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Bài 2: 

Ta có: \(11^{1979}< 11^{1980}=1331^{660}\)

\(37^{1320}=37^{2\cdot660}=1369^{660}\)

mà \(1331^{660}< 1369^{660}\)

nên \(11^{1979}< 37^{1320}\)

a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)

=1-2/4=1/2

b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)

c: x-y=0 nên x=y

\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)

=2019

10 tháng 4 2021

a) Quy luật là gì ??

b) 

Đặt

 \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2020}}\\\Rightarrow2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2019}}\\ \Rightarrow2A-A=1-\dfrac{1}{2^{2020}}\Rightarrow A=1-\dfrac{1}{2^{2020}}\)

Suy ra , phương trình trở thành :

213 -x  =13

<=> x=200

NV
22 tháng 4 2022

\(y=\dfrac{1}{3x^2-x-2}=\dfrac{1}{\left(x-1\right)\left(3x+2\right)}=\dfrac{1}{5}.\dfrac{1}{x-1}-\dfrac{3}{5}.\dfrac{1}{3x+2}\)

\(y'=\dfrac{1}{5}.\dfrac{\left(-1\right)^1.1!}{\left(x-1\right)^2}-\dfrac{3}{5}.\dfrac{\left(-1\right)^1.3^1.1!}{\left(3x+2\right)^2}\)

\(y''=\dfrac{1}{5}.\dfrac{\left(-1\right)^2.2!}{\left(x-1\right)^3}-\dfrac{3}{5}.\dfrac{\left(-1\right)^2.3^2.2!}{\left(3x+2\right)^3}\)

\(\Rightarrow y^{\left(n\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^n.n!}{\left(x-1\right)^{n+1}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^n.3^n.n!}{\left(3x+2\right)^{n+1}}\)

\(\Rightarrow y^{\left(2019\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x-1\right)^{2020}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^{2019}.3^{2019}.2019!}{\left(3x+2\right)^{2019}}\)

\(=\dfrac{2019!}{5}\left(\dfrac{3^{2020}}{\left(3x+2\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)

NV
11 tháng 3 2022

\(F\left(x\right)=\int\left(e^x.ln\left(ax\right)+\dfrac{e^x}{x}\right)dx=\int e^xln\left(ax\right)dx+\int\dfrac{e^x}{x}dx=\int e^xlnxdx+\int\dfrac{e^x}{x}dx+\int e^x.lna.dx\)

Xét \(I=\int e^xlnxdx\)

Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=e^x\end{matrix}\right.\)

\(\Rightarrow I=lnx.e^x-\int\dfrac{e^x}{x}dx\)

\(\Rightarrow F\left(x\right)=e^x.lnx+e^x.lna+C\)

\(F\left(\dfrac{1}{a}\right)=e^{\dfrac{1}{a}}ln\left(\dfrac{1}{a}\right)+e^{\dfrac{1}{a}}.lna+C=0\Rightarrow C=0\)

\(F\left(2020\right)=e^{2020}ln\left(2020\right)+e^{2020}.lna=e^{2020}\)

\(\Rightarrow ln\left(2020a\right)=1\Rightarrow a=\dfrac{e}{2020}\)

Khẳng định a là khẳng định đúng

12 tháng 7 2021

Bài 1.

Ta có:\(\left(x+\sqrt{x^2+2020}\right)\left(\sqrt{x^2+2020}-x\right)=x^2+2020-x^2=2020\)

\(\Rightarrow\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=\left(x+\sqrt{x^2+2020}\right)\left(\sqrt{x^2+2020}-x\right)\)

\(\Rightarrow y+\sqrt{y^2+2020}=\sqrt{x^2+2020}-x\)

\(\Rightarrow x+y=\sqrt{x^2+2020}-\sqrt{y^2+2020}\)   (1)

Ta có:\(\left(y+\sqrt{y^2+2020}\right)\left(\sqrt{y^2+2020}-y\right)=y^2+2020-y^2=2020\)

\(\Rightarrow\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=\left(y+\sqrt{y^2+2020}\right)\left(\sqrt{y^2+2020}-y\right)\)

\(\Rightarrow x+\sqrt{x^2+2020}=\sqrt{y^2+2020}-y\)

