1}{1999}B=1-\dfrac{1998}{1999^{2000}+1999}\)

Vì  \(\dfrac{1998}{1999^{1999}+1999}>\dfrac{1998}{1999^{2000}+1999}=>\dfrac{1}{1999}A< \dfrac{1}{1999}B=>A< B\)

\(A=\dfrac{1999^{1999}+1}{1999^{1998}+1}=\dfrac{\left(1999^{1999}+1\right)^2}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)

\(A=\dfrac{\left(1999^{1999}\right)^2+2.1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\left(1\right)\)

\(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}=\dfrac{\left(1999^{2000}+1\right)\left(1999^{1998}+1\right)}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)

\(B=\dfrac{\left(1999.1999^{1999}+1\right)\left(\dfrac{1}{1999}.1999^{1999}+1\right)}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)

\(B=\dfrac{\left(1999^{1999}\right)^2+1999.1999^{1999}+\dfrac{1}{1999}.1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)

\(B=\dfrac{\left(1999^{1999}\right)^2+\left(1999+\dfrac{1}{1999}\right).1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\left(2\right)\)

mà \(\left(1999+\dfrac{1}{1999}\right)>2\)

\(\left(1\right).\left(2\right)\Rightarrow A< B\)

Sửa dòng cuối chỗ ''Vì phần mẫu của \(A< B\)'' thành ''Vì phần mẫu của \(\dfrac{1998}{1999^{1999}+1999}< \dfrac{1998}{1999^{2000}+1999}\)'' nhé.

So sánh

\(A=\dfrac{1999^{1999}+1}{1999^{1998}+1}\) ; \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}\)

Ta có: \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}>1\) ( vì tử > mẫu )

Do đó: \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}>\dfrac{1999^{2000}+1+1998}{1999^{1999}+1+1998}=\dfrac{1999^{2000}+1999}{1999^{1999}+1999}=\dfrac{1999.\left(1999^{1999}+1\right)}{1999.\left(1999^{1998}+1\right)}=\dfrac{1999^{1999}+1}{1999^{1998}+1}=A\)

Vậy B > A

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