Câu 1
BIẾt rằng: \(2^2+3^2+4^2+...+13^2=818\)
Tính: \(A=1^2+3^2+6^2+9^2+12^2+...+39\)
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\(A=1^2+3^2+6^2+9^2+12^2+...+39^2\)
\(A=1^2+3^2+3^2\left(2^2+3^2+4^2+....+13^2\right)\)
\(A=1+9+9.818\)
\(A=10+7362=7372\)
\(2^2+3^2+...+13^2=818\\ \Rightarrow3^2\left(2^2+3^2+...+39^2\right)=9.818=7362\\ \Rightarrow1^2+3^2+...+39^2=7363\)
A = 1 + 3^2+(6^2+9^2+....+39^2)
= 10 + 3^2.(2^2+3^2+....+13^2) = 10 + 9. 818 = 7372
a: \(B=\frac12-\left\lbrack\frac38+\left(-\frac74\right)\right\rbrack\)
\(=\frac12-\frac38+\frac74\)
\(=\frac48-\frac38+\frac{14}{8}=\frac{15}{8}\)
b: \(-\frac{4}{12}-\left(-\frac{13}{39}-0,25\right)+0,75\)
\(=-\frac13+\frac13+0,25+0,75\)
=0,25+0,75
=1
c: \(\frac12-\frac13+\frac{1}{23}+\frac16\)
\(=\left(\frac12-\frac13+\frac16\right)+\frac{1}{23}\)
\(=\left(\frac16+\frac16\right)+\frac{1}{23}=\frac13+\frac{1}{23}=\frac{26}{69}\)
d: \(\left(-\frac{13}{7}-\frac49\right)-\left(-\frac{10}{7}-\frac49\right)\)
\(=-\frac{13}{7}-\frac49+\frac{10}{7}+\frac49\)
\(=-\frac{13}{7}+\frac{10}{7}=-\frac37\)
e: \(\left(\frac78-\frac52+\frac47\right)-\left(-\frac37+1-\frac{13}{8}\right)\)
\(=\frac78-\frac52+\frac47+\frac37-1+\frac{13}{8}\)
\(=\frac{20}{8}-\frac52=0\)
f: \(-\frac37+\left(3-\frac34\right)-\left(2,25-\frac{10}{7}\right)\)
\(=-\frac37+3-\frac34-2,25+\frac{10}{7}\)
\(=\frac77+0,75-\frac34\)
=1
g: \(\left(\frac53-\frac37+9\right)-\left(2+\frac57-\frac23\right)+\left(\frac87-\frac43-10\right)\)
\(=\frac53-\frac37+9-2-\frac57+\frac23+\frac87-\frac43-10\)
\(=\left(\frac53+\frac23-\frac43\right)+\left(-\frac37-\frac57+\frac87\right)+\left(9-2-10\right)\)
\(=\frac33+7-10=1-3=-2\)