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\(\frac{x+12}{7}+\frac{x+4}{15}+\frac{x+6}{13}=\frac{x+8}{11}+\frac{x+10}{9}+\frac{x+12}{7}\)
=>\(\left(\frac{x+12}{7}+1\right)+\left(\frac{x+4}{15}+1\right)+\left(\frac{x+6}{13}+1\right)=\left(\frac{x+8}{11}+1\right)+\left(\frac{x+10}{9}+1\right)+\left(\frac{x+12}{7}+1\right)\)
=> \(\frac{x+19}{7}+\frac{x+19}{15}+\frac{x+19}{13}=\frac{x+19}{11}+\frac{x+19}{9}+\frac{x+19}{7}\)
=> \(\frac{x+19}{7}+\frac{x+19}{15}+\frac{x+19}{13}-\frac{x+19}{11}-\frac{x+19}{9}-\frac{x+19}{7}=0\)
\(\Rightarrow\left(x+19\right)\left(\frac{1}{7}+\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}-\frac{1}{7}\right)=0\)
=> x + 19 = 0 Vì \(\frac{1}{7}+\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}-\frac{1}{7}\ne0\)
=> x = - 19
Bài làm:
Ta có: \(\frac{x+12}{7}+\frac{x+4}{15}+\frac{x+6}{13}=\frac{x+8}{11}+\frac{x+10}{9}+\frac{x+12}{7}\)
\(\Leftrightarrow\left(\frac{x+12}{7}+1\right)+\left(\frac{x+4}{15}+1\right)+\left(\frac{x+6}{13}+1\right)-\left(\frac{x+8}{11}+1\right)-\left(\frac{x+10}{9}+1\right)-\left(\frac{x+12}{7}+1\right)=0\)
\(\Leftrightarrow\frac{x+19}{7}+\frac{x+19}{15}+\frac{x+19}{13}-\frac{x+19}{11}-\frac{x+19}{9}-\frac{x+19}{7}=0\)
\(\Leftrightarrow\left(x+19\right)\left(\frac{1}{7}+\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}-\frac{1}{7}\right)=0\)
\(\Leftrightarrow\left(x+19\right)\left(\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}\right)=0\)
Mà \(\hept{\begin{cases}\frac{1}{15}< \frac{1}{11}\\\frac{1}{13}< \frac{1}{9}\end{cases}\Rightarrow}\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}< 0\)
\(\Rightarrow x+19=0\)
\(\Rightarrow x=-19\)
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\(\frac{15}{x-9}=\frac{12}{y-12}=\frac{40}{z-24}\)
=> \(\frac{x-9}{15}=\frac{y-12}{12}=\frac{z-24}{40}\)
Đặt \(\frac{x-9}{15}=\frac{y-12}{12}=\frac{z-24}{40}=k\Rightarrow\hept{\begin{cases}x-9=15k\\y-12=12k\\z-24=40k\end{cases}}\)
=> \(\hept{\begin{cases}x=15k+9\\y=12k+12\\z=40k+24\end{cases}}\)
Mà xy = 200
=> \(\left(15k+9\right)\left(12k+12\right)=200\)
=> 15(12k + 12) + 9(12k + 12) = 200
=> 180k + 180 + 108k + 108 = 200
=> 288k + 216 = 200
=> 288k = -16
Đề của bạn chắc chắn đúng chứ , mình thấy sai rồi đấy :v
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Do x = 11 => x + 1 = 12
Ta có :
D = \(x^{17}-\left(x+1\right)x^{16}+\left(x+1\right).x^{15}+...+\left(x+1\right).1\)
= \(x^{17}-x^{17}-x^{16}+x^{16}+x^{15}+...+x+1\)
= \(0+0+0+...+1\)
= \(1\)
Vậy D = 1
Tham khảo nha !!! Chúc học tốt !!!
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\(\dfrac{34-x}{x+12}=\dfrac{8}{15}\\ \Rightarrow15.\left(34-x\right)=8.\left(x+12\right)\\ \Rightarrow510-15x=8x+96\\ \Rightarrow8x+15x=510-96\\ \Rightarrow23x=414\\ \Rightarrow x=18\)
Vậy \(x=18\)
\(U\left(15\right)=\left\{1;3;5;15\right\}\)
mà \(12\ge x\le15\)
\(\Rightarrow x\in\left\{1;3;5\right\}\)
Đề bài yêu cầu gì bạn ?