\(\Rightarrow x+y=\sqrt{y^2+2020}-\sqrt{x^2+2020}\)          (2)

Cộng vế với vế của (1) và (2) ta có:

\(2\left(x+y\right)=\sqrt{y^2+2020}-\sqrt{x^2+2020}+\sqrt{x^2+2020}-\sqrt{y^2+2020}\)

\(\Rightarrow2\left(x+y\right)=0\Rightarrow x+y=0\)

Bài 2: 

Ta có: (2a+1)(2b+1)=9

nên \(2b+1=\dfrac{9}{2a+1}\)

\(\Leftrightarrow2b=\dfrac{9}{2a+1}-\dfrac{2a+1}{2a+1}=\dfrac{8-2a}{2a+1}\)

\(\Leftrightarrow b=\dfrac{8-2a}{4a+2}=\dfrac{4-a}{2a+1}\)

\(\Leftrightarrow b+2=\dfrac{4-a+4a+2}{2a+1}=\dfrac{3a+6}{2a+1}\)

Ta có: \(A=\dfrac{1}{a+2}+\dfrac{1}{b+2}\)

\(=\dfrac{1}{a+2}+\dfrac{2a+1}{3a+6}\)

\(=\dfrac{3+2a+1}{3a+6}\)

\(=\dfrac{2a+4}{3a+6}=\dfrac{2}{3}\)

15 tháng 4 2023

\(A=\dfrac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)}{1\times2020+2\times2019+3\times2018+...+2020\times1}\)

Ta có: \(1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)\)

\(=\left(1+1+1+...+1\right)+\left(2+2+2+...+2\right)+\left(3+3+3+...+3\right)+...+\left(2019+2019\right)+2020\)

Trong đó có: 2020 số 1, 2019 số 2, 2018 số 3,..., 2 số 2019, 1 số 2020

Vậy: \(\left(1+1+...+1\right)+\left(2+2+...+2\right)+\left(3+3+...+3\right)+...+2020\)

\(=1\times2020+2\times2019+3\times2018+...+2020\times1\)

\(\Rightarrow A=\dfrac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)}{1\times2020+2\times2019+3\times2018+...+2020\times1}\)

\(A=\dfrac{1\times2020+2\times2019+3\times2018+...+2020\times1}{1\times2020+2\times2019+3\times2018+...+2020\times1}=1\)

NV
22 tháng 4 2022

\(y=\dfrac{1}{2x^2+x-1}=\dfrac{1}{\left(x+1\right)\left(2x-1\right)}=\dfrac{2}{3}.\dfrac{1}{2x-1}-\dfrac{1}{3}.\dfrac{1}{x+1}\)

\(y'=\dfrac{2}{3}.\dfrac{-2}{\left(2x-1\right)^2}-\dfrac{1}{3}.\dfrac{-1}{\left(x+1\right)^2}=\dfrac{2}{3}.\dfrac{\left(-1\right)^1.2^1.1!}{\left(2x-1\right)^2}-\dfrac{1}{3}.\dfrac{\left(-1\right)^1.1!}{\left(x+1\right)^2}\)

\(y''=\dfrac{2}{3}.\dfrac{\left(-1\right)^2.2^2.2!}{\left(2x-1\right)^3}-\dfrac{1}{3}.\dfrac{\left(-1\right)^2.2!}{\left(x+1\right)^3}\)

\(\Rightarrow y^{\left(n\right)}=\dfrac{2}{3}.\dfrac{\left(-1\right)^n.2^n.n!}{\left(2x-1\right)^{n+1}}-\dfrac{1}{3}.\dfrac{\left(-1\right)^n.n!}{\left(x+1\right)^{n+1}}\)

\(\Rightarrow y^{\left(2019\right)}=\dfrac{2}{3}.\dfrac{\left(-1\right)^{2019}.2^{2019}.2019!}{\left(2x-1\right)^{2020}}-\dfrac{1}{3}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x+1\right)^{2020}}\)

\(=\dfrac{2019!}{3}\left(\dfrac{1}{\left(x+1\right)^{2020}}-\dfrac{2^{2020}}{\left(2x-1\right)^{2020}}\right)\)