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\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
=> \(\frac{2\left(x-1\right)}{4}=\frac{3\left(y-2\right)}{9}=\frac{z-3}{4}\)
=> \(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{\left(2x+3y-z\right)-2-6+3}{9}=\frac{50-5}{9}=\frac{45}{9}\)= 5
=> x-1/2 = 5 => x-1=5 => x=6
y-2/3 = 5 => y-2 = 15 => y =17
z-3/4=5 => z-3=20 => z=23
Lời giải:
Ta có: \(\frac{19}{x+y}+\frac{19}{y+z}+\frac{19}{z+x}=\frac{133}{10}\)
\(\Rightarrow \frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{7}{10}(*)\)
Lại có:
\(\frac{7x}{y+z}+\frac{7y}{z+x}+\frac{7z}{x+y}=\frac{133}{10}\)
\(\Rightarrow \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=\frac{19}{10}\)
\(\Rightarrow \frac{x}{y+z}+1+\frac{y}{z+x}+1+\frac{z}{x+y}+1=\frac{19}{10}+3=\frac{49}{10}\)
\(\Leftrightarrow \frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{x+y}=\frac{49}{10}\)
\(\Leftrightarrow (x+y+z)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{49}{10}(**)\)
Từ \((*);(**)\Rightarrow M=x+y+z=7\)
Lời giải:
Ta có:
\(\frac{19}{x+y}+\frac{19}{y+z}+\frac{19}{z+x}=\frac{133}{10}\)
\(\Leftrightarrow \frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{7}{10}(*)\)
Và: \(\frac{7x}{y+z}+\frac{7y}{z+x}+\frac{7z}{x+y}=\frac{133}{10}\)
\(\Leftrightarrow \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=\frac{19}{10}\)
\(\Rightarrow \frac{x}{y+z}+1+\frac{y}{z+x}+1+\frac{z}{x+y}+1=\frac{49}{10}\)
\(\Leftrightarrow \frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{x+y}=\frac{49}{10}\)
\(\Leftrightarrow (x+y+z)\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)=\frac{49}{10}(**)\)
Từ \((*); (**)\Rightarrow x+y+z=\frac{49}{10}:\frac{7}{10}=7\)
Vậy $M=7$
\(\dfrac{19}{x+y}+\dfrac{19}{y+z}+\dfrac{19}{x+z}=\dfrac{133}{10}\\ \Rightarrow19.\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}\right)=\dfrac{133}{10}\\ \Rightarrow\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}=\dfrac{7}{10}\)
\(\dfrac{7x}{y+z}+\dfrac{7z}{x+y}+\dfrac{7y}{x+z}=\dfrac{133}{10}\\ \Rightarrow\dfrac{x}{y+z}+\dfrac{z}{x+y}+\dfrac{y}{x+z}=\dfrac{133}{10}:7=\dfrac{19}{10}\\ \Rightarrow\left(\dfrac{x}{y+z}+1\right)+\left(\dfrac{z}{x+y}+1\right)+\left(\dfrac{y}{x+z}+1\right)=\dfrac{49}{10}\\ \Rightarrow\left(x+y+z\right)\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right)=\dfrac{49}{10}\\ \Rightarrow\left(x+y+z\right).\dfrac{7}{10}=\dfrac{49}{10}\\ \Rightarrow x+y+z=7\)