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Đặt \(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=-4k\\y=-7k\\z=3k\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{-2.\left(-4k\right)+\left(-7k\right)+5.3k}{-4k-3.\left(-7k\right)-6.3k}=\dfrac{16k}{-1k}=-16\)
a) \(\Rightarrow\frac{2x}{3}.\frac{1}{12}=\frac{3y}{4}.\frac{1}{12}=\frac{4z}{5}.\frac{1}{12}\)
\(\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}\)
Ánh dụng tính chất của dãy tỉ số bằng nhau :
\(\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y+z}{18+16+15}=\frac{49}{49}=1\)
\(\Rightarrow\) x = 1 . 18 = 18
y = 1 . 16 = 16
z = 1 . 15 = 15
b)
Từ 4x = 3y ; 7y=5z => \(\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\)
\(\Rightarrow\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\left(1\right)\)
\(\Rightarrow\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{20}=\frac{z}{28}\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}\)
Áp dụng tính chất của dãy tỉ số bằng nhau :
\(\Rightarrow\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}=\frac{2x+3y-z}{30+60-28}=\frac{124}{62}=2\)
\(\Rightarrow\) x = 2 . 15 = 30
y = 2 . 20 = 40
z = 2 . 28 = 56
c) từ 10x=6y \(\Rightarrow\) \(\frac{x}{6}=\frac{y}{10}\) \(\left(\frac{x}{6}\right)^2\)=\(\left(\frac{y}{10}\right)^2\) \(\Rightarrow\frac{x^2}{36}\)=\(\frac{y^2}{100}\) \(\Rightarrow\frac{2x^2}{72}=\frac{y^2}{100}\)
áp dụng tính chất của dãy tỉ số bằng nhau :
\(\frac{2x^2-y^2}{72-100}\) = \(\frac{-28}{-28}\) = 1
\(\Rightarrow\frac{x}{6}=1\) ; \(\frac{y}{10}=1\)
\(\Rightarrow x=6;y=10\)
hoặc \(\Rightarrow\frac{x}{6}=-1;\frac{y}{10}=-1\)
\(\Rightarrow x=-6;y=-10\)
Chúc bạn học tốt
Bài 1:
a) \(\frac{x-1}{0-2}=\frac{1,2}{1,5}\)
\(\Leftrightarrow\frac{1-x}{2}=\frac{4}{5}\)
\(\Leftrightarrow5-5x=8\)
\(\Leftrightarrow x=-\frac{3}{5}\)
b) Ta có: \(x=\frac{y}{2}=\frac{z}{3}=\frac{4x-3y+2z}{4-6+6}=\frac{16}{4}=4\)
\(\Rightarrow\hept{\begin{cases}x=4\\y=8\\z=12\end{cases}}\)
Bài 1:
c) \(2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\Leftrightarrow\frac{x}{21}=\frac{y}{14}\)
\(5y=7z\Leftrightarrow\frac{y}{7}=\frac{z}{5}\Leftrightarrow\frac{y}{14}=\frac{z}{10}\)
\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2\)
\(\Rightarrow\hept{\begin{cases}x=42\\y=28\\z=20\end{cases}}\)
d) \(x:y:z=3:5:2\Leftrightarrow\frac{x}{3}=\frac{y}{5}=\frac{z}{2}=\frac{5x-7y+5z}{15-35+10}=\frac{124}{-10}\)
\(\Rightarrow\hept{\begin{cases}x=-\frac{186}{5}\\y=-62\\z=-\frac{124}{5}\end{cases}}\)
1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)
2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)
3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)
Áp dụng t/c dtsbn:
\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)
Theo bài ra ta cs
\(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\)
\(4y=5z\Rightarrow\frac{y}{5}=\frac{z}{4}\)
T lại cs
\(\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{15}=\frac{y}{10}\left(1\right)\)
\(\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{x}{10}=\frac{z}{8}\left(2\right)\)
Từ (1);(2) \(\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{8}\)
ADTC dãy tỉ số bằng nhau ta cs
\(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}=\frac{2x+3y-4z}{2.15+3.10-4.8}=\frac{56}{28}=2\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{15}=2\\\frac{y}{10}=2\\\frac{z}{8}=2\end{cases}\Rightarrow\hept{\begin{cases}x=30\\y=20\\z=16\end{cases}}}\)
\(2x=3y;4y=5z\) => \(8x=12y;12y=15z\)
=> \(\frac{8x}{120}=\frac{12y}{120}=\frac{15z}{120}\)=> \(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}\)
=> \(\frac{2x}{30}=\frac{3y}{30}=\frac{4z}{32}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{2x}{30}=\frac{3y}{30}=\frac{4z}{32}=\frac{2x+3y-4z}{30+30-32}=\frac{56}{28}\)
=> \(\frac{2x}{30}=2=>2x=60=>x=30\)
\(\frac{3y}{30}=2=>3y=60=>y=20\)
\(\frac{4z}{32}=2=>4z=64=>z=16\)
a) Vì \(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{21}=\frac{y}{14}\)
\(3y=7z\Rightarrow\frac{y}{7}=\frac{z}{3}\Rightarrow\frac{y}{14}=\frac{z}{6}\)
\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{6}\) và x+y-z=58
APa dụng TC dãy TSBN ta có
\(\frac{x}{21}=\frac{y}{14}=\frac{z}{6}=\frac{x+y-z}{21+14-6}=\frac{58}{29}=2\)
\(\Rightarrow x=42;y=28;z=12\)
Các câu còn lại tương tự
\(2x=3y=5z\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x-y+z}{15-10+6}=\dfrac{-33}{11}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).15=-45\\y=\left(-3\right).10=-30\\z=\left(-3\right).6=-18\end{matrix}\right.\)
Theo tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{5}}=\dfrac{x-y+z}{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{5}}=\dfrac{-33}{\dfrac{11}{30}}=-90\)
Do đó: x=-45; y=-30; z=-18