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Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{x+y+1+x+z+2+x+y-3}{x+y+z}=\dfrac{\left(x+y+z\right)+\left(x+y+z\right)+\left(1+2-3\right)}{x+y+z}=\dfrac{2.\left(x+y+z\right)}{x+y+z}=2\)
Lại có:
\(\dfrac{y+z+1}{x}+\dfrac{x+z+2}{y}+\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
\(\Rightarrow2=\dfrac{1}{x+y+z}\)
\(\Rightarrow2.\left(x+y+z\right)=1\)
\(\Rightarrow x+y+z=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{y+z+1}{x}=2\\\dfrac{x+z+2}{y}=2\\\dfrac{x+y-3}{z}=2\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+y+z+1=3x\\x+y+z+2=3y\\x+y+z-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}+1=3x\\\dfrac{1}{2}+2=3y\\\dfrac{1}{2}-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\dfrac{1}{2}}{3}\\y=\dfrac{\dfrac{1}{2}+2}{3}\\z=\dfrac{\dfrac{1}{2}-3}{3}\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=-\dfrac{5}{6}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=-\dfrac{5}{6}\end{matrix}\right.\) .
1)
Ta có:
\(2x=3y=4z\Leftrightarrow\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x-y-z}{\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}}=-420\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-420.\dfrac{1}{2}=-210\\y=-420.\dfrac{1}{3}=-140\\z=-420.\dfrac{1}{4}=-105\end{matrix}\right.\)
Vậy....
1: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{0,3}=\dfrac{y}{0.2}=\dfrac{z}{0.1}=\dfrac{x-y}{0.3-0.2}=\dfrac{1}{0.1}=10\)
Do đó: x=3; y=2; z=1
TH1: x+y+z=0
\(\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\z+x=-y\end{matrix}\right.\Rightarrow\dfrac{x+y}{z}=\dfrac{y+z}{x}=\dfrac{x+z}{y}=-1\)
TH2:\(x+y+z\ne0\)
Áp dụng tc dãy tỉ số bằng nhau ta có:
\(\dfrac{x+y}{z}=\dfrac{y+z}{x}=\dfrac{z+x}{y}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)Vậy.....
1 a) \(\dfrac{\left(-2\right)}{5}\)= \(\dfrac{-6}{15}\); \(\dfrac{15}{-6}\)= \(\dfrac{5}{-2}\); \(\dfrac{-6}{-2}\)= \(\dfrac{15}{5}\); \(\dfrac{-2}{-6}\)= \(\dfrac{5}{15}\)
a) \(\dfrac{x}{5}=\dfrac{y}{6};\dfrac{y}{8}=\dfrac{z}{7}\)và \(x+y-z=69\)
Theo đề bài, ta có:
\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{5}\times\dfrac{1}{8}=\dfrac{y}{6}\times\dfrac{1}{8}\Rightarrow\dfrac{x}{40}=\dfrac{y}{48}\)(1)
\(\dfrac{y}{8}=\dfrac{z}{7}\Rightarrow\dfrac{y}{8}\times\dfrac{1}{6}=\dfrac{z}{7}\times\dfrac{1}{6}\Rightarrow\dfrac{y}{48}=\dfrac{z}{42}\)(2)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\Rightarrow\dfrac{x}{40}=\dfrac{y}{48}=\dfrac{z}{42}=\dfrac{x+y-z}{40+48-42}=\dfrac{69}{46}=\dfrac{3}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{40}=\dfrac{3}{2}\Rightarrow x=\dfrac{40\times3}{2}=60\\\dfrac{y}{48}=\dfrac{3}{2}\Rightarrow y=\dfrac{48\times3}{2}=72\\\dfrac{z}{42}=\dfrac{3}{2}\Rightarrow z=\dfrac{42\times3}{2}=63\end{matrix}\right.\)
Vậy \(\Rightarrow\left\{{}\begin{matrix}x=60\\y=72\\z=63\end{matrix}\right.\)
Ta có:\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}\)(Nhân 2 vế với \(\dfrac{1}{4}\))
\(\dfrac{y}{8}=\dfrac{x}{7}\Rightarrow\dfrac{y}{24}=\dfrac{z}{21}\)(Nhân 2 vế với \(\dfrac{1}{3}\))
\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)và x+y-z=6
Áp dụng tính chất dãy tỉ số bằng nhau. Ta có:
\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=\dfrac{x+y-z}{20+24-21}=\dfrac{69}{23}=3\)
Vì \(\dfrac{x}{20}=3\Rightarrow x=20.3=60\)
\(\dfrac{y}{24}=3\Rightarrow y=24.3=72\)
\(\dfrac{z}{21}=3\Rightarrow z=3.21=63\)
Vậy x=60; y=72; z=63
Bài 1:
\(3^{-1}.3^n+4.3^n=13.3^5\)
\(\Rightarrow3^{n-1}+4.3.3^{n-1}=13.3^5\)
\(\Rightarrow3^{n-1}\left(1+4.3\right)=13.3^5\)
\(\Rightarrow3^{n-1}.13=13.3^5\)
\(\Rightarrow3^{n-1}=3^5\)
\(\Rightarrow n-1=5\)
\(\Rightarrow n=6\)
Vậy n = 6
Bài 2a: Câu hỏi của Nguyễn Trọng Phúc - Toán lớp 7 | Học trực tuyến
\(a,A=\dfrac{\dfrac{3}{4}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\\ A=\dfrac{\dfrac{405}{572}}{\dfrac{645}{1001}}+\dfrac{\dfrac{5}{12}}{\dfrac{25}{24}}\\ A=\dfrac{189}{172}+\dfrac{2}{5}\\ A=\dfrac{1289}{860}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{y-z}{6-7}=\dfrac{39}{-1}=-39\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-39\right).5=-195\\y=\left(-39\right).6=-234\\z=\left(-39\right).7=-273\end{matrix}\right.\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{y-z}{6-7}=\dfrac{39}{-1}=-39\)
Do đó: x=-195; y=-234; z=-273