\(F=1\dfrac{1}{5}\times1\dfrac{1}{6}\times1\dfrac{1}{7}\times\cdot\cdot\cdot\times1\dfrac{1}{2019}\times1\dfrac{1}{2020}\)

\(F=\dfrac{6}{5}\times\dfrac{7}{6}\times\dfrac{8}{7}\times\cdot\cdot\cdot\times\dfrac{2020}{2019}\times\dfrac{2021}{2020}\)

\(F=\dfrac{6\times7\times8\times\cdot\cdot\cdot\times2020\times2021}{5\times6\times7\times\cdot\cdot\cdot\times2019\times2020}\)

\(F=\dfrac{2021}{5}\)

\(f=1^1_5\times1^1_6\times1^1_7\times......\times1^1_{2019}\times1^1_{2022}\)

\(f=\dfrac{6}{5}\times\dfrac{7}{6}\times\dfrac{8}{7}\times....\times\dfrac{2020}{2019}\times\dfrac{2021}{2020}\)

\(f=\dfrac{6\times7\times8\times....\times2020\times2021}{5\times6\times7\times.....\times2019\times2020}\)

\(f=\dfrac{2021}{5}\)

\(#Tarus\)

F=20215

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(A=1-\frac{1}{2020}\)

\(A=\frac{2019}{2020}\)

\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)

\(2B=\frac{2}{1.3}+\frac{2}{3.5}=\frac{2}{5.7}+...+\frac{2}{2017.2019}\)

\(2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}=\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)

\(2B=1-\frac{1}{2019}\)

\(2B=\frac{2018}{2019}\)

\(B=\frac{2018}{2019}:2=\frac{1009}{2019}\)

\(C=3^0+3^1+3^2+...+3^{100}\)

\(3C=3^1+3^2+3^3+...+3^{101}\)

\(3C-C=\left(3^1+3^2+3^3+...+3^{101}\right)-\left(1+3^1+3^2+...+3^{100}\right)\)

\(2C=3^{101}-1\)

\(C=\frac{3^{101}-1}{2}\)

\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)

\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)

\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)

\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)

\(=>x+1=0\)

\(=>x=-1\)

b,

\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)

\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)

\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)

\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)

\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)

Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)

\(=>x+2021=0\)

\(=>x=-2021\)

c,

\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)

\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)

\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)

\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)

Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)

\(=>x+329=0\)

\(=>x=-329\)

C

C

C

\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{2019}{2020}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}=\dfrac{2019}{2020}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+4}=\dfrac{2019}{2020}\)

\(\Leftrightarrow\dfrac{x+4-x-1}{\left(x+1\right)\left(x+4\right)}=\dfrac{2019}{2020}\)

\(\Leftrightarrow\dfrac{3}{\left(x+1\right)\left(x+4\right)}=\dfrac{2019}{2020}\)

\(\Leftrightarrow\left(x+1\right)\left(x+4\right).2019=6060\)

<=> x = - 0,208387929

P/s: Số lạ zậy?Đề sai ko

6060 nha bạn

\(A=-\left|2x-3\right|+1< =1\)

Dấu = xảy ra khi x=3/2

\(C=-\left|5x+2\right|-\left|3y+12\right|+4< =4\)

Dấu = xảy ra khi x=-2/5 và y=-4

\(D=-3\left(x+1\right)^2+5< =5\)

Dấu = xảy ra khi x=-1

\(E=\dfrac{1}{2}\left(x+1\right)^2+3>=3\)

Dấu = xảy ra khi x=-1

\(F=\dfrac{15}{4}+3\left|x-1\right|>=\dfrac{15}{4}\)

Dấu = xảy ra khi x=1

a) (x - 1)³ - 1 = 0

<=> (x - 1)³ = 0 + 1

<=> (x - 1)³ = 1

<=> (x - 1)³ = 1³

<=> x - 1 = 1

<=> x = 1 + 1

<=> x = 2

=> x = 2

b) (x - 4)²⁰¹⁹ = 1

<=> (x - 4)²⁰¹⁹ = 1²⁰¹⁹

<=> x - 4 = 1

<=> x = 1 + 4

<=> x = 5

=> x = 5

c) (x - 2019)²⁰²⁰ = 0

<=> (x - 2019)²⁰²⁰ = 0²⁰²⁰

<=> x - 2019 = 0

<=> x = 0 + 2019

<=> x = 2019

=> x = 2019

d) (x - 1)² = (x - 1)³

<=> x² - 2x + 1 = x³ - 2x² + x - x² + 2x - 1

<=> x² - 2x + 1 = x³ - 3x² + 3 - 1

<=> x² - 2x + 1 - x³ + 3x² - 3 + 1 = 0

<=> 4x² - 5x + 2 - x³ = 0

<=> (-x² + 3x - 2)(x - 1) = 0

<=> (x² - 3x + 2)(x - 1) = 0

<=> (x - 2)(x - 1)(x - 1) = 0

<=> x - 2 = 0 hoặc x - 1 = 0

       x = 0 + 2         x = 0 + 1

       x = 2               x = 1

=> x = 1 hoặc x = 2