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Đặt \(\frac{x-2}{6}=\frac{y+3}{9}=\frac{z-7}{10}=k\Rightarrow\hept{\begin{cases}x=6k+2\\y=9k-3\\z=10k+7\end{cases}}\)
Theo đề bài: x+y+z=106
<=>\(6k+2+9k-3+10k+7=106\)
<=>\(25k+6=106\)
<=> 25k = 100
<=> k = 4
=> \(\hept{\begin{cases}x=6.4+2=26\\y=9.4-3=33\\z=10.4+7=47\end{cases}}\)
Vậy .........................
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
\(\frac{1}{2}x+\frac{3}{5}x=\frac{-33}{10}\)\(\Leftrightarrow\left(\frac{1}{2}+\frac{3}{5}\right)x=\frac{-33}{10}\)
\(\Leftrightarrow\frac{11}{10}x=\frac{-33}{10}\)\(\Leftrightarrow x=\frac{-33}{10}:\frac{11}{10}=\frac{-33}{10}.\frac{10}{11}=-3\)
Vậy \(x=-3\)
\(\frac{1}{2}\cdot x+\frac{3}{5}\cdot x=-\frac{33}{10}\)
\(\left(\frac{1}{2}+\frac{3}{5}\right)\cdot x=-\frac{33}{10}\)
\(\left(\frac{5}{10}+\frac{6}{10}\right)\cdot x=-\frac{33}{10}\)
\(\frac{11}{10}\cdot x=-\frac{33}{10}\)
\(x=-\frac{33}{10}:\frac{11}{10}=-\frac{33}{10}\cdot\frac{10}{11}\)
\(x=-\frac{33}{11}=-3\)
Ta có : \(\frac{x-5}{5x-1}=\frac{4x-10}{20x+4}\)
=> \(\frac{x-5}{5x-1}=\frac{2x-5}{10x+2}\)
=> (x - 5)(10x + 2) = (2x - 5)(5x - 1)
=> 10x2 + 2x - 50x - 10 = 10x2 - 2x - 25x + 5
=> 10x2 - 48x - 10x2 + 27x = 5 + 10
=> -21x = 15
=> x = 15 : (-21) = -5/7
Thay x = -5/7 vào \(\frac{x-5}{5x-1}=\frac{y}{3}\)
=> \(\frac{-\frac{5}{7}-5}{5.\left(-\frac{5}{7}\right)-1}=\frac{y}{3}\)
=> \(\frac{-\frac{40}{7}}{-\frac{32}{7}}=\frac{y}{3}\)
=> \(\frac{5}{4}=\frac{y}{3}\)
=> 4y = 15
=> y = 15/4
Vậy ...
Ta có: \(\frac{5}{y}=\frac{3}{x}\) => \(\frac{x}{3}=\frac{y}{5}\) => \(\frac{x^2}{9}=\frac{y^2}{25}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x^2}{9}=\frac{y^2}{25}=\frac{y^2+x^2}{25+9}=\frac{125}{34}\)
=> \(\hept{\begin{cases}\frac{x^2}{9}=\frac{125}{34}\\\frac{y^2}{25}=\frac{125}{34}\end{cases}}\) => \(\hept{\begin{cases}x^2=\frac{125}{34}.9=\frac{1125}{34}\\y^2=\frac{125}{34}.25=\frac{3125}{34}\end{cases}}\) => \(\hept{\begin{cases}x=\pm\frac{15\sqrt{170}}{34}\\y=\pm\frac{25\sqrt{170}}{34}\end{cases}}\)
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
\(\dfrac{3}{x}=2:\dfrac{9}{10}\Rightarrow\dfrac{3}{x}=\dfrac{20}{9}\Rightarrow x=3:\dfrac{20}{9}=\dfrac{27}{20}\)
\(\dfrac{-2}{\dfrac{3}{x}}=\dfrac{9}{10}\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{9}{10}\Leftrightarrow x=\dfrac{-20}{27}\)