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Dựa theo tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}=\frac{2x+3y+4z}{3+4+5}=\frac{2x+3y+4z}{12}\)
Rút gọn đi, ta có:
\(\frac{2x+3y+4z}{12}=\frac{x+3y+4z}{6}=\frac{x+y+4z}{2}=\frac{x+y+z}{\left(\frac{2}{4}\right)}=\frac{48}{\left(\frac{2}{4}\right)}=96\) (1)
Từ (1), ta có: \(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}=96\Rightarrow\hept{\begin{cases}2x=96.3\\3y=96.4\\4z=96.5\end{cases}}\Rightarrow\hept{\begin{cases}x=144\\y=128\\z=120\end{cases}}\)
Kết luận: .....
Đặt \(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}=k\)
\(\Rightarrow x=\frac{3}{2}k;y=\frac{4}{3}k;z=\frac{5}{4}k\)
Có: \(x+y+z=49\)
\(\Rightarrow\frac{3}{2}k+\frac{4}{3}k+\frac{5}{4}k=49\)
\(k.\left(\frac{3}{2}+\frac{4}{3}+\frac{5}{4}\right)=49\)
\(k.\frac{49}{12}=49\)
\(\Rightarrow k=12\)
\(\Rightarrow\hept{\begin{cases}x=\frac{3}{2}.12=18\\y=\frac{4}{3}.12=16\\z=\frac{5}{4}.12=15\end{cases}}\)
Vậy \(\hept{\begin{cases}x=18\\y=16\\z=15\end{cases}}\)
Tham khảo nhé~
a. Có \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}\) => \(\dfrac{x}{4}=\dfrac{3x}{9}=\dfrac{4z}{36}\) và x-3y+4z=62
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\dfrac{x}{4}=\dfrac{3y}{9}=\dfrac{4z}{36}\)= \(\dfrac{x-3y+4z}{4-9+36}=\dfrac{62}{31}=2\)
=> x=8
3y=18=>y=6
4z=72=>z=18
Vậy x=8 ; y=6 ; z=18
b, Ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{5z}{20}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{5z}{20}\\ =\dfrac{2x+3y-5z}{4+9-20}=\dfrac{-21}{-7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=3\cdot2=6\\y=3\cdot3=9\\z=3\cdot4=12\end{matrix}\right.\\ vậy...\)
Câu c bạn làm tương tự nhé!
d, Ta có : \(\left|x+y-z\right|=95\Rightarrow\left[{}\begin{matrix}x+y-z=95\\x+y-z=-95\end{matrix}\right.\)
\(2x=3y=5z=\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}=\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(2x=3y=5z=\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}=\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\\ =\dfrac{x+y-z}{15+10-6}=\dfrac{x+y-z}{19}\\ \Rightarrow\left[{}\begin{matrix}x+y-z=95\\x+y-z=-95\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=15\cdot5=75\\y=10\cdot5=50\\z=6\cdot5=30\end{matrix}\right.\\\left\{{}\begin{matrix}x=-5\cdot15=-75\\y=-5\cdot10=-50\\z=-5\cdot6=-30\end{matrix}\right.\end{matrix}\right.\)
Vậy...
\(a,Đặt\dfrac{x}{y}=\dfrac{2}{3}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=k\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\\ A=\dfrac{2x-3y}{x-5y}=\dfrac{2\cdot2k-3\cdot3k}{2k-5\cdot3k}\\ =\dfrac{4k-9k}{2k-15k} \\ =\dfrac{5k}{13k}\\ =\dfrac{5}{13}\)
\(b,Thayx-y=7vàoB,tacó:\\ B=\dfrac{2x+7}{3x-y}+\dfrac{2y-7}{3y-x}\\ =\dfrac{2x+x-y}{3x-y}+\dfrac{2y-x+y}{3y-x}\\ =\dfrac{3x-y}{3x-y}+\dfrac{3y-x}{3y-x}\\ =1+1\\ =2\)
\(c,Đặt\dfrac{x}{3}=\dfrac{y}{5}=k\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\\ C=\dfrac{5x^2+3y^2}{10x^2-3y^2}\\ =\dfrac{5\left(3k\right)^2+3\left(5k\right)^2}{10\left(3k\right)^2-3\left(5k\right)^2}\\ =\dfrac{45k^2+75k^2}{90k^2-75k^2}\\ =\dfrac{120k^2}{15k^2}\\ =8\)
