\(M=\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2020^2}\right)X\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)-\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)X\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)\)Làm nhanh và ngắn gọn nhất có thể nhé ! mình tik cho 10 tik
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Tính giá trị của :
D=\(\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}\right)x\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)-\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)x\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}\right)\)
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 : \(D=ab-\left(b+1\right)\left(a-1\right)\)
\(\Rightarrow D=ab-\left(ab+a-b-1\right)\)
\(\Rightarrow D=b-a+1=\frac{1}{2020^2}-1+1=\frac{1}{2020^2}\)
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a) \(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)
b) \(\frac{x-2018}{2}+\frac{x-2020}{4}=\frac{x-2040}{8}+\frac{x-2030}{14}\)
\(a,\left(\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|\right):10=\left(1-\frac{1}{2}\right)....\left(1-\frac{1}{10}\right)\)
\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\Leftrightarrow\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|=1\)
\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.|x-2|=1\Leftrightarrow|x-2|.\frac{2}{3}=1\Leftrightarrow|x-2|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
\(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)
\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\)
\(\Leftrightarrow\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|=1\)
\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.\left|x-2\right|=1\)
\(\Leftrightarrow\left|x-2\right|.\frac{2}{3}=1\Leftrightarrow\left|x-2\right|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
Mình làm tiếp câu b nha !
b, Bài giải
\(\frac{x-2018}{2}+\frac{x-2020}{4}=\frac{x-2040}{8}+\frac{x-2030}{14}\)
\(\left(\frac{x-2018}{2}+1\right)+\left(\frac{x-2020}{4}+1\right)=\left(\frac{x-2040}{8}+1\right)+\left(\frac{x-2030}{14}+1\right)\)
\(\frac{x-2016}{2}+\frac{x-2016}{4}=\frac{x-2032}{8}+\frac{x-2016}{14}\)
\(\left(x-2016\right)\left(\frac{1}{2}+\frac{1}{4}\right)=\frac{x-2016}{8}-2+\frac{x-2016}{14}\)
\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\left(\frac{1}{8}+\frac{1}{14}\right)-2\)
\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\cdot\frac{11}{56}-2\)
\(\left(x-2016\right)\cdot\frac{3}{4}-\left(x-2016\right)\cdot\frac{11}{56}=-2\)
\(\left(x-2016\right)\left(\frac{3}{4}-\frac{11}{56}\right)=-2\)
\(\left(x-2016\right)\cdot\frac{31}{56}=-2\)
\(x-2016=-2\text{ : }\frac{31}{56}\)
\(x-2016=-\frac{112}{31}\)
\(x=-\frac{112}{31}+2016\)
\(x=\frac{62384}{31}\)
Cho \(\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\\\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=12\end{cases}}\)tính \(\left(\frac{1}{a}-3\right)^{2020}+\left(\frac{1}{b}-3\right)^{2020}+\left(\frac{1}{c}-3\right)^{2020}\)
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
Tính :
A = \(\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}\right).\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)-\)\(\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right).\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}\right)\)
Giúp em với ạ @Nguyễn Việt Lâm,@Akai Haruma
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:
\(A=a(a-1+\frac{1}{2020^2})-(a+\frac{1}{2020^2})(a-1)\)
\(=a(a-1)+\frac{a}{2020^2}-[a(a-1)+\frac{a-1}{2020^2}]\)
\(=\frac{1}{2020^2}\)
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Tính giá trị biểu thức\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{2020}\left(1+2+3+..+2020\right)\)
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{2020}\left(1+2+3+...+2020\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{2020}.\frac{2020.2021}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{2021}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{2021}{2}\)
\(=\frac{\left[\left(2021-2\right)+1\right]\left(2021+2\right)}{2}:2\)
\(=1021615\)
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Giải các phương trình sau:
a) left(frac{x-2}{x-1}right)^2-5left(frac{x+2}{x+1}right)^2+4left(frac{x^2-4}{x^2-1}right)1
b) left(frac{x-1}{x}right)^2+left(frac{x-1}{x-2}right)^2frac{40}{9}
c) x.frac{4-x}{x+2}.left(frac{8-2x}{x+2}right)3
d) frac{1}{3x-2020}+frac{1}{4x-2018}+frac{1}{5x-2017}frac{1}{12x-2019}
Đọc tiếp
Giải các phương trình sau:
a) \(\left(\frac{x-2}{x-1}\right)^2-5\left(\frac{x+2}{x+1}\right)^2+4\left(\frac{x^2-4}{x^2-1}\right)=1\)
b) \(\left(\frac{x-1}{x}\right)^2+\left(\frac{x-1}{x-2}\right)^2=\frac{40}{9}\)
c) \(x.\frac{4-x}{x+2}.\left(\frac{8-2x}{x+2}\right)=3\)
d) \(\frac{1}{3x-2020}+\frac{1}{4x-2018}+\frac{1}{5x-2017}=\frac{1}{12x-2019}\)
a,\(\frac{1}{3}+\frac{-4}{3}:x=\frac{-5}{6}\)
b, \(\frac{1}{2}x\frac{3}{5}.\left(x-2\right)=3\)
c,\(\left(2-\frac{1}{3}x\right)2018=\left(2-\frac{1}{3}x\right)2020\)
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
B1 : Tìm GTNN :
\(\left(x+2020\right)^4+\left|y-2019\right|-2018\)
B2 : Tính :
\(P=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{2019}.\left(1+2+3+...+2019\right)\)
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 \(x=-2020,y=2019\)