Bạn A sẽ bốc sao cho:2014 sẽ chia hết cho số bi mà bạn bốc

\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Rightarrow\left(2x-1\right)^8-\left(2x-1\right)^6=0\)

\(\Rightarrow\left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\\left(2x-1\right)^2-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=0\\\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=1\\\left[{}\begin{matrix}2x=2\\2x=0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\end{matrix}\right.\)

bn tự đăng câu hỏi rồi tự trả lời à

\(S_n=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

\(S_n=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)+1\left(n+2\right)}\right)\)

\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n^2+2n+n+2}\right)\)

\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n^2+3n+2}\right)\)

\(S_n=\dfrac{1}{4}-\dfrac{1}{2\left(n^2+3n+2\right)}\)

\(S_n=\dfrac{1}{4}-\dfrac{1}{2n^2+6n+4}\)

\(S_n=\dfrac{2n^2+6n+4}{4\left(2n^2+6n+4\right)}-\dfrac{4}{4\left(2n^2+6n+4\right)}\)

\(S_n=\dfrac{2n^2+6n+4}{8n^2+48n+16}-\dfrac{4}{8n^2+48n+16}\)

\(S_n=\dfrac{2n^2+6n}{8n^2+48n+16}\)

\(S_n=\dfrac{2\left(n^2+3n\right)}{2\left(4n^2+24n+8\right)}=\dfrac{n^2+3n}{4n^2+24n+8}\)

\(A=\left|x-2017\right|+\left|x-2018\right|\)

\(A=\left|x-2017\right|+\left|2018-x\right|\)

\(A\ge\left|x-2017+2018-x\right|\)

\(A\ge1\)

Dấu "=" xảy ra khi:

\(2017\le x\le2018\)

\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)

\(\Rightarrow4\left(3x-y\right)=3\left(x+y\right)\)

\(\Rightarrow12x-4y=3x+3y\)

\(\Rightarrow12x-4y-3y=3x\)

\(\Rightarrow12x-7y=3x\)

\(\Rightarrow12x-3x=7y\)

\(\Rightarrow9x=7y\)

\(\Rightarrow\dfrac{x}{7}=\dfrac{y}{9}\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)

\(\sqrt{\left(3-\sqrt{5}\right)^2}=\left|3-\sqrt{5}\right|=3-\sqrt{5}\)

\(\sqrt{\left(\sqrt{3-\sqrt{4}}\right)^2}=\left|\sqrt{3-\sqrt{4}}\right|=\sqrt{3-\sqrt{4}}\)

\(\sqrt{\left(7-\sqrt{34}\right)^2}=\left|7-\sqrt{34}\right|=7-\sqrt{34}\)

Ta có hình vẽ:

Ta có:

Vì \(Ot\) là tia phân giác của \(\widehat{xOy}\) nên: \(\widehat{tOx}=\widehat{tOy}=\dfrac{1}{2}\widehat{xOy}\)

Vì \(Ot'\)là tia phân giác của \(\widehat{x'Oy'}\) nên: \(\widehat{t'Ox'}=\widehat{t'Oy'}=\dfrac{1}{2}\widehat{x'Oy'}\)

Vì \(\widehat{xOy}=\widehat{x'Oy'}\)(đối đỉnh) nên \(\dfrac{1}{2}\widehat{xOy}=\dfrac{1}{2}\widehat{x'Oy'}\)

Nên \(\widehat{x'Ot'}=\widehat{xOt}\)(1)

Vì \(Ot\) và \(Oy\) nằm giữa \(\widehat{xOx'}\) nên:

\(\widehat{xOt}+\) \(\widehat{tOy}+\widehat{x'Oy}=\widehat{xOx'}\)

\(\Rightarrow\widehat{xOy}+\widehat{x'Oy}=\widehat{xOx'}\)(kề bù)

Nên \(\widehat{xOx'}=180^o\) nên \(Ox\) đối \(Ox'\)(2)

Vì \(Oy\) và \(Ox'\) nằm giữa \(\widehat{tOt'}\) nên:

\(\widehat{tOy}+\widehat{yOx'}+\widehat{x'Ot'}=\widehat{tOt'}\)

\(\Rightarrow\dfrac{1}{2}\widehat{xOy}+\widehat{yOx'}+\dfrac{1}{2}\widehat{x'Oy'}=\widehat{tOt'}\)

\(\Rightarrow\dfrac{1}{2}\widehat{xOy}+\widehat{yOx'}+\dfrac{1}{2}\widehat{xOy}=\widehat{tOt'}\)

\(\Rightarrow\widehat{tOt'}=\widehat{xOy}+\widehat{yOx'}\)(kề bù)

Nên \(\widehat{tOt'}=180^o\) suy ra \(Ot\) đối \(Ot'\)(3)

Từ (1);(2) và (3) ta có:

\(\widehat{xOt}\) và \(\widehat{x'Ot'}\) đối đỉnh

Ta có hình vẽ:

\(\widehat{A3}=\widehat{A1}=130^o\)( đối đỉnh)

Mà \(\widehat{A1}+\widehat{B2}=130^o+50^o=180^o\)(2 góc trong cùng phía)

Nên:\(a//b\)

\(\Rightarrowđpcm\)

Ta có:

\(\widehat{B3}+\widehat{B2}=180^o\)(kề bù)

\(\Rightarrow\widehat{B3}+50^o=180^o\Rightarrow\widehat{B3}=130^o\)

Vì \(\widehat{A1}=\widehat{B3}=130^o\)(so le trong) nên \(a//b\)

\(\Rightarrowđpcm\)

Ta có:

\(\widehat{A3}=\widehat{B3}=180^o\)(đồng vị) nên \(a//b\)

\(\Rightarrowđpcm\)