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2.
Ta có: \(2a=3b=4c\Leftrightarrow\dfrac{12a}{6}=\dfrac{12b}{4}=\dfrac{12c}{3}\Rightarrow\dfrac{a}{6}=\dfrac{b}{4}=\dfrac{c}{3}\) và \(a+b-c=14\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{6}=\dfrac{b}{4}=\dfrac{c}{3}=\dfrac{a+b-c}{6+4-3}=\dfrac{14}{7}=2\)
+) \(\dfrac{a}{6}=2\Rightarrow a=6\cdot2=12\)
+) \(\dfrac{b}{4}=2\Rightarrow b=2\cdot4=8\)
+) \(\dfrac{c}{3}=2\Rightarrow c=3\cdot2=6\)
Vậy...
1 thiếu đề
2. \(2a=3b=4c\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{2}=\dfrac{a+b-c}{3+4-2}=\dfrac{14}{5}\)
Từ đây có thể tính a,b và c
1)
Ta có: \(\dfrac{a}{2}=\dfrac{b}{3};\dfrac{a}{3}=\dfrac{c}{4}\Leftrightarrow\dfrac{a}{6}=\dfrac{b}{9};\dfrac{a}{6}=\dfrac{c}{8}\Rightarrow\dfrac{a}{6}=\dfrac{b}{9}=\dfrac{c}{8}\)và \(a-b-c=22\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{6}=\dfrac{b}{9}=\dfrac{c}{8}=\dfrac{a-b-c}{6-9-8}=\dfrac{22}{-11}=-2\)
\(\dfrac{a}{6}=-2\Rightarrow a=-2.6=-12\)
\(\dfrac{b}{9}=-2\Rightarrow b=-2.9=-18\)
\(\dfrac{c}{8}=-2\Rightarrow c=-2.8=-16\)
Vậy \(a=-12\) ; \(b=-18\) ; \(c=-16\)
câu 2
\(\dfrac{2a}{1}\)=\(\dfrac{3b}{1}\)=\(\dfrac{4c}{1}\)=\(\dfrac{2a}{12}\)=\(\dfrac{3b}{12}\)=\(\dfrac{4c}{12}\)
\(\Rightarrow\)\(\dfrac{a}{6}\)=\(\dfrac{b}{4}\)=\(\dfrac{c}{3}\)
Theo tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{6}\)=\(\dfrac{b}{4}\)=\(\dfrac{c}{3}\)=\(\dfrac{a+b-c}{6+4-3}\)=\(\dfrac{14}{7}\)=2
\(\)\(\dfrac{a}{6}\)=\(\dfrac{2}{1}\)thì a=2.6:1=12
\(\dfrac{b}{4}\)=\(\dfrac{2}{1}\)thì b=2.4:1=8
\(\dfrac{c}{3}\)=\(\dfrac{2}{1}\)thì c=2.3:1=6
