a) ( 100   –   1 ) 3  = 970299.                          b) ( 91   +   9 ) 3   =   100 3 .

c) ( 1000   +   1 ) 3  = 1003003001.                d) ( 102   –   2 ) 3   =   100 3 .

\(\left(7x-11\right)^3=2^5.5^2+2.10^2\)

\(\Rightarrow\left(7x-11\right)^3=32.25+2.100\)

\(\Rightarrow\left(7x-11\right)^3=800+200\)

\(\Rightarrow\left(7x-11\right)^3=1000\)

\(\Rightarrow\left(7x-11\right)^3=10^3\)

\(\Rightarrow7x-11=10\)

\(\Rightarrow7x=21\)

\(\Rightarrow x=3\)

(7\(x\) - 11)³ = 2⁵.5² + 2.10²

(7\(x\) - 11)³ = 1000

(7\(x\) - 11)³ = 10³

7\(x\) - 11 = 10

7\(x\)        = 10 + 11

7\(x\)        = 21

   \(x\)       = 21:7

   \(x\)        = 3

\(\left(7x-11\right)^3=2^5.5^2+2.10^2\)

\(\left(7x-11\right)^3=2^5.5^2+2.\left(2.5\right)^2\)

\(\left(7x-11\right)^3=2^5.5^2+2.2^2.5^2\)

\(\left(7x-11\right)^3=2^5.5^2+2^3.5^2\)

\(\left(7x-11\right)^3=5^2\left(2^5+2^3\right)\)

\(\left(7x-11\right)^3=5^2\left(32+8\right)\)

\(\left(7x-11\right)^3=5^2.40\)

\(\left(7x-11\right)^3=25.40\)

\(\left(7x-11\right)^3=1000\)

\(\left(7x-11\right)^3=10^3\)

\(\Rightarrow7x-11=10\)

\(\Rightarrow7x=10+11\)

\(\Rightarrow7x=21\)

\(\Rightarrow x=21:7\)

\(\Rightarrow x=3\)

a)2

b)63

a) ( - 2 . 10^2 - 6 . 10^2 + 10^3 ) : 100

= ( - 2 . 10^2 - 6 . 10^2 + 10 . 10^2 ) : 100

= [ 10^2 . ( - 2 - 6 + 10 ) ] : 100

= [ 100 . 2 ] : 100

= 200 : 100

= 2

b) ( 2 . 27^2 + 3^8 - 4 . 9^3 ) : 9^2

= [ 2 . ( 3^3 )^2 + 3^6 . 3^2 - 4 . ( 3^2 )^3 ) : 9^2

= [ 2 . 3^6 + 3^6 . 3^2 - 4 . 3^6 ] : 9^2

= [ 3^6 . ( 2 + 3^2 - 4 ) ] : 9^2

= [ 3^6 . 7 ] : 81

= 5103 : 81

= 63

a)  2 3

b)  1 100

\(\dfrac{101}{2}.\dfrac{102}{2}.\dfrac{103}{2}.\dfrac{104}{2}.....\dfrac{200}{2}\\ =\dfrac{101.102.103.104.....200}{2^{100}}\\ =\dfrac{\left(101.102.103.....200\right)\left(1.2.3.....100\right)}{2^{100}.\left(1.2.3.....100\right)}\\ =\dfrac{1.2.3.....200}{\left(2.1\right)\left(2.2\right)\left(2.3\right).....\left(2.100\right)}\\ =\dfrac{\left(1.3.5.....199\right)\left(2.4.6.....200\right)}{4.6.8.....200}\\ =1.3.5.7.....197.199\)

=> Điều phải chứng minh

\(\frac{101}{2}\times\frac{102}{2}\times\frac{103}{2}\times...\times\frac{200}{2}\)

\(=\frac{1.2.3.....100.101.102.103.....200}{1.2.3.....100.2^{100}}\)

\(=\frac{\left(1.3.5.....199\right).\left(2.4.6.....200\right)}{\left(1.2\right).\left(2.2\right).\left(3.2\right).....\left(100.2\right)}\)

\(=1.3.5.....199\)

sai toét

sai rồi

c: \(\Leftrightarrow2x+1=3\)

hay x=1