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

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

=2744-33856 +1

=-31111

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

=2744-183.185

=2744-33855

=-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

Bài 2

a) x(x - 2012) - 2013x + 2012.2013 = 0

⇔ x(x - 2012) - 2013(x - 2012) = 0

⇔ (x - 2012)(x - 2013) = 0

⇔ x - 2012 = 0 hoặc x - 2013 = 0

*) x - 2012 = 0

⇔ x = 2012

*) x - 2013 = 0

⇔ x = 2013

Vậy x = 2012; x = 2013

b) (x - 1)³ + 1 + 3x(x - 4) = 0

⇔ x³ - 3x² + 3x - 1 + 1 + 3x² - 12x = 0

⇔ x³ - 9x = 0

⇔ x(x² - 9) = 0

⇔ x(x - 3)(x + 3) = 0

⇔ x = 0 hoặc x - 3 = 0 hoặc x + 3 = 0

*) x - 3 = 0

⇔ x = 3

*) x + 3 = 0

⇔ x = -3

Vậy x = -3; x = 0; x = 3

c) (x + 4)² - 16 = 0

⇔ (x + 4)² - 4² = 0

⇔ (x + 4 - 4)(x + 4 + 4) = 0x

⇔ (x + 8) = 0

⇔ x = 0 hoặc x + 8 = 0

*) x + 8 = 0

⇔ x = -8

Vậy x = -8; x = 0

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

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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:\(\frac{-1}{183}<\frac{1}{184}\)

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