Tính B=\(2020.\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).....\left(\frac{1}{2019}-1\right)\left(\frac{1}{2020}-1\right)\)
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21 tháng 10 2020
Xét \(f\left(x\right)+f\left(1-x\right)=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3+3x+3-6x+3x^2}\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3x+3x^2}\)
\(=\frac{1-3x+3x^2}{1-3x+3x^2}=1\)
Thay vào ta tính được:
\(A=\left[f\left(\frac{1}{2020}\right)+f\left(\frac{2019}{2020}\right)\right]+...+\left[f\left(\frac{1009}{2020}\right)+f\left(\frac{1011}{2020}\right)\right]+f\left(\frac{1010}{2020}\right)\)
\(A=1+...+1+f\left(\frac{1010}{2020}\right)\) (với 1009 số 1)
\(A=1009+f\left(\frac{1}{2}\right)=1009+\frac{\left(\frac{1}{2}\right)^3}{1-3\cdot\frac{1}{2}+3\cdot\left(\frac{1}{2}\right)^2}\)
\(A=1009+\frac{1}{2}=\frac{2019}{2}\)
Vậy \(A=\frac{2019}{2}\)
12 tháng 8 2020
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+2019\right)\left(x+2020\right)}\)
( ĐKXĐ : \(x\ne\left\{0;-1;-2;...;-2019;-2020\right\}\))
\(=\frac{1}{x}-\frac{1}{\left(x+1\right)}+\frac{1}{\left(x+1\right)}-\frac{1}{\left(x+2\right)}+\frac{1}{\left(x+2\right)}-\frac{1}{\left(x+3\right)}+...+\frac{1}{\left(x+2019\right)}-\frac{1}{\left(x+2020\right)}\)
\(=\frac{1}{x}-\frac{1}{x+2020}\)
\(=\frac{x+2020}{x\left(x+2020\right)}-\frac{x}{x\left(x+2020\right)}\)
\(=\frac{x+2020-x}{x\left(x+2020\right)}\)
\(=\frac{2020}{x\left(x+2020\right)}\)
F
12 tháng 8 2020
Bài giải
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2019\right)\left(x+2020\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2019}-\frac{1}{x+2020}\)
\(=\frac{1}{x}-\frac{1}{x+2020}\)
\(=\frac{x+2020}{x\left(x+2020\right)}-\frac{x}{x+2020}=\frac{2020}{x\left(x+2020\right)}\)
10 tháng 2 2020
Thật ra tui cũng không rõ lắm đâu. Cậu thử nhân A với \(\dfrac{2019}{2020}\)rồi lại cộng lại với A thử coi nào <Chú Ý : chưa chắc đã đúng >
NB
19 tháng 12 2019
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 \(x=-2020,y=2019\)
HT
18 tháng 10 2020
iem chỉ biết làm câu đầu , NHƯNG KO BÍT có ĐUG HAY KO
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2019}\right)\left(1-\frac{1}{2020}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2018}{2019}\cdot\frac{2019}{2020}\)
\(A=\frac{1\cdot2\cdot3\cdot...\cdot2018\cdot2019}{2\cdot3\cdot4\cdot..\cdot2019\cdot2020}\)
\(A=\frac{1}{2020}\)
TA
18 tháng 10 2020
Với \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{2019}\right)\left(1-\frac{1}{2020}\right)\) , ta có : \(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{2018}{2019}\cdot\frac{2019}{2020}=\frac{1}{2020}\)
Ta có \(7A=\frac{7}{2020}\) , \(9A=\frac{9}{2020}\) , \(1+A=\frac{2021}{2020}\)
\(\frac{1+7A}{1+9A}=\frac{1+\frac{7}{2020}}{1+\frac{9}{2020}}=\frac{\frac{2027}{2020}}{\frac{2029}{2020}}\)
Ta thấy \(\frac{\frac{2027}{2020}}{\frac{2029}{2020}}\)có tử kém mẫu \(\frac{2}{2020}\)đơn vị và không thể rút gọn được nữa .
\(\Rightarrow\frac{1+7A}{1+9A}\)là p/s tối giản.
XO
25 tháng 8 2020
Ta có :\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=36\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=36\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=12\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
\(\Rightarrow\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}=\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\)
=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{ab}-\frac{2}{bc}-\frac{2}{ca}=0\)
=> \(\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\left(\frac{1}{b^2}-\frac{2}{bc}+\frac{1}{c^2}\right)+\left(\frac{1}{c^2}-\frac{2}{ac}+\frac{1}{a^2}\right)=0\)
=> \(\left(\frac{1}{a}-\frac{1}{b}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2+\left(\frac{1}{c}-\frac{1}{a}\right)^2=0\)
=> \(\hept{\begin{cases}\frac{1}{a}-\frac{1}{b}=0\\\frac{1}{b}-\frac{1}{c}=0\\\frac{1}{c}-\frac{1}{a}=0\end{cases}}\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)
Khi đó \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\Leftrightarrow3\frac{1}{a}=6\Rightarrow\frac{1}{a}=2\Leftrightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=2\)
Khi đó Đặt P = \(\left(\frac{1}{a}-3\right)^{2020}+\left(\frac{1}{b}-3\right)^{2020}+\left(\frac{1}{c}-3\right)^{2020}\)
= (2 - 3)2020 + (2 - 3)2020 + (2 - 3)2020
= 1 + 1 + 1 = 3
Vậy P = 3