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\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{2020}\left(1+2+3+...+2020\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{2020}.\frac{2020.2021}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{2021}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{2021}{2}\)
\(=\frac{\left[\left(2021-2\right)+1\right]\left(2021+2\right)}{2}:2\)
\(=1021615\)
\(1.\)\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{6}-\frac{1}{7}\)
\(M=1-\frac{1}{7}=\frac{6}{7}\)
Mình làm câu 1 thoi nha!
1.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
B1:
\(A=\left(x+2020\right)^4+\left|y-2019\right|-2018\)
+Có: \(\left(x+2020\right)^4\ge0với\forall x\\\left|y-2019\right|\ge0với\forall y\\\Rightarrow \left(x+2020\right)^4+\left|y-2019\right|-2018\ge-2018\\ \Leftrightarrow A\ge-2018 \)
+Dấu "=" xảy ra khi
\(\left(x+2020\right)^4=0\\ \Leftrightarrow x=-2020\)
\(\left|y-2019\right|=0\\ \Leftrightarrow y=2019\)
+Vậy \(A_{min}=-2018\) khi \(x=-2020,y=2019\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+2019\right)\left(x+2020\right)}\)
( ĐKXĐ : \(x\ne\left\{0;-1;-2;...;-2019;-2020\right\}\))
\(=\frac{1}{x}-\frac{1}{\left(x+1\right)}+\frac{1}{\left(x+1\right)}-\frac{1}{\left(x+2\right)}+\frac{1}{\left(x+2\right)}-\frac{1}{\left(x+3\right)}+...+\frac{1}{\left(x+2019\right)}-\frac{1}{\left(x+2020\right)}\)
\(=\frac{1}{x}-\frac{1}{x+2020}\)
\(=\frac{x+2020}{x\left(x+2020\right)}-\frac{x}{x\left(x+2020\right)}\)
\(=\frac{x+2020-x}{x\left(x+2020\right)}\)
\(=\frac{2020}{x\left(x+2020\right)}\)
Bài giải
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2019\right)\left(x+2020\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2019}-\frac{1}{x+2020}\)
\(=\frac{1}{x}-\frac{1}{x+2020}\)
\(=\frac{x+2020}{x\left(x+2020\right)}-\frac{x}{x+2020}=\frac{2020}{x\left(x+2020\right)}\)
Lời giải:
Đặt: \(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{2019^2}=a\).
Biểu thức $A$ lúc đó được biểu diễn như sau:
\(A=a(a-1+\frac{1}{2020^2})-(a+\frac{1}{2020^2})(a-1)\)
\(=a(a-1)+\frac{a}{2020^2}-[a(a-1)+\frac{a-1}{2020^2}]\)
\(=\frac{1}{2020^2}\)
Đặt \(a=\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{2019^2}\)
\(b=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\)
Khi đó : \(D=ab-\left(b+1\right)\left(a-1\right)\)
\(\Rightarrow D=ab-\left(ab+a-b-1\right)\)
\(\Rightarrow D=b-a+1=\frac{1}{2020^2}-1+1=\frac{1}{2020^2}\)