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\(2x=3y\text{⇒}\dfrac{x}{3}=\dfrac{y}{2}\text{⇒}\dfrac{x}{21}=\dfrac{y}{14}\)
\(5y=7z\text{⇒}\dfrac{y}{7}=\dfrac{z}{5}\text{⇒}\dfrac{y}{14}=\dfrac{z}{10}\)
⇒\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)⇒\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x-7y+5z}{63-98+50}=\dfrac{30}{15}=2\)
⇒x=42,y=28,z=20
\(\dfrac{x}{3}=\dfrac{y}{2}\)⇒\(\dfrac{x}{15}=\dfrac{y}{10}\)
\(\dfrac{x}{5}=\dfrac{z}{7}\text{⇒}\dfrac{x}{15}=\dfrac{z}{21}\)
⇒\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{21}\)⇒\(\dfrac{x}{15}=\dfrac{2y}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{15}=\dfrac{2y}{20}=\dfrac{x+2y}{15+20}=\dfrac{-112}{35}=\dfrac{-16}{5}\)
⇒x=48,y=32,z=336/5
a: \(\Leftrightarrow x^3=-216\)
=>x=-6
b: \(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{\dfrac{5}{2}}=\dfrac{z}{\dfrac{7}{4}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{\dfrac{5}{2}}=\dfrac{z}{\dfrac{7}{4}}=\dfrac{3x+5y+7z}{3\cdot2+5\cdot\dfrac{5}{2}+7\cdot\dfrac{7}{4}}=\dfrac{123}{\dfrac{123}{4}}=4\)
=>x=8; y=10; z=7
a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-49}{7}=-7\)
Do đó: x=-70; y=-135; z=-84
a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
x10=y15=z12=x−y+z10−15+12=−497=−7x10=y15=z12=x−y+z10−15+12=−497=−7
Do đó: x=-70; y=-135; z=-84
\(\dfrac{x}{y}=\dfrac{3}{7}.\\ \Rightarrow x=\dfrac{3}{7}y.\\ x-y=16.\\\Rightarrow\dfrac{3}{7}y-y=16.\\ \Rightarrow y=-28.\\ \Rightarrow x=-12.\)
\(\dfrac{x}{1,8}=\dfrac{y}{3,2}.\\ \Rightarrow\dfrac{x}{y}=\dfrac{1,8}{3,2}=\dfrac{9}{16}.\\ \Rightarrow x=\dfrac{9}{16}y.\\ y-x=7.\\ \Rightarrow y-\dfrac{9}{16}y=7.\\ \Leftrightarrow y=16.\\ \Leftrightarrow x=9.\)
\(\dfrac{x}{5}=\dfrac{y}{8}.\\ \Rightarrow\dfrac{x}{y}=\dfrac{5}{8}.\\ \Rightarrow x=\dfrac{5}{8}y.\\ x+2y=42.\\ \Rightarrow\dfrac{5}{8}y+2y=42.\\ \Leftrightarrow y=16.\\ \Rightarrow x=10.\)
\(\dfrac{x}{5}=\dfrac{y}{7}.\\ \Rightarrow\dfrac{x}{y}=\dfrac{5}{7}.\\ \Rightarrow x=\dfrac{5}{7}y.\\ x.y=35.\\ \Rightarrow\dfrac{5}{7}y.y=35.\\ \Leftrightarrow y^2=49.\\ \Leftrightarrow u=\pm7.\\ \Rightarrow x=\pm5.\)
\(\dfrac{x}{2}=\dfrac{y}{3}\text{⇒}\dfrac{x}{10}=\dfrac{y}{15}\)
\(\dfrac{y}{5}=\dfrac{z}{4}\text{⇒}\dfrac{y}{15}=\dfrac{z}{12}\)
⇒\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-21}{-3}=7\)
⇒x=70;y=105;z=84
\(\dfrac{x+1}{3}=\dfrac{3x+3}{9}\)
\(\dfrac{y+2}{-4}=\dfrac{2y+4}{-8}\)
\(\dfrac{t+3}{5}=\dfrac{4t+12}{20}\)
Do đó : \(\dfrac{3x+3}{9}=\dfrac{2y+4}{-8}=\dfrac{4t+12}{20}=\dfrac{3x+2y+4t+19}{21}=\dfrac{47+19}{21}=\dfrac{22}{7}\)
Suy ra : x = \(\dfrac{59}{7}\); y = \(-\dfrac{102}{7}\); z = \(\dfrac{89}{7}\)
1/ Đặt: \(\dfrac{x}{2}=\dfrac{2y}{3}=\dfrac{3t}{4}=k\)
=> \(x=2k;y=\dfrac{3k}{2};t=\dfrac{4k}{3}\)
=> \(xyt=2k\cdot\dfrac{3k}{2}\cdot\dfrac{4k}{3}=4k^3=-108\)
=> \(k^3=-27\Rightarrow k=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k=2\cdot\left(-3\right)=-6\\y=\dfrac{3k}{2}=\dfrac{3\cdot\left(-3\right)}{2}=-\dfrac{9}{2}\\t=\dfrac{4k}{3}=\dfrac{4\cdot\left(-3\right)}{3}=-4\end{matrix}\right.\)
Vậy ...........
2/ Sửa đề: 3x + 5y+7t = 123
Ta có: \(\dfrac{x}{2}=\dfrac{2y}{5}=\dfrac{4t}{7}\)
\(\Rightarrow\dfrac{3x}{6}=\dfrac{5y}{12,5}=\dfrac{7t}{12,25}\)
A/dung t/c của dãy tỉ số bằng nhau ta có:
\(\dfrac{3x}{6}=\dfrac{5y}{12,5}=\dfrac{7t}{12,25}=\dfrac{3x+5y+7t}{6+12,5+12,25}=\dfrac{123}{30,75}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4\cdot6}{3}=8\\y=\dfrac{4\cdot12,5}{5}=10\\t=\dfrac{4\cdot12,25}{7}=7\end{matrix}\right.\)
Vậy............
Thanks bạn