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a) Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\)
\(\Leftrightarrow\dfrac{a}{8}=\dfrac{b}{12}\)(1)
Ta có: \(\dfrac{b}{4}=\dfrac{c}{5}\)
nên \(\dfrac{b}{12}=\dfrac{c}{15}\)(2)
Từ (1) và (2) suy ra \(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}\)
mà a+b+c=2
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}=\dfrac{a+b+c}{8+12+15}=\dfrac{2}{35}\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{a}{8}=\dfrac{2}{35}\\\dfrac{b}{12}=\dfrac{2}{35}\\\dfrac{c}{15}=\dfrac{2}{35}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{16}{35}\\b=\dfrac{24}{35}\\c=\dfrac{30}{35}=\dfrac{6}{7}\end{matrix}\right.\)
Vậy: \(a=\dfrac{16}{35}\); \(b=\dfrac{24}{35}\); \(c=\dfrac{6}{7}\)
b) Ta có: 2a=3b=5c
nên \(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}\)
mà a+b-c=3
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}=\dfrac{a+b-c}{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}=\dfrac{3}{\dfrac{19}{30}}=\dfrac{90}{19}\)
Do đó:
\(\left\{{}\begin{matrix}2a=\dfrac{90}{19}\\3b=\dfrac{90}{19}\\5c=\dfrac{90}{19}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{45}{19}\\b=\dfrac{30}{19}\\c=\dfrac{18}{19}\end{matrix}\right.\)
Vậy: \(a=\dfrac{45}{19}\); \(b=\dfrac{30}{19}\); \(c=\dfrac{18}{19}\)
a,3x=2y;7y=5z
=>\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta co:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x-y+z}{10-15+21}=\dfrac{32}{16}=2\\ \Rightarrow x=2.10=20\\ y=2.15=30\\ z=2.21=42\)
Các câu sau tương tự
b,\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\),\(\dfrac{y}{3}\)=\(\dfrac{z}{5}\) và 2x-3y+z=6
Từ đề bài ta có:
\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)\(\Rightarrow\)\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)(1)
\(\dfrac{y}{3}\)=\(\dfrac{z}{5}\)\(\Rightarrow\)\(\dfrac{y}{12}\)=\(\dfrac{z}{20}\)(2)
từ (1) và (2)\(\Rightarrow\)\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)=\(\dfrac{z}{20}\)\(\Rightarrow\)\(\dfrac{2x}{18}\)=\(\dfrac{3y}{36}\)=\(\dfrac{z}{20}\)
Áp dụng t/c 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
\(\Rightarrow\)x=3.9=27
y=3.12=36
z=3.20=60
Vậy.....
chúc bạn học tốt,nhớ tick cho mình nha
b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)
Đặt \(x=15k;y=20k;z=24k\)
Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
1. Ta có: \(\dfrac{x}{-7}=\dfrac{y}{4}\Rightarrow\dfrac{2x}{-14}=\dfrac{3y}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x-3y}{-14-12}=\dfrac{-78}{-26}=3\)
=> \(\left\{{}\begin{matrix}x=-21\\y=12\end{matrix}\right.\)
2. Ta có:
- \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)
- \(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)
=> \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)
=> \(\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)
1.
Đặt \(\dfrac{x}{5}=\dfrac{y}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\end{matrix}\right.\)
\(\Rightarrow x^2-y^2=\left(5k\right)^2-\left(4k\right)^2=25k^2-16k^2=9k^2=4\)
\(\Rightarrow k^2=\dfrac{4}{9}\Rightarrow k=\pm\dfrac{2}{3}\)
\(\circledast k=\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=\dfrac{8}{3}\end{matrix}\right.\)
\(\circledast k=-\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{3}\\y=-\dfrac{8}{3}\end{matrix}\right.\)
2.
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+1+3y-2}{5+7}=\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)
\(\Rightarrow6x=12\Rightarrow x=2\)
\(\Rightarrow y=\dfrac{\dfrac{2\cdot2+1}{5}\cdot7+2}{3}=3\)
3.
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Leftrightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-\left(z-3\right)}{4+9-4}=\dfrac{95-8+3}{9}=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10\cdot4+2}{2}=21\\y=\dfrac{10\cdot9+6}{3}=32\\z=10\cdot4+3=43\end{matrix}\right.\)
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
Bài 1:
a) \(4\dfrac{1}{3}:\dfrac{x}{4}=6:0,3\)
\(\Rightarrow\dfrac{13}{3}.\dfrac{4}{x}=20\)
\(\Rightarrow\dfrac{52}{3x}=20\)
\(\Rightarrow52=20.3x\)
\(\Rightarrow60x=52\)
\(\Rightarrow x=\dfrac{13}{15}\)
b) \(\left(2^3:2^4\right).2^{x+1}=64\)
\(\Rightarrow2^{3-4}.2^{x+1}=64\)
\(\Rightarrow2^{-1}.2^{x+1}=64\)
\(\Rightarrow2^{-1+x+1}=64\)
\(\Rightarrow2^x=64\)
\(\Rightarrow2^x=2^6\)
\(\Rightarrow x=6\)
c) \(\left(x-1\right)^5=-32\)
\(\Rightarrow\left(x-1\right)^5=\left(-2\right)^5\)
\(\Rightarrow x-1=-2\)
\(\Rightarrow x=-2+1=-1\)
d) \(|3-2x|-3=-3\)
\(\Rightarrow|3-2x|=-3+3=0\)
\(\Rightarrow3-2x=0\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
e) \(|x+\dfrac{4}{5}|-\dfrac{1}{7}=0\)
\(\Rightarrow|x+\dfrac{4}{5}|=\dfrac{1}{7}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{1}{7}\\x+\dfrac{4}{5}=-\dfrac{1}{7}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}-\dfrac{4}{5}\\x=-\dfrac{1}{7}-\dfrac{4}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{23}{35}\\x=-\dfrac{33}{35}\end{matrix}\right.\)
Bài 2:
Ta có:
\(2x=3y=6z\)
\(=\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{6}}\)
\(=\dfrac{x+y+z}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}}\) ( Áp dụng tính chất dãy tỉ số bằng nhau )
\(=\dfrac{1830}{1}=1830\)
Với \(\left\{{}\begin{matrix}2x=1830\\3y=1830\\6z=1830\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=915\\y=610\\z=305\end{matrix}\right.\)