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Ta có :\(\frac{3}{x}+\frac{4}{y}+\frac{5}{z}=6\)
\(\Leftrightarrow\frac{6}{2x}+\frac{12}{3y}+\frac{20}{4z}=6\)
\(\Leftrightarrow\frac{6}{2x}+\frac{12}{2x}+\frac{20}{2x}=6\)
\(\Leftrightarrow\frac{6+12+20}{2x}=6\)
\(\Leftrightarrow\frac{19}{x}=6\)
\(\Leftrightarrow x=\frac{19}{6}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{2}{3}.\frac{19}{6}=\frac{19}{9}=y\)
\(\Leftrightarrow\frac{3}{4}y=\frac{3}{4}.\frac{19}{9}=\frac{19}{12}=z\)
Vậy \(\hept{\begin{cases}x=\frac{19}{6}\\y=\frac{19}{9}\\z=\frac{19}{12}\end{cases}}\)
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\(\frac{z}{4}=\frac{y}{3}\Rightarrow\frac{z}{20}=\frac{y}{15}^{\left(1\right)}\)
\(\frac{x}{2}=\frac{z}{5}\Rightarrow\frac{x}{8}=\frac{z}{20}^{\left(2\right)}\)
\(\left(1\right),\left(2\right)\Rightarrow\frac{x}{8}=\frac{z}{20}=\frac{y}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x}{8}=\frac{z}{20}=\frac{y}{15}=\frac{x+y+z}{8+20+15}=\frac{51}{43}\)
..... ( tới bước này bạn tự làm tiếp nhá )
\(\frac{z}{4}=\frac{y}{3}\Rightarrow\frac{z}{20}=\frac{y}{15}\)
\(\frac{x}{2}=\frac{z}{5}\Rightarrow\frac{x}{8}=\frac{z}{20}\)
Suy ra: \(\frac{x}{8}=\frac{y}{15}=\frac{z}{20}=\frac{x+y+z}{8+15+20}=\frac{51}{43}\)
Vậy \(x=8.\frac{51}{43}=\frac{408}{43}\)
\(y=15.\frac{51}{43}=\frac{765}{43}\)
\(z=20.\frac{51}{43}=\frac{1020}{43}\)
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Đặt \(\frac{x-2}{6}=\frac{y+3}{9}=\frac{z-7}{10}=k\Rightarrow\hept{\begin{cases}x=6k+2\\y=9k-3\\z=10k+7\end{cases}}\)
Theo đề bài: x+y+z=106
<=>\(6k+2+9k-3+10k+7=106\)
<=>\(25k+6=106\)
<=> 25k = 100
<=> k = 4
=> \(\hept{\begin{cases}x=6.4+2=26\\y=9.4-3=33\\z=10.4+7=47\end{cases}}\)
Vậy .........................
Tim x , y, z , biet
\(\frac{x}{3}\)=\(\frac{y}{4}\); \(\frac{y}{3}\)=\(\frac{Z}{5}\) Va 2x - 3y + z=6
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Ta có:\(\frac{x}{3}=\frac{y}{4}\)\(\Rightarrow\frac{x}{9}=\frac{y}{12}\)
\(\frac{y}{3}=\frac{z}{5}\)\(\Rightarrow\frac{y}{12}=\frac{z}{20}\)
Suy ra:\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\)
Đặt\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=k\)
\(\Rightarrow\hept{\begin{cases}x=9k\\y=12k\\z=20k\end{cases}}\)
Mà\(2x-3y+z=6\)
\(\Rightarrow2.9k-3.12k+20k=6\)
\(\Leftrightarrow18k-36k+20k=6\)
\(\Leftrightarrow2k=6\)
\(\Leftrightarrow k=3\)
\(\Rightarrow\hept{\begin{cases}x=3.9=27\\y=3.12=36\\z=3.20=60\end{cases}}\)(Thỏa mãn)
Vậy\(\hept{\begin{cases}x=27\\y=36\\z=60\end{cases}}\)
Linz
Ta có : \(\hept{\begin{cases}\frac{x}{3}=\frac{y}{4}\\\frac{y}{3}=\frac{z}{5}\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{9}=\frac{y}{12}\\\frac{y}{12}=\frac{z}{20}\end{cases}\Rightarrow\frac{x}{9}=\frac{y}{12}=\frac{z}{20}}\)
=> \(\frac{2x}{18}=\frac{3y}{36}=\frac{z}{20}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)(dãy tỉ số bằng nhau)
=> x = 27 ; y = 36 ; z = 60