\(d,\dfrac{a}{b}=\dfrac{5}{7}\Leftrightarrow\dfrac{a}{5}=\dfrac{b}{7}=k\Leftrightarrow\left\{{}\begin{matrix}a=5k\\b=7k\end{matrix}\right.\\ D=\dfrac{5a-b}{3a-2b}\\ =\dfrac{5\cdot5k-7k}{3\cdot5k-2\cdot7k}\\ =\dfrac{25k-7k}{15k-14k}\\ =\dfrac{18k}{k}=18\)
\(e,Thayx-y=5vàoE,tacó:\\ E=\dfrac{3x-5}{2x+y}-\dfrac{4y+5}{x+3y}\\ =\dfrac{3x-x+y}{2x+y}-\dfrac{4y+x-y}{x+3y}\\ =\dfrac{2x+y}{2x+y}-\dfrac{3y+x}{x+3y}\\ =1-1=0\)
a,
\(\dfrac{2x}{3y}=\dfrac{-1}{3}\\ \Rightarrow\dfrac{2x}{-1}=\dfrac{3y}{3}\\ \Leftrightarrow\dfrac{-2x}{1}=\dfrac{3y}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{-2x}{1}=\dfrac{3y}{3}=\dfrac{-2x+3y}{1+3}=\dfrac{7}{4}\)
\(\dfrac{-2x}{1}=\dfrac{7}{4}\Rightarrow-2x=\dfrac{7}{4}\Rightarrow x=\dfrac{7}{4}:\left(-2\right)=\dfrac{-7}{8}\\ \dfrac{3y}{3}=\dfrac{7}{4}\Rightarrow y=\dfrac{7}{4}\)
Vậy \(x=\dfrac{-7}{8};y=\dfrac{7}{4}\)
b,
\(\dfrac{x}{3}=\dfrac{y}{4}\\ \Leftrightarrow\dfrac{2x}{6}=\dfrac{5y}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{6}=\dfrac{5y}{20}=\dfrac{2x+5y}{6+20}=\dfrac{10}{26}=\dfrac{5}{13}\\ \dfrac{x}{3}=\dfrac{2x}{6}=\dfrac{5}{13}\Rightarrow x=\dfrac{5}{13}\cdot3=\dfrac{15}{13}\\ \dfrac{y}{4}=\dfrac{5y}{20}=\dfrac{5}{13}\Rightarrow y=\dfrac{5}{13}\cdot4=\dfrac{20}{13}\)
Vậy \(x=\dfrac{15}{13};y=\dfrac{20}{13}\)
c,
\(7x=3y\\ \Rightarrow\dfrac{x}{3}=\dfrac{y}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{16}{-4}=-4\\ \dfrac{x}{3}=-4\Rightarrow x=\left(-4\right)\cdot3=-12\\ \dfrac{y}{7}=-4\Rightarrow y=\left(-4\right)\cdot7=-28\)
Vậy \(x=-12;y=-28\)
d,
\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{z}{-2}\\ \Leftrightarrow\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{-2z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{-2z}{4}=\dfrac{x+y+\left(-2z\right)}{5+1+4}=\dfrac{x+y-2z}{10}=\dfrac{160}{10}=16\\ \dfrac{x}{5}=16\Rightarrow x=16\cdot5=80\\ \dfrac{y}{1}=16\Rightarrow y=16\\ \dfrac{z}{-2}=\dfrac{-2z}{4}=16\Rightarrow z=16\cdot\left(-2\right)=-32\)
Vậy \(x=80;y=16;z=-32\)
e,
\(\dfrac{x}{10}=\dfrac{y}{5}\Rightarrow\dfrac{x}{20}=\dfrac{y}{10};\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{y}{10}=\dfrac{z}{15}\\ \Rightarrow\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{15}\\ \Leftrightarrow\dfrac{2x}{40}=\dfrac{3y}{30}=\dfrac{4z}{60}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{40}=\dfrac{3y}{30}=\dfrac{4z}{60}=\dfrac{2x-3y+4z}{40-30+60}=\dfrac{330}{70}=\dfrac{33}{7}\)
\(\dfrac{x}{20}=\dfrac{2x}{40}=\dfrac{33}{7}\Rightarrow x=\dfrac{33}{7}\cdot20=\dfrac{660}{7}\\ \dfrac{y}{10}=\dfrac{3y}{30}=\dfrac{33}{7}\Rightarrow y=\dfrac{33}{7}\cdot10=\dfrac{330}{7}\\ \dfrac{z}{15}=\dfrac{4z}{60}=\dfrac{33}{7}\Rightarrow z=\dfrac{33}{7}\cdot15=\dfrac{495}{7}\)
Vậy \(x=\dfrac{660}{7};y=\dfrac{330}{7};z=\dfrac{495}{7}\)
f,
\(\dfrac{x}{-2}=\dfrac{-y}{4}=\dfrac{z}{5}\\ \Leftrightarrow\dfrac{x}{-2}=\dfrac{-2y}{8}=\dfrac{3z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{-2}=\dfrac{-2y}{8}=\dfrac{3z}{15}=\dfrac{x+\left(-2y\right)+3z}{\left(-2\right)+8+15}=\dfrac{x-2y+3z}{21}=\dfrac{1200}{21}=\dfrac{400}{7}\)
