tìm các số a b c biết 2a=3b 5b=7c va 3a - 7b + 5c = -30
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do A=2024 => \(\left|x+2023\right|=2024\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2023=2024\\x+2023=-2024\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2024-2023\\x=-2024-2023\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4047\end{matrix}\right.\)
vậy x \(\in\left\{-4047;1\right\}\)
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\(1+\dfrac{7}{n\left(n+8\right)}=\dfrac{n^2+8n+7}{n\left(n+8\right)}=\dfrac{\left(n+1\right)\left(n+7\right)}{n\left(n+8\right)}\)
\(\Rightarrow P=\left(1+\dfrac{7}{1.\left(1+8\right)}\right)\left(1+\dfrac{7}{2.\left(2+8\right)}\right)\left(1+\dfrac{7}{3.\left(3+8\right)}\right)...\left(1+\dfrac{7}{50.\left(50+8\right)}\right)\)
\(=\left(\dfrac{2.8}{1.9}\right).\left(\dfrac{3.9}{2.10}\right).\left(\dfrac{4.10}{3.11}\right)...\left(\dfrac{51.57}{50.58}\right)\)
\(=\dfrac{2.3.4...51}{1.2.3...50}.\dfrac{8.9.10...57}{9.10.11...58}=\dfrac{51}{1}.\dfrac{8}{58}=\dfrac{204}{29}\)
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\(2^x\) = \(4^{y-4}\)
x = 2 (y - 4)
y = \(\dfrac{x+8}{2}\)
\(27^y\) = \(3^{x+8}\)
x = 3y - 8
y = \(\dfrac{x+8}{3}\)
cảm ơn bạn đã trở lời giúp mình nhưng mình cần tìm x và y
\(2a=3b;5b=7c\)
\(\Rightarrow\dfrac{a}{3}=\dfrac{b}{2};\dfrac{b}{7}=\dfrac{c}{5}\)
\(\Rightarrow\dfrac{a}{21}=\dfrac{b}{14};\dfrac{b}{14}=\dfrac{c}{10}\)
\(\Rightarrow\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}\)
\(\Rightarrow\dfrac{3a}{63}=\dfrac{7b}{98}=\dfrac{5c}{50}\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\dfrac{3a}{63}=\dfrac{7b}{98}=\dfrac{5c}{50}=\dfrac{3a-7b+5c}{63-98+50}=-\dfrac{30}{15}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{21}=-2\Leftrightarrow a=-42\\\dfrac{b}{14}=-2\Leftrightarrow b=-28\\\dfrac{c}{10}=-2\Leftrightarrow c=20\end{matrix}\right.\)