Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{3}=\dfrac{z}{5}\)
\(\Rightarrow\dfrac{x}{3}.\dfrac{1}{3}=\dfrac{y}{4}.\dfrac{1}{3};\dfrac{y}{3}.\dfrac{1}{4}=\dfrac{z}{5}.\dfrac{1}{4}\)
\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)
Áp dụng tính chất của dãy tỉ số bằng nhau :
\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}=\dfrac{2x-3y+z}{18-36+20}=\dfrac{6}{2}=3\)
\(\Rightarrow x=3.9=27\)
\(y=12.3=36\)
\(z=3.20=60\)
Theo bài ra ta có:
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}\\\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{20}\end{matrix}\right.\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}=\dfrac{2x-3y+z}{2.9-3.12+20}=\dfrac{6}{2}=3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=3\Rightarrow x=27\\\dfrac{y}{12}=3\Rightarrow y=36\\\dfrac{z}{20}=3\Rightarrow z=60\end{matrix}\right.\)
Vậy ..............................
Theo bài ra ta có:
⇒⎧⎪ ⎪⎨⎪ ⎪⎩x3=y4⇒x9=y12y3=z5⇒y12=z20⇒x9=y12=z20⇒{x3=y4⇒x9=y12y3=z5⇒y12=z20⇒x9=y12=z20
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
x9=y12=z20=2x−3y+z2.9−3.12+20=62=3x9=y12=z20=2x−3y+z2.9−3.12+20=62=3
⇒⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩x9=3⇒x=27y12=3⇒y=36z20=3⇒z=60⇒{x9=3⇒x=27y12=3⇒y=36z20=3⇒z=60
Vậy ..............................
\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{3}=\dfrac{z}{5}\)và 2x-3x+z=6
\(\dfrac{x}{3.3}=\dfrac{y}{4.3};\dfrac{y}{3.4}=\dfrac{z}{5.4}\)
\(\dfrac{x}{9}=\dfrac{y}{12};\dfrac{y}{12}=\dfrac{z}{20}\)
\(=>\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)và 2x-3x+z=6
Áp dụng t/c dãy tỉ số bằng nhau,ta có:
\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}=\dfrac{2x-3x+z}{2.9-3.9+20}=\dfrac{6}{11}\)
\(\dfrac{x}{9}=\dfrac{6}{11}=>x.11=6.9\)
\(x.11=54\)
\(x=54:11\)
\(x=\dfrac{54}{11}\)
\(\dfrac{y}{12}=\dfrac{6}{11}=>y.11=6.12\)
\(y.11=72\)
\(y=72:11\)
\(y=\dfrac{72}{11}\)
\(\dfrac{z}{20}=\dfrac{6}{11}=>z.11=6.20\)
\(z.11=120\)
\(z=120:11\)
\(z=\dfrac{120}{11}\)
