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Đặt 20212020=x
=>\(A=\dfrac{3\left(x+1\right)\left(x+3\right)-5x-2\cdot\left(x+1\right)^2-5}{\left(x+1\right)}\)
\(=\dfrac{3\left(x^2+4x+3\right)-5x-2x^2-4x-2-5}{\left(x+1\right)}\)
\(=\dfrac{3x^2+12x+9-2x^2-9x-7}{x+1}=\dfrac{x^2+3x+2}{x+1}=x+2\)
=20212022
\(\left(5+5^2+5^3+...+5^{10}\right)+4x-1=\frac{1}{4}5^{11}+\frac{1}{2}x+3\)
\(\Leftrightarrow\left(1+5+5^2+5^3+...+5^{10}\right)+4x-2=\frac{1}{4}5^{11}+\frac{1}{2}x+3\)(1)
1./ Trước tiên, ta tính:
\(S=1+5+5^2+5^3+...+5^{10}\)
\(\left(5-1\right)\cdot S=\left(5-1\right)\left(1+5+5^2+5^3+...+5^{10}\right)\)
\(\Leftrightarrow4S=5^{11}-5^{10}+5^{10}-5^9+...+5-1=5^{11}-1\)
\(\Leftrightarrow S=\frac{5^{11}-1}{4}=\frac{1}{4}5^{11}-\frac{1}{4}\)
2./ (1) trở thành:
\(\Leftrightarrow\frac{1}{4}5^{11}-\frac{1}{4}+4x-2=\frac{1}{4}5^{11}+\frac{1}{2}x+3\)
\(\Leftrightarrow4x-\frac{1}{2}x=5+\frac{1}{4}\Leftrightarrow\frac{7}{2}x=\frac{21}{4}\)
\(\Leftrightarrow x=\frac{3}{2}\).
\(\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)
\(\Leftrightarrow\)\(\frac{3}{35}-\frac{3}{5}-x=\frac{2}{7}\)
\(\Leftrightarrow\)\(-x=\frac{2}{7}-\frac{3}{35}+\frac{3}{5}\)
\(\Leftrightarrow-x=\frac{4}{5}\)
\(\Leftrightarrow x=-\frac{4}{5}\)
\(\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)
\(\frac{3}{5}+x=\frac{3}{35}-\frac{2}{7}\)
\(\frac{3}{5}+x=\frac{-1}{5}\)
\(x=\frac{-1}{5}-\frac{3}{5}\)
\(x=\frac{-4}{5}\)
vậy \(x=\frac{-4}{5}\)
a)\(\frac{3}{7}\left(-\frac{5}{2}\right)+\left(-\frac{3}{7}\right)\)
\(=\left(-\frac{15}{17}\right)+\left(-\frac{3}{7}\right)\)
\(=-\frac{156}{119}\)
b) \(\frac{2}{3}+\frac{3}{4}\left(-\frac{4}{9}\right)\)
\(=\frac{2}{3}+-\frac{1}{3}=\frac{1}{3}\)
áp dụng công thức \(\frac{n\left(n-1\right)}{2}\)
<=>\(\frac{114\cdot\left(114-1\right)}{2}\)
<=> A =6441
A=1+2-3-4+5+6-7-8+...-111-112+113+114
A=1+(2-3-4+5)+(6-7-8+9)+...+(110-111-112+113)+114
A=1+ 0 +0 +.........+0+114
A=115