Question

Tìm các số x,y,z biết

a) \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\) và \(2x+3y+5z=86\)

b) \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{6}=\dfrac{z}{8}\) và \(3x-2y-z=13\)

c) \(x:y:z=2:5:7\) và \(3x+2y-z=27\)

d) \(\dfrac{x}{3}=\dfrac{y}{7}=2x^2+y^2+3z^2=316\)

Answer 1

a)vì\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=\(\dfrac{z}{5}\)=>\(\dfrac{2x}{6}\)=\(\dfrac{3y}{12}\)=\(\dfrac{5z}{25}\)và 2x+3y+5z=86

áp dụng tính chất của dãy tỉ số bằng nhau ta có

\(\dfrac{2x}{6}\)=\(\dfrac{3y}{12}\)=\(\dfrac{5z}{25}\)=\(\dfrac{2x+3y+5z}{6+12+25}\)\(\dfrac{86}{43}\)=2

vì\(\dfrac{2x}{6}\)=2=>2x=2.6=12=>x=12:2=6

\(\dfrac{3y}{12}\)=2=>3y=12.2=24=>y=24:3=8

\(\dfrac{5z}{25}\)=2=>5z=25.2=50=>z=50:5=10

vậy x=6,y=8,z=10

Answer 2

vì\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=>\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)(1)

\(\dfrac{y}{6}\)=\(\dfrac{z}{8}\)=>\(\dfrac{y}{12}\)=\(\dfrac{z}{16}\)(2)

từ (1)(2)=>\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)=\(\dfrac{z}{16}\)=>\(\dfrac{3x}{27}\)=\(\dfrac{2y}{24}\)=\(\dfrac{z}{16}\)và 3x-2y-z=13

áp dụng tính chất của dãy tỉ số bằng nhau ta có

\(\dfrac{3x}{27}\)=\(\dfrac{2y}{24}\)=\(\dfrac{z}{16}\)=\(\dfrac{3x-2y-z}{27-24-16}\)=\(\dfrac{13}{-13}\)=-1

vì\(\dfrac{3x}{27}\)=-1=>3x=-1.27=-27=>x=-27x;3=-9

\(\dfrac{2y}{24}\)=-1=>2y=-1.24=-24=>y=-24:2=-12

\(\dfrac{z}{16}\)=-1=>z=-1.16=-16

vậy...

Answer 3

\(a,\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\Leftrightarrow\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{5z}{25}\) và \(2x+3y+5z=86\)

Áp dụng tính chất dãy tỉ số bằng nhau có:

\(\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{5z}{25}=\dfrac{2x+3y+5z}{6+12+25}=\dfrac{86}{43}=2\)

+) \(\dfrac{2x}{6}=2\Rightarrow2x=2\cdot6=12\Rightarrow x=12:2=6\)

+) \(\dfrac{3y}{12}=2\Rightarrow3y=2\cdot12=24\Rightarrow y=24:3=8\)

+) \(\dfrac{5z}{25}=2\Rightarrow5z=2\cdot25=50\Rightarrow5z=50:5=10\)

Vậy ....

\(b,\dfrac{x}{3}=\dfrac{y}{4}\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{12}\left(1\right)\)

\(\dfrac{y}{6}=\dfrac{z}{8}\Leftrightarrow\dfrac{y}{12}=\dfrac{z}{16}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\Leftrightarrow\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}\) và \(3x-2y-z=13\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}=\dfrac{3x-2y-z}{27-24-16}=\dfrac{13}{-13}=-1\)

+) \(\dfrac{3x}{27}=-1\Rightarrow3x=-27\Rightarrow x=-27:3=-9\)

+) \(\dfrac{2y}{24}=-1\Rightarrow2y=-24\Rightarrow y=-24:2=-12\)

+) \(\dfrac{z}{16}=-1\Rightarrow x=-16\)

Vậy .....

\(c,x:y:z=2:5:7\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}\Leftrightarrow\dfrac{3x}{6}=\dfrac{2y}{10}=\dfrac{z}{7}\) và \(3x+2y-z=27\)

Áp dụng tính chất dãy tỉ số bằng nhau có:

\(\dfrac{3x}{6}=\dfrac{2y}{10}=\dfrac{z}{7}=\dfrac{3x+2y-z}{6+10-7}=\dfrac{27}{9}=3\)

+) \(\dfrac{3x}{6}=3\Rightarrow3x=3\cdot6=18\Rightarrow x=18:3=6\)

+) \(\dfrac{2y}{10}=3\Rightarrow2y=3\cdot10=30\Rightarrow y=30:2=15\)

+) \(\dfrac{z}{7}=3\Rightarrow z=3\cdot7=21\)

Vậy ....

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