1+2+3+...+2022=[(2022-1):1+1]x(2022+1):2=2022x2023:2=2045253

\(A=\dfrac{-1}{5}x^3\cdot\dfrac{1}{32}x^{20}y^5\cdot\dfrac{64}{27}x^3y^9\cdot z^{2022}=-\dfrac{2}{135}x^{26}y^{14}z^{2022}\)

`A=\frac{-1}{5}x^3 \times \frac{1}{32}x^{20}y^5 \times \frac{64}{27}x^3y^9 \times z^{2022}=-\frac{2}{135}x^{26}y^{14}z^{2022}`

2022

2022

2022

cực gấpppppp

có đáp án chx để mik giải

\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

x²+4x+4=0
(x+2)²=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)²+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0           Th2: x+1=0
            x=-3                      x=-1
vậy ...

\(x:\dfrac{2}{3}=\dfrac{5}{4}\)

\(x=\dfrac{5}{4}\times\dfrac{2}{3}\)

\(x=\dfrac{5}{6}\)

x : 2/3 = 1/2 + 3/4 

x : 2/3 = 5/4

x = 5/4 x 2/3

x = 5/6 

\(x:\dfrac{2}{3}=\dfrac{1}{2}+\dfrac{3}{4}\)

\(x:\dfrac{2}{3}=\dfrac{2}{4}+\dfrac{3}{4}\)

\(x:\dfrac{2}{3}=\dfrac{5}{4}\)

\(x=\dfrac{5}{4}\times\dfrac{2}{3}\)

\(x=\dfrac{10}{12}\)

\(x=\dfrac{5}{6}\)

cái cuối là \(R\left(2023\right)\) hay 2.2023 vậy bạn ?

Sửa đề: 1/R(2023)

R(3)=1*3

R(4)=2*4

R(5)=3*5

...

R(2022)=2020*2022

R(2023)=2021*2023

=>\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{2021\cdot2023}+\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{2020\cdot2022}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2021\cdot2023}+\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2020\cdot2022}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2021}-\dfrac{1}{2023}+\dfrac{1}{2}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{2022}{2023}+\dfrac{505}{1011}\right)\simeq0.7496\)

Cuối đề là \(\dfrac{1}{2.2023}\) đó , ko pk là \(\dfrac{1}{R\left(2023\right)}\) đâu

\(\Leftrightarrow3x+6+x^2-3x+2=9\)

\(\Leftrightarrow x^2+8=9\)

hay \(x\in\left\{1;-1\right\}\)

ĐKXĐ:\(x\ne\pm2\)

\(\dfrac{3}{x-2}+\dfrac{x-1}{x+2}=\dfrac{9}{x^2-4}\\ \Leftrightarrow\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{9}{\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow\dfrac{3\left(x+2\right)+\left(x-1\right)\left(x-2\right)-9}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow3\left(x+2\right)+\left(x-1\right)\left(x-2\right)-9=0\\ \Leftrightarrow3x+6+x^2-x-2x+2-9=0\\ \Leftrightarrow x^2-1=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)