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\(\frac{1+7y}{7x}=\frac{1+9y}{2x}\) \(\Leftrightarrow\frac{1+7y}{7}=\frac{1+9y}{2}\)
\(\Leftrightarrow\left(1+7y\right)2=7\left(1+9y\right)\)
\(\Leftrightarrow2+14y=7+63y\)
\(\Leftrightarrow63y-14y=2-7\)
\(\Leftrightarrow y=-\frac{5}{49}\)
Thay \(x=-\frac{5}{49}\) vào biểu thức ta có :
\(\frac{1+7.\frac{-5}{49}}{7.x}=\frac{1+9.\frac{-5}{49}}{2x}\)
\(\Leftrightarrow x=2\)
Vậy..
b) Áp dụng t/c dãy tỉ số bằng nhau,ta có:
\(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}=\frac{1+3y+1+5y+1+7y}{12+5x+4x}=\frac{3+15y}{12+5x+4x}=\frac{3\left(1+5y\right)}{2.3.2+5x+4x}=\frac{1+5y}{4+9x}=\frac{1+5y}{5x}\)<=> 4 + 9x = 5x
....
a/ Từ giả thiêt ta có \(\frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}\Leftrightarrow\frac{x}{15}-\frac{3}{5}=\frac{y}{20}-\frac{3}{5}=\frac{z}{40}-\frac{3}{5}\)
\(\Leftrightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{40}\). Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{40}=k\)
\(\Rightarrow\begin{cases}x=15k\\y=20k\\z=40k\end{cases}\)
Theo đề bài : \(xy=1200\Leftrightarrow15k.20k=1200\Leftrightarrow k^2=4\Leftrightarrow k=\pm2\)
Tới đây dễ rồi nhé :)
b/ \(\frac{1+5y}{5x}=\frac{1+7y}{4x}\Leftrightarrow\frac{1+5y}{5}=\frac{1+7y}{4}\Leftrightarrow\frac{7+35y}{35}=\frac{5+35y}{20}=\frac{7+35y-5-35y}{35-20}=\frac{2}{15}\)
\(\Rightarrow y=-\frac{1}{15}\)
Thay y vào \(\frac{1+3y}{12}=\frac{1+5y}{5x}\) tìm được x = 2
a )
Ta có :
\(\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
\(\Rightarrow\frac{4\left(1+5y\right)}{20x}=\frac{5\left(1+7y\right)}{20x}\)
\(\Rightarrow\frac{4+20y}{20x}=\frac{5+35y}{20x}\)
\(\Rightarrow4+20y=5+35y\)
\(\Rightarrow35y-20y=4-5\)
\(\Rightarrow15y=4-5\)
\(\Rightarrow15y=-1\)
\(\Rightarrow y=-\frac{1}{15}\)
Lại có :
\(\frac{1+3y}{12}=\frac{1+5y}{5x}\)
\(\Rightarrow\frac{1+3.-\frac{1}{15}}{12}=\frac{1+5.-\frac{1}{15}}{5x}\)
\(\Rightarrow\frac{1-\frac{1}{5}}{12}=\frac{1-\frac{1}{3}}{5x}\)
\(\Rightarrow\frac{4}{5}:12=\frac{4}{3}:5x\)
\(\Rightarrow\frac{1}{15}=\frac{4}{3}:5x\)
\(\Rightarrow5x=\frac{4}{3}:\frac{1}{15}\)
\(\Rightarrow5x=20\)
\(\Rightarrow x=4\)
Vậy \(x=4;y=-\frac{1}{15}\)
a) Xét \(\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
\(\Rightarrow\frac{4x\left(1+5y\right)}{20x}=\frac{5\left(1+7y\right)}{20x}\)
\(\Rightarrow4x\left(1+5y\right)=5\left(1+7y\right)\)
\(\Rightarrow4+20y=5+35y\)
\(\Rightarrow35y-20y=4-5\)
\(\Rightarrow15y=-1\)
\(\Rightarrow y=\frac{-1}{15}\)
Xét \(\frac{1+3y}{12}=\frac{1+5y}{5x}\)
\(\Rightarrow\frac{1+3.\frac{-1}{15}}{12}=\frac{1+5.\frac{-1}{15}}{5x}\)
\(\Rightarrow\frac{1+\frac{-1}{5}}{12}=\frac{1+\frac{-1}{3}}{5x}\)
\(\Rightarrow\frac{\frac{4}{5}}{12}=\frac{\frac{2}{3}}{5x}\)
\(\Rightarrow\frac{4}{5}:12=\frac{2}{3}:5x\)
\(\Rightarrow\frac{1}{15}=\frac{2}{3}:5x\)
\(\Rightarrow5x=\frac{2}{3}:\frac{1}{15}\)
\(\Rightarrow5x=\frac{30}{3}\)
\(\Rightarrow x=\frac{30}{3}:5\)
\(\Rightarrow x=\frac{30}{3}.\frac{1}{5}\)
\(\Rightarrow x=2\)
Vậy x = 2 ; y = \(\frac{-1}{15}\)