\(\dfrac{x}{-2}=\dfrac{400}{7}\Rightarrow x=\dfrac{400}{7}\cdot\left(-2\right)=\dfrac{-800}{7}\\ \dfrac{-y}{4}=\dfrac{-2y}{8}=\dfrac{400}{7}\Rightarrow-y=\dfrac{400}{7}\cdot4=\dfrac{1600}{7}\Rightarrow y=\dfrac{-1600}{7}\\ \dfrac{z}{5}=\dfrac{3z}{15}=\dfrac{400}{7}\Rightarrow z=\dfrac{400}{7}\cdot5=\dfrac{2000}{7}\)
Vậy \(x=\dfrac{-800}{7};y=\dfrac{-1600}{7};z=\dfrac{2000}{7}\)
g,
\(\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}\\ \Leftrightarrow\dfrac{2x}{6}=\dfrac{3y}{24}=\dfrac{z}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{6}=\dfrac{3y}{24}=\dfrac{z}{5}=\dfrac{2x+3y-z}{6+24-5}=\dfrac{50}{25}=2\)
\(\dfrac{x}{3}=\dfrac{2x}{6}=2\Rightarrow x=2\cdot3=6\\ \dfrac{y}{8}=\dfrac{3y}{24}=2\Rightarrow y=2\cdot8=16\\ \dfrac{z}{5}=2\Rightarrow z=2\cdot5=10\)
Vậy \(x=6;y=16;z=10\)
Làm gấp nên k có kiểm tra, bn bấm máy tính dò lại nhé
\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{3}=\dfrac{z}{5}\)
\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12};\dfrac{y}{12}=\dfrac{z}{20}\)
\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)
\(\Rightarrow\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}\)
\(=\dfrac{2x-3y+z}{18-36+20}\)
\(=\dfrac{6}{2}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.9=27\\y=3.12=36\\z=3.20=60\end{matrix}\right.\)
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\)
\(\Rightarrow x.\dfrac{2}{3}=y.\dfrac{3}{4}=z.\dfrac{4}{5}\)
\(\Rightarrow x:\dfrac{3}{2}=y:\dfrac{4}{3}=z:\dfrac{5}{4}\)
\(\Rightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)
\(=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}\)
\(=\dfrac{49}{\dfrac{49}{12}}=12\)
\(\Rightarrow\left\{{}\begin{matrix}x=12.\dfrac{3}{2}=18\\y=12.\dfrac{4}{3}=16\\z=12.\dfrac{5}{4}=15\end{matrix}\right.\)
Ta có :
\(\dfrac{x}{3}=\dfrac{y}{4}=>\dfrac{x}{9}=\dfrac{y}{12}\left(1\right)\)
\(\dfrac{y}{3}=\dfrac{z}{5}=>\dfrac{y}{12}=\dfrac{z}{20}\left(2\right)\)
Từ (1),(2)=>\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)=\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{2x-3y+z}{18-36+20}=\dfrac{6}{2}=3\)
=>\(\left\{{}\begin{matrix}x=27\\y=36\\z=60\end{matrix}\right.\)
\(\frac{x}{3}=\frac{y}{5}=\frac{z}{7}\)
\(\Rightarrow\frac{2x}{6}=\frac{3y}{15}=\frac{z}{7}=\frac{2x+3y-z}{6+15-7}=-1\)
\(\Rightarrow\frac{2x}{6}=-1\Rightarrow2x=-6\Rightarrow x=-3\)
\(\Rightarrow\frac{3y}{15}=-1\Rightarrow3y=-15\Rightarrow y=-5\)
\(\Rightarrow\frac{z}{7}=-1\Rightarrow z=-7\)
theo đề ta có: \(\frac{x}{3}=\frac{y}{5}=\frac{z}{7}\) và 2x + 3y - z = -14
=> \(\frac{2x}{6}=\frac{3y}{15}=\frac{z}{7}\)
Áp dụng t/c DTSBN ta có:
\(\frac{2x}{6}=\frac{3y}{15}=\frac{z}{7}=\frac{2x+3y-z}{6+15-7}=\frac{-14}{14}\) = \(-1\)
=> \(\frac{x}{3}=-1=>x=-3\)
\(\frac{y}{5}=-1=>y=-5\)
\(\frac{z}{7}=-1=>z=-7\)
t i c k nha!! 4354565475677687978873535752456465465765786876897978
\(\dfrac{2x}{3y}=-\dfrac{1}{3}\Rightarrow\dfrac{2x}{-1}=\dfrac{3y}{3}\)
áp dụng t/c của dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{-1}=\dfrac{3y}{3}=\dfrac{2x-3y}{-1-3}=\dfrac{7}{-4}\)
\(\dfrac{2x}{-1}=\dfrac{7}{-4}\Rightarrow x=\dfrac{7}{8}\\ \dfrac{3y}{3}=\dfrac{7}{-4}\Rightarrow y=-\dfrac{7}{4}\)