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\(21.[(1 245 + 987) : 2^3 - 15.12] + 21\)
= 21.(2232 : 8 - 180) + 21
= 21.(279 - 180) + 21
= 21. 99 + 21
= 21. (99+1)
= 21. 100
= 2 100.
M = 1 + 5 + 52 +....+ 521
5M = 5 + 52 + .... + 522
5M - M = 522 - 1
4M = 522 - 1
4M + 4 = 522 - 1 + 4
4M + 4 = 522 + 3
Ta có : M = 1 + 5 + 52 + 53 + ..... + 521
=> 5M = 5 + 52 + 53 + ..... + 522
=> 5M - M = 522 - 1
=> 4M = 522 - 1
=> 4M + 4 = 522 - 1 + 4
=> 4M + 4 = 522 + 3
Bài 1:
\(A=\dfrac{-1}{3}+1+\dfrac{1}{3}=1\)
\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)
\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)
Bài 2:
a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)
\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)
\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)
b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)
Bài 3:
\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)
\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)
Bài 4:
\(\dfrac{3}{4}-x=1\)
\(\Rightarrow-x=1-\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
\(x+4=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{5}-4\)
\(\Rightarrow x=-\dfrac{19}{5}\)
Vậy: \(x=-\dfrac{19}{5}\)
\(x-\dfrac{1}{5}=2\)
\(\Rightarrow x=2+\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{11}{5}\)
Vậy: \(x=\dfrac{11}{5}\)
\(x+\dfrac{5}{3}=\dfrac{1}{81}\)
\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)
\(\Rightarrow x=-\dfrac{134}{81}\)
Vậy: \(x=-\dfrac{134}{81}\)
\(a,\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-\frac{20}{15}+\frac{3}{7}\)
\(=>\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac{7}{21}+\frac{3}{7}\right)-\frac{20}{15}\)
\(=>1+\frac{16}{21}-\frac{20}{15}\)
\(=>\frac{37}{21}-\frac{20}{15}\)
\(=>\frac{3}{7}\)
\(b,12-8\cdot\left(\frac{3}{2}\right)^3\)
\(=>12-8\cdot\frac{27}{8}\)
\(=>12-27\)
\(=>-15\)
\(c,\left(\frac{1}{9}\right)^{2005}\cdot9^{2005}-96^2:24^2\)
\(=>\left(\frac{1^{2005}^{ }}{9^{2005}}\cdot9^{2005}\right)-\left(96^2:24^2\right)\)
\(=>\left(1^{2005}\right)-16\)
\(=>1-16\)
\(=>-15\)
a) ( 12 + 21 - 23 ) - ( 23 - 21 + 10 )
= 12 + 21 - 23 - 23 - 21 + 10
= ( 12 + 10 ) - ( 21 - 21 ) - ( 23 - 23 )
= 22 - 0 - 0
= 22
b) ( 55 + 45 + 15 ) - ( 15 - 55 + 45 )
= 55 + 45 + 15 - 15 - 55 + 45
= ( 55 + 45 ) + ( 55 + 45 ) + ( 15 - 15)
= 100 + 100 + 0
= 200
\(a)\)\(-\left(12+21-23\right)-\left(23-21+10\right)\)
\(=-12-21+23-23+21-10\)
\(=\left(-21+21\right)+23-23-12-10\)
\(=0+0-12-10\)
\(=-22\)
\(b)\)\(\left(55+45+15\right)-\left(15-55+45\right)\)
\(=55+45+15-15+55-45\)
\(=\left(15-15\right)+\left(45-45\right)+55+55\)
\(=0+0+55+55\)
\(=110\)
=21 .[ 2232 : 8 - 180 ] + 21
=21.99 +21
=2100
\(21\cdot\left[\left(1245+987\right):2^3-15\cdot12\right]+21\)
\(=21\left(279-180+1\right)\)
\(=21\cdot100=2100\)