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Ta có: \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\Rightarrow\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}\) và x2 - y2 + 2x2 = 108
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}=\frac{x^2-y^2+2z^2}{4-9+32}=\frac{108}{27}=4\)
\(\Rightarrow\frac{x^2}{4}=4\Rightarrow x=4\)
\(\Rightarrow\frac{y^2}{9}=4\Rightarrow y=6\)
\(\Rightarrow\frac{2z^2}{32}=4\Rightarrow z=8\)
theo dãy tỉ số bằng nhau ta có :
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)<=> \(\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}\)<=> \(\frac{x^2-y^2+2z^2}{4+9+32}=\frac{108}{45}=\frac{12}{5}\)
=> x=245
y=36/5
z= 48/5
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nguyen tran phuong vy: vt sai kìa, phải là I don't know
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1) Áp dụng tích chất dãy tỉ số bằng nhau ta có:
\(\frac{x+y}{2015}=\frac{xy}{2016}=\frac{x-y}{2017}=\frac{x+y-x+y}{2015-2017}=\frac{2y}{-2}\)
\(=-y\)
\(\Rightarrow xy=-2016y;x+y=-2015y;\)
\(x-y=-2017y\)
\(\Rightarrow-2016y-xy=0\)
\(\Rightarrow y\left(-2016-x\right)=0\)
\(\Rightarrow\orbr{\orbr{\begin{cases}y=0\\-2016-x=0\end{cases}\Rightarrow}}\orbr{\begin{cases}y=0\\x=-2016\end{cases}}\)
\(+) \)\(y=0\Rightarrow0+x=-2015.0=0\Rightarrow x=0\)
\(+) \)\(x=-2016\Rightarrow-2016-y=-2017y\Rightarrow-2016\)
Vậy +) x=y=0
+) x=-2016;y=1
2) Có: \(\frac{2x+2}{3}=\frac{x+1}{1,5};\frac{4z+2}{5}=\frac{z+0,5}{1,25};\frac{3y-1}{4}=\frac{y-\frac{1}{3}}{\frac{4}{3}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x+1}{1,5}=\frac{y-\frac{1}{3}}{\frac{4}{3}}=\frac{z+0,5}{1,25}=\frac{x+y+z+\left(1-\frac{1}{3}+0,5\right)}{1,5+\frac{4}{3}+1,25}=\frac{7+\frac{7}{6}}{\frac{49}{12}}=2\)
Suy ra: \(x+1=2.1,5=3\Rightarrow x=2\)
\(y-\frac{1}{3}=2.\frac{4}{3}=\frac{8}{3}\Rightarrow y=3\)
\(z+0,5=2.1,25=2,5\Rightarrow z=2\)
Vậy x=2;y=3;z=2.
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a) Xem lại đề
b) Ta có: \(2x=4y=5z\)=> \(\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{4}}=\frac{z}{\frac{1}{5}}\) => \(\frac{2x}{1}=\frac{3y}{\frac{3}{4}}=\frac{z}{\frac{1}{5}}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{2x}{1}=\frac{3y}{\frac{3}{4}}=\frac{z}{\frac{1}{5}}=\frac{2x-3y-z}{1-\frac{3}{4}-\frac{1}{5}}=\frac{1}{\frac{1}{20}}=20\)
=> \(\hept{\begin{cases}\frac{x}{\frac{1}{2}}=20\\\frac{y}{\frac{1}{4}}=20\\\frac{z}{\frac{1}{5}}=20\end{cases}}\) => \(\hept{\begin{cases}x=20.\frac{1}{2}=10\\y=20.\frac{1}{4}=5\\z=20.\frac{1}{5}=4\end{cases}}\)
Vậy x = 10; y = 5 và z = 4
a)\(\frac{x}{5}=\frac{y}{6};\frac{y}{2}=\frac{z}{3}\)va \(x^3-2x^2y+z^3\)
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\(\frac{x}{2}-2=\frac{y}{3}-2=\frac{z}{4}-2\)
\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+y+z}{2+3+4}=\frac{27}{9}=3\)
\(\Rightarrow x=6,y=9,z=12\)
Áp dụng tính chất dãy tỉ số bằng nhau ,ta có:
\(\frac{x-4}{2}=\frac{y-6}{3}=\frac{z-8}{4}=\frac{x+y+z-18}{2+3+4}=1\)
Ta có:\(\frac{x-4}{2}=1\Rightarrow x=6\)
\(\frac{y-6}{3}=1\Rightarrow y=9\)
\(\frac{z-8}{4}=1\Rightarrow z=12\)