\(\frac{1+2013x}{60}=\frac{1+2015x}{5y}=\frac{1+2017x}{4y}\)
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CC
1 tháng 3 2020
Ta có: \(\frac{1+2013x}{60}=\frac{1+2017x}{4y}=\frac{1+2013x+1+2017x}{60+4y}=\frac{2+4030x}{60+4y}\)
\(=\frac{2\left(1+2015x\right)}{2\left(30+2y\right)}=\frac{1+2015x}{30+2y}\)
mà \(\frac{1+2013x}{60}=\frac{1+2015x}{5y}=\frac{1+2017x}{4y}\)\(\Rightarrow\frac{1+2015x}{5y}=\frac{1+2015x}{30+2y}\)
\(\Rightarrow5y=30+2y\)\(\Leftrightarrow5y-2y=30\)\(\Leftrightarrow3y=30\)\(\Leftrightarrow y=10\)
Thay \(y=10\)vào biểu thức ta được:\(\frac{1+2013x}{60}=\frac{1+2015x}{5.10}=\frac{1+2015x}{50}\)
\(\Rightarrow50\left(1+2013x\right)=60\left(1+2015x\right)\)
\(\Leftrightarrow50+100650x=60+120900x\)\(\Leftrightarrow120900x-100650x=50-60\)
\(\Leftrightarrow20250=-10\)\(\Leftrightarrow x=\frac{-10}{20250}=\frac{-1}{2025}\)
Vậy \(x=\frac{-1}{2025}\)và \(y=10\)
TA
0
27 tháng 7 2017
1)
a) 3x = 4y \(\Rightarrow\frac{x}{4}=\frac{y}{3}\)\(\Rightarrow\frac{x}{8}=\frac{y}{6}\)( 1 )
5y = 6z \(\Rightarrow\frac{y}{6}=\frac{z}{5}\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=\frac{x+y+z}{8+6+5}=\frac{1}{19}\)
\(\Rightarrow x=\frac{8}{19};y=\frac{6}{19};z=\frac{5}{19}\)
27 tháng 7 2017
b) \(\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}\Rightarrow\frac{3x-3}{9}=\frac{4y-8}{16}=\frac{5z-15}{25}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\frac{3x-3}{9}=\frac{4y-8}{16}=\frac{5z-15}{25}=\frac{\left(3x-3\right)+\left(4y-8\right)+\left(5z-15\right)}{9+16+25}=\frac{-25}{50}=\frac{-1}{2}\)
\(\Rightarrow x=\frac{-1}{2};y=0;z=\frac{1}{2}\)
29 tháng 12 2018
a) \(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+15}=0\)
\(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+5}\cdot\left(x-3\right)^{10}=0\)
\(\left(x-3\right)^{x+5}\cdot\left[1-\left(x-3\right)^{10}\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^{x+5}=0\\1-\left(x-3\right)^{10}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^{10}=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\\left(x-3\right)^{10}=\left(\pm1\right)^{10}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=\left\{4;2\right\}\end{cases}}\)
Vậy........
16 tháng 12 2019
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{\left(1+2y\right)}{18}=\frac{\left(1+4y\right)}{24}\)\(=\frac{\left(1+4y-1-2y\right)}{24-18}\)\(=\frac{2y}{6}=\frac{y}{3}\)
\(\Rightarrow\frac{\left(1+2y\right)}{18}=\frac{y}{3}\)
\(\Rightarrow3.\left(1+2y\right)=18y\)
\(\Rightarrow3+6y=18y\)
\(\Rightarrow18y-6y=3\)
\(\Rightarrow12y=3\)
\(\Rightarrow y=\frac{1}{4}\)
Thay \(y=\frac{1}{4}\)vào \(\frac{\left(1+2y\right)}{18}=\frac{\left(1+6y\right)}{6x}\)ta có:
\(6x.\left(1+2y\right)=18.\left(1+6y\right)\)
\(\Rightarrow6x.\left(\frac{3}{2}\right)=18.\left(\frac{5}{2}\right)\)
\(\Rightarrow6x=30\)
\(\Rightarrow x=5\)
Vậy \(y=\frac{1}{4};x=5\)
AK
12 tháng 7 2018
\(\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1,6+\frac{3}{5}\right|\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1,6+0,6\right|\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1\right|\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\frac{3}{4}=1\)
\(\Rightarrow\left|x-\frac{1}{2}\right|=1-\frac{3}{4}\)
\(\Rightarrow\left|x-\frac{1}{2}\right|=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{4}\\x-\frac{1}{2}=-\frac{1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{4}\end{cases}}}\)
Vậy ...
12 tháng 7 2018
\(1)\) Ta có :
\(3x=4y\)\(\Leftrightarrow\)\(\frac{x}{4}=\frac{y}{3}\)\(\Leftrightarrow\)\(\frac{x}{8}=\frac{y}{6}\)
\(5y=6z\)\(\Leftrightarrow\)\(\frac{y}{6}=\frac{z}{5}\)
\(\Rightarrow\)\(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}\)
Đặt \(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=k\)\(\Rightarrow\)\(\hept{\begin{cases}x=8k\\y=6k\\z=5k\end{cases}}\) \(\left(1\right)\)
Thay \(\left(1\right)\) vào \(xyz=30\) ta được :
\(8k.6k.5k=30\)
\(\Leftrightarrow\)\(240k^3=30\)
\(\Leftrightarrow\)\(k^3=\frac{30}{240}\)
\(\Leftrightarrow\)\(k^3=\frac{1}{8}\)
\(\Leftrightarrow\)\(k^3=\left(\frac{1}{2}\right)^3\)
\(\Leftrightarrow\)\(k=\frac{1}{2}\)
Suy ra :
\(x=8k=8.\frac{1}{2}=\frac{8}{2}=4\)
\(y=6k=6.\frac{1}{2}=\frac{6}{2}=3\)
\(z=5k=5.\frac{1}{2}=\frac{5}{2}\)
Vậy \(x=4\)\(;\)\(y=3\) và \(z=\frac{5}{2}\)
Chúc bạn học tốt ~
\(\frac{1+2013x}{60}=\frac{1+2017x}{4y}=\frac{1+2013x+1+2017x}{60+4y}=\frac{2+4030x}{2\left(30+2y\right)}\)
\(=\frac{2\left(1+2015x\right)}{2\left(30+2y\right)}=\frac{1+2015x}{30+2y}=\frac{1+2015x}{5y}\)
\(\Leftrightarrow30+2y=5y\)\(\Leftrightarrow5y-2y=30\)\(\Leftrightarrow3y=30\)\(\Leftrightarrow y=10\)
Ta có: \(\frac{1+2013x}{60}=\frac{1+2015x}{50}\)\(\Rightarrow50\left(1+2013x\right)=60\left(1+2015x\right)\)
\(\Leftrightarrow5\left(1+2013x\right)=6\left(1+2015x\right)\)\(\Leftrightarrow5+10065x=6+12090x\)
\(\Leftrightarrow12090x-10065x=5-6\)\(\Leftrightarrow2025x=-1\)\(\Leftrightarrow x=\frac{-1}{2025}\)
Vậy \(x=\frac{-1}{2025}\)