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Ta có: \(\frac{4x}{-5}=\frac{6y}{7}=\frac{-3z}{8}\)(1) và x + 3y - 2z = -273
(1) => \(\frac{x}{\frac{-5}{4}}=\frac{3y}{\frac{7}{2}}=\frac{-z}{\frac{8}{3}}\)=> \(\frac{x}{\frac{-5}{4}}=\frac{3y}{\frac{7}{2}}=\frac{-2z}{\frac{16}{3}}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x}{\frac{-5}{4}}=\frac{3y}{\frac{7}{2}}=\frac{-2z}{\frac{16}{3}}=\frac{x+3y-2z}{\frac{-5}{4}+\frac{7}{2}-\frac{16}{3}}=\frac{-273}{\frac{-37}{12}}=\frac{3276}{37}\)
=> \(\frac{x}{\frac{-5}{4}}=\frac{3276}{37}\)=> \(37x=3276\left(\frac{-5}{4}\right)\)=> x = \(\frac{-4095}{37}\)
và \(\frac{3y}{\frac{7}{2}}=\frac{3276}{37}\)=> \(111y=3276.\frac{7}{2}\)=> y = \(\frac{3822}{37}\)
và \(\frac{-2z}{\frac{16}{3}}=\frac{3276}{37}\)=> \(-74z=3276.\frac{16}{3}\)=> z = \(\frac{-8736}{37}\)
=> A = x + y + z + 1 = \(\frac{-4095}{37}\)+ \(\frac{3822}{37}\)+ \(\frac{-8736}{37}\)+ 1 = \(\frac{-8972}{37}\).
c1:Thay số
Q=\(\frac{5+2.4-3.3}{5-2.4+3.3}\)
O=\(\frac{4}{6}\)=\(\frac{2}{3}\)
\(\frac{4x}{6y}=\frac{2x+8}{3y+11}\)
\(4x\left(3y+1\right)=6y\left(2x+8\right)\)
\(12xy+4x=12xy+48y\)
\(4x-48y=0\)
\(4x=48y\)
Ta có:\(\frac{4x}{48y}\)
\(\Leftrightarrow\)\(\frac{x}{y}=\frac{1}{12}\)
x:y:z=5:4:3=>x/5=y/4=z/3
\(\frac{x+2y-3z}{5+4.2-3.3}=\frac{x-2y+3z}{5-4.2+3.3}\Leftrightarrow\frac{x+2y-3z}{5+8-9}=\frac{x-2y+3z}{5-8+9}\)
\(\frac{x+2y-3z}{4}=\frac{x-2y+3z}{6}\Leftrightarrow\frac{x+2y-3z}{x-2y+3z}=\frac{4}{6}=\frac{2}{3}\)
\(\Rightarrow P=\frac{x+2y-3z}{x-2y+3z}+\frac{1}{3}=\frac{2}{3}+\frac{1}{3}=\frac{3}{3}=1\)
vay P=1
nhớ tick
Ta có : \(\frac{3x-2y}{4}=\frac{4y-3z}{2}=\frac{2z-4x}{3}\)
\(\Leftrightarrow\frac{12x-8y}{16}=\frac{8y-6z}{4}=\frac{6z-12x}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{12x-8y}{16}=\frac{8y-6z}{4}=\frac{6z-12x}{9}=\frac{12x-8y+8y-6z+6z-12x}{16+4+9}=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{3x-2y}{4}=0\\\frac{4y-3z}{2}=0\\\frac{2z-4x}{3}=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}3x=2y\\4y=3z\\2z=4x\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{3}=\frac{z}{4}\\\frac{x}{2}=\frac{z}{4}\end{cases}}\) \(\Leftrightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)
\(\Leftrightarrow\frac{x}{2}=\frac{2y}{6}=\frac{3z}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{2}=\frac{2y}{6}=\frac{3z}{12}=\frac{x-2y+3z}{2-6+12}=\frac{8}{8}=1\)
\(\Leftrightarrow\hept{\begin{cases}\frac{x}{2}=1\\\frac{y}{3}=1\\\frac{z}{4}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=3\\z=4\end{cases}}\)
Vậy : \(\left(x,y,z\right)=\left(2,3,4\right)\)
a, \(3x=5y=7z\Rightarrow\frac{x}{\frac{1}{3}}=\frac{y}{\frac{1}{5}}=\frac{z}{\frac{1}{7}}\)
\(\Rightarrow\frac{2x}{\frac{2}{3}}=\frac{y}{\frac{1}{5}}=\frac{3z}{\frac{3}{7}}\)
Áp dụng t/c
\(\Rightarrow\frac{2x}{\frac{2}{3}}=\frac{y}{\frac{1}{5}}=\frac{3z}{\frac{3}{7}}=\frac{2x-y+3z}{\frac{2}{3}-\frac{1}{5}+\frac{3}{7}}=\frac{188}{\frac{105}{94}}=210\)
\(\frac{x}{\frac{1}{3}}=210\Rightarrow x=70\)
\(\frac{y}{\frac{1}{5}}=210\Rightarrow y=42\)
\(\frac{z}{\frac{1}{7}}=210\Rightarrow z=30\)
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