Tính 6^3^1;3^2^3;7^1^2^3^4;2003^2^0^1^0
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Lời giải chi tiết:
6 – 3 – 1 = 2 | 1 + 3 + 2 = 6 | 6 – 1 – 2 = 3 |
6 – 3 – 2 = 1 | 3 + 1 + 2 = 6 | 6 – 1 – 3 = 2 |
6 – 3 – 1 = 2… | 1 + 3 + 2 = 6… | 6 – 1 – 2 = 3… |
6 – 3 – 2 = 1… | 3 + 1 + 2 = 6… | 6 – 1 – 3 = 2… |
1 + 3 + 2 = 6 6 - 3 -1 = 2 6 - 1 - 2 = 3
3 + 1 + 2 = 6 6 - 3 -2 = 1 6 - 1 - 3 = 2
1 + 3 + 2 =6 6 - 3 -1 =2 6 - 1 - 2 =3
3 + 1 + 2 = 6 6 - 3 -2 = 1 6 - 1 - 3 =2
5 +1 = 6 4 + 2 = 6 3 + 3 = 6
6 - 5 = 1 6 - 2 = 4 6 - 3 = 3
6 - 1 = 5 6 - 4 = 2 6 - 6 = 0
5 +1 = 6 4 + 2 = 6 3 + 3 =6
6 - 5 =1 6 - 2 =4 6 - 3 =2
6 - 1 = 5 6 - 4 = 2 6 - 6 =0
4 + 5 = 9 6 + 3 = 9 1 + 8 = 9
4 + 1 + 4 = 9 6 + 1 + 2 = 9 1 + 2 + 6 = 9
4 + 2 + 3 = 9 6 + 3 + 0 = 9 1 + 5 + 3 = 9
4+5 =9 . 6+3=9 1+8=9 4+1+4=9 6+1+2=10 1+2+6=9 4+2+3=9 6+3+0=9 1+5+3=9 Đúng thì cho chứ đừng cho sai nha
Lời giải chi tiết:
6 – 5 – 1 = 0 | 6 – 4 – 2 = 0 | 6 – 3 – 3 = 0 |
6 – 1 – 5 = 0 | 6 – 2 – 4 = 0 | 6 – 6 = 0 |
\(A=\frac{1}{3}+\frac{1}{3+6}+\frac{1}{3+6+9}+...+\frac{1}{3+6+9+...+2013}\)
\(A=\frac{1}{3}.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...671}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{\left(1+0\right).2:2}+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+671\right).671:2}\right)\)
\(A=\frac{1}{3}.\left(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{671.672}\right)\)
\(A=\frac{1}{3}.2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{671.672}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{671}-\frac{1}{672}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{672}\right)\)
\(A=\frac{2}{3}.\frac{671}{672}=\frac{671}{1008}\)
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: \(x=-\dfrac{134}{81}\)
\(\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{4}{12}+\dfrac{9}{12}+\dfrac{6}{12}=\dfrac{19}{12}\)
\(\dfrac{6}{8}+\dfrac{2}{4}+\dfrac{6}{24}=\dfrac{3}{4}+\dfrac{2}{4}+\dfrac{1}{4}=\dfrac{6}{4}=\dfrac{3}{2}\)
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{5}{6}=\dfrac{10}{6}=\dfrac{5}{3}\)
\(\dfrac{3}{5}+\dfrac{3}{2}+2=\dfrac{6}{10}+\dfrac{15}{10}+\dfrac{20}{10}=\dfrac{41}{10}\)
\(\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{13}{12}+\dfrac{1}{2}=\dfrac{19}{12}\\ \dfrac{6}{8}+\dfrac{2}{4}+\dfrac{6}{24}=\dfrac{5}{4}+\dfrac{6}{24}=\dfrac{3}{2}\\ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{5}{6}+\dfrac{5}{6}=\dfrac{10}{6}\\ \dfrac{3}{5}+\dfrac{3}{2}+2=\dfrac{21}{10}+\dfrac{20}{10}=\dfrac{41}{10}\)