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

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: \(347\cdot2^2-2^2\cdot\left(216+184\right):8\)

\(=347\cdot4-4\cdot400:8\)

\(=347\cdot4-4\cdot50=4\cdot\left(347-50\right)=4\cdot297=1188\)

2: \(132-\left\lbrack116-\left(132-128\right)^2\right\rbrack\)

\(=132-\left\lbrack116-4^2\right\rbrack\)

=132-(116-16)

=132-100

=32

3: \(16:\left\lbrace400:\left\lbrack200-\left(37+46\cdot3\right)\right\rbrack\right\rbrace\)

=16:{400:[200-(37+138)]}

=16:{400:[200-175]}

=16:{400:25}

=16:16

=1

4: \(\left\lbrace184:\left\lbrack96-124:31\right\rbrack-2\right\rbrace\cdot3651\)

\(=\left\lbrace\frac{184}{96-4}-2\right\rbrace\cdot3651\)

=(184:92-2)*3651

=0

5: \(46-\left\lbrack\left(16+71\cdot4\right):15\right\rbrack-2\)

=46-[(16+284):15]-2

=46-300:15-2

=46-20-2

=46-22

=24

6: \(3^3\cdot18+72\cdot4^2-41\cdot18\)

\(=18\left(3^3-41\right)+72\cdot16\)

\(=18\cdot\left(27-41\right)+18\cdot64\)

=18(-14+64)

=18*50

=900

7: \(\left(56\cdot46-25\cdot23\right):23\)

\(=56\cdot\frac{46}{23}-25\)

=112-25

=87

8: \(\left(28\cdot54+56\cdot36\right):21:2\)

\(=18\cdot28\left(3+2\cdot2\right):42\)

\(=18\cdot\frac{28}{42}\cdot7=18\cdot\frac23\cdot7=12\cdot7=84\)

9: \(\left(76\cdot34-19\cdot64\right):\left(38\cdot9\right)\)

\(=\frac{19\cdot\left(4\cdot34-64\right)}{38\cdot9}=\frac{4\cdot34-64}{9\cdot2}=\frac{4\cdot\left(34-16\right)}{18}=4\)

10: \(\left(2+4+6+\cdots+100\right)\cdot\left(36\cdot333-108\cdot111\right)\)

\(=\left(2+4+6+\cdots+100\right)\cdot36\cdot111\left(3-3\right)\)

=0

11: \(\left(5\cdot4^{11}-3\cdot16^5\right):4^{10}\)

\(=\left(5\cdot4^{11}-3\cdot4^{10}\right):4^{10}\)

\(=5\cdot\frac{4^{11}}{4^{10}}-3\cdot\frac{4^{10}}{4^{10}}\)

=20-3

=17

12: \(\frac{7256\cdot4375-725}{3650+4375\cdot7255}=\frac{7255\cdot4375+4375-725}{7255\cdot4375+3650}\)

\(=\frac{7255\cdot4375+3650}{7255\cdot4375+3650}\)

=1

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

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