cho 3 số x,y,z thỏa mãn \(\dfrac{19}{x+y}\)+\(\dfrac{19}{y+z}\)+\(\dfrac{19}{z+x}\)=\(\dfrac{7x}{y+z}\)+\(\dfrac{7y}{z+x}\)+\(\dfrac{7z}{x+y}\)=\(\dfrac{133}{10}\)
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AH
Akai Haruma
Giáo viên
13 tháng 6 2018
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 \((*);(**)\Rightarrow M=x+y+z=7\)
AH
Akai Haruma
Giáo viên
19 tháng 4 2018
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 $M=7$
NT
25 tháng 10 2017
\(\dfrac{19}{x+y}+\dfrac{19}{y+z}+\dfrac{19}{x+z}=\dfrac{133}{10}\\ \Rightarrow19.\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}\right)=\dfrac{133}{10}\\ \Rightarrow\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}=\dfrac{7}{10}\)
\(\dfrac{7x}{y+z}+\dfrac{7z}{x+y}+\dfrac{7y}{x+z}=\dfrac{133}{10}\\ \Rightarrow\dfrac{x}{y+z}+\dfrac{z}{x+y}+\dfrac{y}{x+z}=\dfrac{133}{10}:7=\dfrac{19}{10}\\ \Rightarrow\left(\dfrac{x}{y+z}+1\right)+\left(\dfrac{z}{x+y}+1\right)+\left(\dfrac{y}{x+z}+1\right)=\dfrac{49}{10}\\ \Rightarrow\left(x+y+z\right)\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right)=\dfrac{49}{10}\\ \Rightarrow\left(x+y+z\right).\dfrac{7}{10}=\dfrac{49}{10}\\ \Rightarrow x+y+z=7\)
17 tháng 8 2018
mình nghĩ bạn chép sai đề bài
dấu ''='' thứ 2 thay bằng dấu ''+''
ta có
\(\dfrac{19}{x+y}+\dfrac{19}{y+z}+\dfrac{19}{x+z}=\dfrac{133}{10}\)
\(\Rightarrow19\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}\right)=\dfrac{133}{10}\)
\(\Rightarrow\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}=\dfrac{7}{10}\)
lại có
\(\dfrac{7x}{y+z}+\dfrac{7y}{x+z}+\dfrac{7z}{x+y}=\dfrac{133}{10}\)
\(\Rightarrow7\left(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\right)=\dfrac{133}{10}\)
\(\Rightarrow\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=\dfrac{19}{10}\)
\(\Rightarrow\dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{x+z}+\dfrac{x+y+z}{x+y}=\dfrac{49}{10}\)
\(\Rightarrow\left(x+y+z\right)\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)=\dfrac{49}{10}\)
\(\Rightarrow\dfrac{7}{10}\left(x+y+z\right)=\dfrac{49}{10}\Rightarrow\left(x+y+z\right)^2=49.\)
13 tháng 6 2018
b) Ta có:
\(\dfrac{19}{x+y}=\dfrac{19}{y+z}=\dfrac{19}{z+x}=\dfrac{133}{10}\)
\(\Rightarrow\dfrac{133}{7\left(x+y\right)}=\dfrac{133}{7\left(y+z\right)}=\dfrac{133}{7\left(z+x\right)}=\dfrac{133}{10}\)
\(\Rightarrow7\left(x+y\right)=7\left(y+z\right)=7\left(z+x\right)=10\)
\(\Rightarrow7\left(x+y\right)+7\left(y+z\right)+7\left(z+x\right)=10\)
\(\Rightarrow7\left[2\left(x+y+z\right)\right]=10\)
\(\Rightarrow14\left(x+y+z\right)=10\)
\(\Leftrightarrow x+y+z=\dfrac{5}{7}\)
tìm x,y,z nha m.n