98.28-(184-1)(184+1)

=98.28-\(184^2\)+1

=2744-33856 +1

=-31111

\(Bài.1:\\ a,104^2-16=104^2-4^2=\left(104+4\right)\left(104-4\right)=108.100=10800\\ b,9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\\ =\left(9.2\right)^8-\left(18^8-1\right)=18^8-18^8+1=1\\ c,999^3+3.999^2+3.999+1\\ =999^3+3.999^2.1+3.999.1^2+1^3=\left(999+1\right)^3=1000^3=1000000000\\ d,42^3-6.42^2+12.42-8\\ =42^3-3.42^2.2+3.42.2^2-2^3\\ =\left(42-2\right)^3=40^3=64000\)

Bài 1

a) 104² - 16

= 104² - 4²

= (104 - 4)(104 + 4)

= 100.108

= 10800

b) 9⁸.2⁸ - (18⁴ - 1)(18⁴ + 1)

= 18⁸ - (18⁸ - 1)

= 18⁸ - 18⁸ + 1

= 1

c) 999³ + 3.999² + 3.999 + 1

= (999 + 1)³

= 1000³

= 1000000000

d) 42³ - 6.42² + 12.42 - 8

= (42 - 2)³

= 40³

= 64000

1/Vì 179/197<1 ; 971/917>1

=>179/197<971/917

2/Vì 183/184<1 ; 184/183>1

=>183/184<184/183

3/Ta có : -3/31=-3*101/31*101=-303/3131

Vì -303>-789 =>-303/3131>-789/3131 =>-3/31>-789/3131

1/>

2/<

3/>

Duyệt lẹ !Duyệt lẹ !Duyệt lẹ !

1) Ta có: \(\frac{179}{197}<1;\frac{971}{917}>1\)

=> \(\frac{179}{197}<1<\frac{971}{917}\)

=> \(\frac{179}{197}<\frac{971}{917}\)

2) Ta có: \(\frac{183}{184}<1;\frac{184}{183}>1\)

=> \(\frac{183}{184}<1<\frac{184}{183}\)

=> \(\frac{183}{184}<\frac{184}{183}\)

Câu 1:

1) 179/197 và 971/917

Ta có:

\(1-\frac{179}{197}=\frac{18}{197}\)

\(1-\frac{971}{917}=\frac{-54}{917}\)

Mà \(\frac{-54}{917}<\frac{18}{197}\)

\(\Rightarrow\frac{971}{917}<\frac{179}{197}\)

Câu 2:

Ta có:

\(1-\frac{183}{184}=\frac{1}{184}\)

\(1-\frac{184}{183}=\frac{-1}{183}\)

Mà:\(\frac{-1}{183}<\frac{1}{184}\)

\(\Rightarrow\frac{184}{183}<\frac{183}{184}\)