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#)Giải :
a)\(9^2\div\left(27^3.81^2\right)=9^2\div\left[\left(3^3\right)^3.\left(9^2\right)^2\right]=9^2\div\left(3^9.9^4\right)\)
Tự lm típ
a,\(9^2:\left(27^3.81^2\right)=3^4:\left(3^9.3^8\right)=3^4:\left(3^{9+8}\right)=3^4:3^{17}=3^{-13}\)
a, 32 . 5 - 2x . 6 + 183
= 9 . 5 - 2x . 6 + 1
= 45 - 2x . 6 + 1
= 45 - 2x + 1 . 3 + 1
= 3( 15 - 2x + 1) + 1
b, 100 : { 280 : [ 450 - ( 480 - 4 . 52 ) ] }
= 100 : { 280 : [ 450 - 380 ] }
= 100 : { 280 : 70 }
= 100 : 4 = 25
c, 8 . { 24 - [ 3 . ( 5 + 25 ) + 15 ] : 15 }
= 8 . { 24 - [ 3 . 30 + 15 ] : 15 }
= 8 . { 24 - 105 : 15 }
= 8 . { 24 - 7 }
= 8 . 17 = 136
b ) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
c ) Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
=> A < 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100= 1 - 1/100 = 99/100 < 1
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)< 1
b, \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\)\(\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
c,Ta thấy
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(.....\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
a) \(4.5^2-81:3^2=4.25-81:9=100-9=91\)
b) \(3^3.23-3^3.19=3^3.\left(23-19\right)=27.4=108\)
c) \(2^4.5-[131-\left(13-4\right)^2]=16.5-\left(131-9^2\right)=80-\left(131-81\right)=80-50=30\)
d) \(100:\left\{250:\left[450-\left(4.5^3-2^2.25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-4.\left(125-25\right)\right]\right\}\)
\(=100:\left[250:\left(450-4.100\right)\right]\)
\(=100:\left[250:\left(450-400\right)\right]=100:\left(250:50\right)=100:5=20\)
a, A = 3 . 42 - 81 : 32
A = 3 . 16 - 81 : 9
A = 48 - 9
A = 39
b, B = 1999 + [ 100 : ( 8 - 6 )2 - 22 . 3 ]
B = 1999 + [ 100 : 22 - 4 . 3 ]
B = 1999 + [ 100 : 4 - 12 ]
B = 1999 + [ 25 - 12 ]
B = 1999 + 13
B = 2012
a=39;
b=2012.
Vậy a=39;b=2012.
k cho mình nhé