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\(\dfrac{7x-3z}{5}=\dfrac{3y-5x}{7}=\dfrac{5z-7y}{3}\)
\(\Rightarrow\dfrac{35x-15z}{25}=\dfrac{21y-35x}{49}=\dfrac{15z-21y}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{35x-15z}{25}=\dfrac{21y-35x}{49}=\dfrac{15z-21y}{9}\)
\(=\dfrac{35x-15z+21y-35x+15z-21y}{25+49+9}\)
\(=\dfrac{0}{25+49+9}=0\)
\(\Rightarrow\left\{{}\begin{matrix}7x=3z\Rightarrow\dfrac{x}{3}=\dfrac{z}{7}\\3y=5x\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\\5z=7y\Rightarrow\dfrac{z}{7}=\dfrac{y}{5}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{3+5+7}=\dfrac{30}{15}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=2.5=10\\z=2.7=14\end{matrix}\right.\)
Tương tự
Bài 1:
Ta có:
+) \(3.4=2.6\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}=\dfrac{6}{4}\\\dfrac{3}{6}=\dfrac{2}{4}\\\dfrac{4}{2}=\dfrac{6}{3}\\\dfrac{4}{6}=\dfrac{2}{3}\end{matrix}\right.\)
+) \(3.6=2.9\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}=\dfrac{9}{6}\\\dfrac{3}{9}=\dfrac{2}{6}\\\dfrac{6}{2}=\dfrac{9}{3}\\\dfrac{6}{9}=\dfrac{2}{3}\end{matrix}\right.\)
Bài 2:
a) Ta có: \(\dfrac{x}{11}=\dfrac{y}{13}\) và \(x-y=6\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{11}=\dfrac{y}{13}=\dfrac{x-y}{11-13}=\dfrac{6}{-2}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=11.\left(-3\right)=-33\\y=13.\left(-3\right)=-39\end{matrix}\right.\)
Vậy \(x=-33;y=-39\)
b) Theo bài ra ta có:
\(x:y:z=1:2:3\)
\(\Rightarrow\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\)
\(\Rightarrow\dfrac{4x}{4}=\dfrac{3y}{6}=\dfrac{2z}{6}\)
và \(4x-3y+2z=36\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{4x}{4}=\dfrac{3y}{6}=\dfrac{2z}{6}=\dfrac{4x-3y+2z}{4-6+6}=\dfrac{36}{4}=9\)
\(\Rightarrow\left\{{}\begin{matrix}4x=4.9=36\Rightarrow x=9\\3y=6.9=54\Rightarrow y=18\\2z=6.9=54\Rightarrow z=27\end{matrix}\right.\)
Vậy \(x=9;y=18;z=27\)
c) Ta có: \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}\)
\(\Rightarrow\dfrac{5x}{15}=\dfrac{y}{5}=\dfrac{3z}{-6}\)
và \(5x-y+3z=124\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{5x}{15}=\dfrac{y}{5}=\dfrac{3z}{-6}=\dfrac{5x-y+3z}{15-5+\left(-6\right)}=\dfrac{124}{4}=31\)
\(\Rightarrow\left\{{}\begin{matrix}5x=15.31=465\Rightarrow x=93\\y=5.31=155\\3z=\left(-6\right).31=-186\Rightarrow z=-62\end{matrix}\right.\)
Vậy \(x=93;y=155;z=-62\)
Theo đề bài, ta có:
\(\dfrac{x}{3}\)=\(\dfrac{y}{5}\)=\(\dfrac{z}{6}\)=\(\dfrac{2x}{6}\)=\(\dfrac{3y}{15}\)=\(\dfrac{4z}{24}\)
\(\dfrac{x}{3}\)=\(\dfrac{y}{5}\)=\(\dfrac{z}{6}\)=\(\dfrac{x}{3}\)=\(\dfrac{11y}{55}\)=\(\dfrac{4z}{24}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{x}{3}\)=\(\dfrac{y}{5}\)=\(\dfrac{z}{6}\)=\(\dfrac{2x}{6}\)=\(\dfrac{3y}{15}\)=\(\dfrac{4z}{24}\)= \(\dfrac{2x-3y+4z}{6-15+24}\)=\(\dfrac{2x-3y+4z}{15}\)(*)
\(\dfrac{x}{3}\)=\(\dfrac{y}{5}\)=\(\dfrac{z}{6}\)=\(\dfrac{x}{3}\)=\(\dfrac{11y}{55}\)=\(\dfrac{4z}{24}\)=\(\dfrac{x-11y-4z}{3-55-24}\)=\(\dfrac{x-11y-4z}{-76}\)(**)
Từ (*) và (**) suy ra:
\(\dfrac{2x-3y+4z}{15}\)=\(\dfrac{x-11y-4z}{-76}\)=\(\dfrac{2x-3y+4z}{x-11y-4z}\)=\(\dfrac{15}{-76}\)
=> m=\(\dfrac{15}{-76}\)
Vậy m=\(\dfrac{15}{-76}\)
Xét \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\) (1)
Thay (1) vào P
=> P = \(\dfrac{3k+2.4k+3.5k}{2.5k+3.4k+4.5k}+\dfrac{2.5k+3.4k+4.5k}{3.3k+4.4k+5.5k}\) + \(\dfrac{3.3k+4.4k+5.5k}{4.3k+5.4k+6.5k}\)
=> P = \(\dfrac{26k}{42k}+\dfrac{42k}{50k}\) + \(\dfrac{50k}{62k}\)
=> P = \(\dfrac{13}{21}+\dfrac{21}{25}+\dfrac{25}{31}\approx2,265499232\)
Sửa: \(\dfrac{2x-3y}{4}=\dfrac{3y-4z}{5}=\dfrac{2z-x}{6}\)
\(\Rightarrow\dfrac{2x-3y}{4}=\dfrac{3y-4z}{5}=\dfrac{4z-2x}{12}=\dfrac{2x-3y+3y-4z+4z-2x}{4+5+12}=0\\ \Rightarrow\left\{{}\begin{matrix}2x-3y=0\\3y-4z=0\\4z-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=3y\\3y=4z\\4z=2x\end{matrix}\right.\Rightarrow2x=3y=4z\)
Vậy x,y,z tỉ lệ nghịch với 2;3;4