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Áp dụng t/c dtsbn:
\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}=\dfrac{-3x+3-4y-12+5z-25}{-6-16+30}=\dfrac{50+3-12-25}{8}=\dfrac{16}{8}=2\\ \Rightarrow\left\{{}\begin{matrix}x-1=4\\y+3=8\\z-5=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\y=5\\z=17\end{matrix}\right.\)
`@` `\text {Ans}`
`\downarrow`
Ta có: \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}\)
\(\Rightarrow\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}\)`=`\(\dfrac{\left(5z-25\right)-\left(3x-3\right)-\left(4y+12\right)}{30-6-16}\)
`=`\(\dfrac{5z-25-3x+3-4y-12}{8}\)
`=`\(\dfrac{\left(5z-3x-4y\right)+\left(-25+3-12\right)}{8}\)
`=`\(\dfrac{50-34}{8}\)`=`\(\dfrac{16}{8}=2\)
`=>`\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=2\)
`=>`\(\left\{{}\begin{matrix}x=2\cdot2+1=5\\y=2\cdot4-3=5\\z=2\cdot6+5=17\end{matrix}\right.\)
Vậy, `x,y,z` lần lượt là `5; 5; 17.`
b: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x-4y+5z+3-12-25}{-3\cdot2-4\cdot4+5\cdot6}=\dfrac{16}{8}=2\)
Do đó: x=5; y=5; z=17
\(a,\dfrac{x^3}{8}=\dfrac{y^3}{27}=\dfrac{z^3}{64}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\)
Áp dụng t/c dtsbn:
\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}=\dfrac{x^2+2y^2-3z^2}{4+18-48}=\dfrac{-650}{-26}=25\\ \Rightarrow\left\{{}\begin{matrix}x^2=100\\y^2=225\\z^2=400\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm10\\y=\pm15\\z=\pm20\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)\) có giá trị là hoán vị của \(\left(\pm10;\pm15;\pm20\right)\)
Có: \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}\)
<=> \(\dfrac{3\left(x-1\right)}{3.2}=\dfrac{4\left(y+3\right)}{4.4}=\dfrac{5\left(z-5\right)}{6.5}\)
<=> \(\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}\)
mà 5z-3x-4y=50
ADTCDTSBN ta có:\(\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}=\dfrac{5z-25-\left(4y+12\right)-\left(3x-3\right)}{30-16-6}=\dfrac{5z-25-4y-12-3x+3}{8}=\dfrac{\left(5z-4y-3x\right)-\left(25+12-3\right)}{8}=\dfrac{50-34}{8}=2\)
Do đó: \(\dfrac{3x-3}{6}=2\) <=> \(\dfrac{x-1}{2}=2\) <=> x-1=4 <=> x=5
\(\dfrac{4y+12}{16}=2\) <=> \(\dfrac{y+3}{4}=2\) <=> y+3=8<=> y=5
\(\dfrac{5z-25}{30}=2\) <=> \(\dfrac{z-5}{6}=2\) <=> z-5=12 <=> z=17
Vậy x=5 , y=5 , z=17
a: 2x-3y-4z=24
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y-4z}{2\cdot1-3\cdot6-4\cdot3}=\dfrac{24}{-28}=\dfrac{-6}{7}\)
=>x=-6/7; y=-36/7; z=-18/7
b: 6x=10y=15z
=>x/10=y/6=z/4=k
=>x=10k; y=6k; z=4k
x+y-z=90
=>10k+6k-4k=90
=>12k=90
=>k=7,5
=>x=75; y=45; z=30
d: x/4=y/3
=>x/20=y/15
y/5=z/3
=>y/15=z/9
=>x/20=y/15=z/9
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{9}=\dfrac{x-y-z}{20-15-9}=\dfrac{-100}{-4}=25\)
=>x=500; y=375; z=225
\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}\Rightarrow\dfrac{5\left(z-5\right)}{30}=\dfrac{3\left(x-1\right)}{6}=\dfrac{4\left(y+3\right)}{16}=\dfrac{5\left(z-5\right)-3\left(x-1\right)-4\left(y+3\right)}{30-6-16}=\dfrac{5z-25-3x+3-4y-12}{8}=\dfrac{\left(5z-3x-4y\right)+\left(-25+3-12\right)}{8}=\dfrac{50+\left(-34\right)}{8}=\dfrac{16}{8}=2\)\(\Rightarrow\dfrac{x-1}{2}=2\Rightarrow x=5\\ \dfrac{y+3}{4}=4\Rightarrow y=13\\ \dfrac{z-5}{6}=2\Rightarrow z=17\)
Theo bài ra ta có :
\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}\\ \Rightarrow\dfrac{3\left(x-1\right)}{6}=\dfrac{4\left(y+3\right)}{16}=\dfrac{5\left(z-5\right)}{30}\\ \Rightarrow\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}\\ \Rightarrow\dfrac{5z-25}{30}=\dfrac{3x-3}{6}=\dfrac{4y+12}{16}\)
\(5z-3x-4y=50\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được :
\(\dfrac{5z-25}{30}=\dfrac{3x-3}{6}=\dfrac{4y+12}{16}\\ =\dfrac{\left(5z-25\right)-\left(3x-3\right)-\left(4y+12\right)}{30-6-16}\\ =\dfrac{5z-25-3x+3-4y-12}{8}\\ =\dfrac{\left(5z-3x-4y\right)-\left(25-3+12\right)}{8}\\ =\dfrac{50-34}{8}\\ =\dfrac{16}{8}\\ =2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-3}{6}=2\Rightarrow3x-3=12\Rightarrow3x=15\Rightarrow x=5\\\dfrac{4y+12}{16}=2\Rightarrow4y+12=32\Rightarrow4y=20\Rightarrow y=5\\\dfrac{5z-25}{30}=2\Rightarrow5z-25=60\Rightarrow5z=85\Rightarrow z=17\end{matrix}\right.\)
\(\text{Vậy }x=5\\ y=5\\ z=17\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}\)
\(\Rightarrow\dfrac{5z-25}{30}=\dfrac{3x-3}{6}=\dfrac{4y+12}{16}\\ =\dfrac{\left(5z-25\right)-\left(3x-3\right)-\left(4y+12\right)}{30-6-16}\\ =\dfrac{5z-25-3x+3-4y-12}{8}\\ =\dfrac{5z-3x-4y-34}{8}\\ \dfrac{50-34}{8}=2\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{5z-25}{30}=2\\\dfrac{3x-3}{6}=2\\\dfrac{4y+12}{16}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}5z=85\\3x=15\\4y=20\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z=17\\x=5\\y=5\end{matrix}\right.\)