(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))

vậy x= 2023

Câu 1: 

1: Ta có: \(P=\left(\dfrac{x^2}{x^2-3}+\dfrac{2x^2-24}{x^4-9}\right)\cdot\dfrac{7}{x^2+8}\)

\(=\left(\dfrac{x^2\left(x^2+3\right)}{\left(x^2-3\right)\left(x^2+3\right)}+\dfrac{2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\right)\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{x^4+3x^2+2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{x^4+5x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{x^4+8x^2-3x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{x^2\left(x^2+8\right)-3\left(x^2+8\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{\left(x^2+8\right)\left(x^2-3\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{7}{x^2+3}\)

Câu 2a đề sai, pt này ko giải được

2b.

\(P\left(x\right)=\left(2x+7\right)\left(x^2-4x+4\right)+\left(a+20\right)x+\left(b-28\right)\)

Do \(\left(2x+7\right)\left(x^2-4x+4\right)⋮\left(x^2-4x+4\right)\)

\(\Rightarrow P\left(x\right)\) chia hết \(Q\left(x\right)\) khi \(\left(a+20\right)x+\left(b-28\right)\) chia hết \(x^2-4x+4\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+20=0\\b-28=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-20\\b=28\end{matrix}\right.\)

3a.

\(VT=\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}=\dfrac{2+x^2+y^2}{1+x^2+y^2+x^2y^2}=1+\dfrac{1-x^2y^2}{1+x^2+y^2+x^2y^2}\le1+\dfrac{1-x^2y^2}{1+2xy+x^2y^2}\)

\(VT\le1+\dfrac{\left(1-xy\right)\left(1+xy\right)}{\left(xy+1\right)^2}=1+\dfrac{1-xy}{1+xy}=\dfrac{2}{1+xy}\) (đpcm)

3b

Ta có: \(n^3-n=n\left(n-1\right)\left(n+1\right)\) là tích 3 số nguyên liên tiếp nên luôn chia hết cho 6

\(\Rightarrow n^3\) luôn đồng dư với n khi chia 6

\(\Rightarrow S\equiv2021^{2022}\left(mod6\right)\)

Mà \(2021\equiv1\left(mod6\right)\Rightarrow2021^{2020}\equiv1\left(mod6\right)\)

\(\Rightarrow2021^{2022}-1⋮6\)

\(\Rightarrow S-1⋮6\)

2a. 

À nãy mình nhìn lộn dấu trừ bên vế phải thành dấu cộng

ĐKXĐ: ...

\(\Leftrightarrow\dfrac{3x+2022+2x-2021}{\left(2x-2021\right)\left(3x+2022\right)}=\dfrac{10x-2024-\left(15x-2023\right)}{\left(15x-2023\right)\left(10x-2024\right)}\)

\(\Leftrightarrow\dfrac{5x-1}{\left(2x-2021\right)\left(3x+2022\right)}=-\dfrac{5x-1}{\left(15x-2023\right)\left(10x-2024\right)}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\Rightarrow x=...\\\dfrac{1}{\left(2x-2021\right)\left(3x+2022\right)}=-\dfrac{1}{\left(15x-2023\right)\left(10x-2024\right)}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left(2x-2021\right)\left(3x+2022\right)+\left(15x-2023\right)\left(10x-2024\right)=0\)

\(\Leftrightarrow\left[12x-4045-\left(10x-2024\right)\right]\left(3x+2022\right)+\left(12x-4045+3x+2022\right)\left(10x-2024\right)=0\)

\(\Leftrightarrow\left(12x-4045\right)\left(3x+2022\right)-\left(10x-2024\right)\left(3x+2022\right)+\left(12x-4045\right)\left(10x-2024\right)+\left(3x+2022\right)\left(10x-2024\right)=0\)

\(\Leftrightarrow\left(12x-4045\right)\left(13x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{13}\\x=\dfrac{4045}{12}\end{matrix}\right.\)

1)\(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{11}{70}\)

\(\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{x\left(x+3\right)}\right):3=\dfrac{11}{70}\)

\(\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+.....+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{11}{70}\cdot3\)

