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Sửa đề: Tìm x, y thuộc Z biết x² + 2x + y = xy

Bài làm:

\(x^2+2x+y=xy\)

\(x^2+2x=xy-y\)

\(x\left(x+2\right)=y\left(x-1\right)\)

\(\dfrac{x}{y}=\dfrac{x-1}{x+2}\)

Đặt xk = x - 1; yk = x + 2; k ≠ 0. Nếu k = 1 thì x = x - 1 hay 0 = -1, vô lí.

Suy ra

xk - x = -1

x(k - 1) = -1

\(x=-\dfrac{1}{k-1}\)

\(yk=2-\dfrac{1}{k-1}\)

\(y=\dfrac{2-\dfrac{1}{k-1}}{k}\)

(từ đoạn này thì phải tìm k để x và y nguyên nhưng chưa xử lí được)

\(S=\dfrac{1}{5.6}+\dfrac{1}{10.9}+\dfrac{1}{15.12}+...+\dfrac{1}{3350.2013}\\ \Rightarrow S=\dfrac{1}{5.3}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{670.671}\right)\\ \Rightarrow S=\dfrac{1}{5.3}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{670}-\dfrac{1}{671}\right)\\ \Rightarrow S=\dfrac{1}{5.3}\left(1-\dfrac{1}{671}\right)\\ \Rightarrow S=\dfrac{1}{5.3}.\dfrac{670}{671}\\ \Rightarrow S=\dfrac{1}{15}.\dfrac{670}{671}=\dfrac{134}{2013}\)

SAI RỐI

a/

AD=AE (gt)

AB=AC (gt)

\(\Rightarrow\dfrac{AD}{AB}=\dfrac{AE}{AC}\) => DE//BC (Talet đảo trong tg)

=> BDEC là hình thang

Ta có

tg ABC cân tại A \(\Rightarrow\widehat{B}=\widehat{C}\) (góc ở đáy tg cân)

=> BDEC là hình thang cân

b/

Xét tg ABC

\(\widehat{B}+\widehat{C}=2\widehat{B}=180^o-\widehat{A}=180^o-50^o=130^o\)

\(\Rightarrow\widehat{B}=\widehat{C}=\dfrac{130^o}{2}=65^o\)

...

\(=\left(3^4\right)^5.\left(2^5\right)^4=3^{20}.2^{20}=6^{20}\)

Kb:0

Số sách giáo khoa:

120 . 2/3 = 80 (quyển)

Số sách tham khảo:

120 - 80 = 40 (quyển)

\(\dfrac{1}{100}\)

\(A=\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}\)

\(\Rightarrow A=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}\)

\(\Rightarrow A=\dfrac{5}{8}\)

\(\left|x-1\right|+\left|x+3\right|\)

\(\left|1-x\right|+\left|x+3\right|\ge\left|1-x+x+3\right|=4\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\x+3\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le1\\x\ge-3\end{matrix}\right.\)

Vậy \(\left|x-1\right|+\left|x+3\right|\ge4\Leftrightarrow-3\le x\le1\)

\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=5x\left(1\right)\)

Ta có :

\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|\ge\left|x+1+x+2+x+3+x+4\right|=\left|4x+10\right|\)

\(pt\left(1\right)\Leftrightarrow\left|4x+10\right|=5x\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+10=5x\\4x+10=-5x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\\9x=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-\dfrac{10}{9}\end{matrix}\right.\) \(\left(thỏa.mãnx\inℚ\right)\)