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\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) \(\Rightarrow\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}=\frac{x-1-2y-4+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(1+4+9\right)}{8}=\frac{14-14}{8}=0\)
\(x=0.2+1=1\) ; \(y=\left(0.6+4\right):2=2\) ; \(z=\left(0.12+9\right):3=3\)
\(\frac{x-5}{3}=\frac{y-4}{4}=\frac{z-3}{5}=\frac{x-5+y-4+z-3}{3+4+5}=\frac{36-12}{12}=\frac{24}{12}=2\)
\(\Rightarrow\hept{\begin{cases}x-5=6\\y-4=8\\z-3=10\end{cases}}\Rightarrow\hept{\begin{cases}x=11\\y=12\\z=13\end{cases}}\)
e, Ta có: \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
\(\Rightarrow\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}=\frac{\left(x-2y+3z\right)+\left(-1+4-9\right)}{8}=\frac{14-6}{8}=1\)
Do đó: \(\frac{x-1}{2}=1\Rightarrow x=2.1+1=3\)
\(\frac{2y-4}{6}=1\Rightarrow y=\frac{6.1+4}{2}=5\)
\(\frac{3z-9}{12}=1\Rightarrow z=\frac{12.1+9}{3}=7\)
Vậy x=3; y=5; z=7
h, Ta có: \(\frac{x}{2}=\frac{y}{3}=\left(\frac{x}{2}\right)^2=\left(\frac{y}{3}\right)^2=\frac{x^2}{4}=\frac{y^2}{9}=\frac{x.y}{2.3}=\frac{54}{6}=9\)
Do đó: \(\frac{x^2}{4}=9\Rightarrow x^2=4.9=36\Rightarrow x=6;x=-6\)
\(\frac{y^2}{9}=9\Rightarrow y^2=9.9=81\Rightarrow y=9;y=-9\)
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=>\frac{x-1}{2}=\frac{2\left(y-2\right)}{6}=\frac{3\left(z-3\right)}{12}=>\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}\)
Theo t/c dãy tỉ số=nhau:
\(\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}=\frac{x-1-\left(2y-4\right)+\left(3z-9\right)}{2-6+12}=\frac{x-1-2y+4+3z-9}{8}\)
\(=\frac{\left(x-2y+3z\right)-\left(1-4+9\right)}{8}=\frac{14-6}{8}=\frac{8}{8}=1\)
Do đó: \(\frac{x-1}{2}=1=>x-1=2=>x=3\)
\(\frac{y-2}{3}=1=>y-2=3=>y=5\)
\(\frac{z-3}{4}=1=>z-3=4=>z=7\)
Vậy x=3;y=5;z=7
b) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) = \(\dfrac{x+y+z}{2+3+5}=\dfrac{-90}{10}=-9\)
\(\dfrac{x}{2}=-9\) => x= -18
\(\dfrac{y}{3}=-9\) => y = -27
\(\dfrac{z}{5}=-9\) => z = -45
a) \(4x=5y\) <=> \(x=\dfrac{5y}{4}\)
\(3\cdot\dfrac{5y}{4}-2y=35\)
=> y = 20
=> x = \(\dfrac{5\cdot20}{4}\)=25
\(\frac{x-1}{2}\)= \(\frac{2y-4}{6}\)=\(\frac{3z-9}{12}\)=\(\frac{x-1-2y+4+3z-9}{2-6+12}\)= \(\frac{14-1+4-9}{8}\)= 1
=> x =2+1=3
y= (6+4) : 2=5
z=(12+9) : 3=7