\(\dfrac{1}{2}-\dfrac{1}{x+3}=\dfrac{33}{70}\)

\(\dfrac{1}{x+3}=\dfrac{1}{2}-\dfrac{33}{70}\)

\(\dfrac{1}{x+3}=\dfrac{2}{70}\)

\(\dfrac{1}{x+3}=\dfrac{1}{35}\)

\(x+3=35\\ x=35-3\\ x=32\)

2) Số góc đc tạo thành từ 2023 tia chung gốc là:\(\dfrac{2023\cdot2022}{2}=2045253\) (góc)

Bài 1 thì bạn Ánh làm đúng rồi

Bài 2 thì giải chi tiết như này em nhé:

Cứ 1 tia tạo với 2023 - 1 tia còn lại là 2023 - 1 góc

Với 2023 tia thì tạo được số góc là:  (2023 - 1)\(\times\) 2023 góc

Theo cách tính trên thì mỗi góc đã được tính hai lần

Vậy số góc tạo được là: (2023-1)\(\times\) 2023: 2 = 2045253 (góc)

Kết luận: ...

1)12⋅5+15⋅8+...+1�(�+3)=11702⋅51​+5⋅81​+...+x(x+3)1​=7011​

(32⋅5+35⋅8+...+3�(�+3)):3=1170(2⋅53​+5⋅83​+...+x(x+3)3​):3=7011​

(12−15+15−18+.....+1�−1�+3)=1170⋅3(21​−51​+51​−81​+.....+x1​−x+31​)=7011​⋅3

12−1�+3=337021​−x+31​=7033​

1�+3=12−3370x+31​=21​−7033​

1�+3=270x+31​=702​

1�+3=135x+3

\(x=\sqrt{\dfrac{2\sqrt{3}+2-6\sqrt{3}}{2\sqrt{3}\left(2\sqrt{3}+2\right)}}=\sqrt{\dfrac{2-4\sqrt{3}}{2\sqrt{3}\left(2\sqrt{3}+2\right)}}\) ko tồn tại vì 2-4căn 3<0

Chia nhỏ ra

a: =>1/2x=7/2-2/3=21/6-4/6=17/6

=>x=17/3

b: =>2/3:x=-7-1/3=-22/3

=>x=2/3:(-22/3)=-1/11

c: =>1/3x+2/5x-2/5=0

=>11/15x=2/5

hay x=6/11

d: =>2x-3=0 hoặc 6-2x=0

=>x=3/2 hoặc x=3

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

    \(2x-\dfrac{1}{3}=\dfrac{1}{3}\)

    \(2x=\dfrac{1}{3}+\dfrac{1}{3}\)

    \(2x=\dfrac{2}{3}\)

      \(x=\dfrac{2}{3}\div2\)

      \(x=\dfrac{1}{3}\)

b) \(\dfrac{1}{3}+\dfrac{2}{3}\cdot\left(x-2\right)=3\)

            \(\dfrac{2}{3}\cdot\left(x-2\right)=3-\dfrac{1}{3}\)

            \(\dfrac{2}{3}\cdot\left(x-2\right)=\dfrac{8}{3}\)

                   \(x-2=\dfrac{8}{3}\div\dfrac{2}{3}\)

                   \(x-2=\dfrac{24}{6}\)

                  \(x-2=4\)

                  \(x=4+2\)

                  \(x=6\)

Đề sai rồi bạn

\(\left[{}\begin{matrix}2x-\dfrac{2}{3}=\dfrac{1}{3}\\2x-\dfrac{2}{3}=\dfrac{-1}{3}\end{matrix}\right.\left[{}\begin{matrix}2x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)

a)\(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{5}{2}\\x+\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c)\(\left[{}\begin{matrix}x-2=2x-1\\-x+2=2x-1\end{matrix}\right.\left[{}\begin{matrix}x-2=2x-1\\-x+2=2x-1\end{matrix}\right.\left[{}\begin{matrix}x-2x=-1+2\\-x-2x=-1-2\end{matrix}\right.\left[{}\begin{matrix}-1x=1\\-3x=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=1:\left(-1\right)\\x=-3:\left(-3\right)\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)