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Ta có:\(\dfrac{x^2}{4}=\dfrac{x}{2};\dfrac{y^2}{9}=\dfrac{y}{3};\dfrac{z^2}{25}=\dfrac{z}{5}\)
Aps dụng tính chất dãy tỉ số bằn nhau:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-y+z}{2-3+5}=\dfrac{4}{4}=1\)
=>\(\dfrac{x}{2}=1=>x=2\)
\(\dfrac{y}{3}=1=>y=3\)
\(\dfrac{z}{5}=1=>z=5\)
Vậy x=2, y=3, z=5
Ta có : \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{25}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-y+z}{2-3+5}=\dfrac{4}{4}=1\)
\(\Leftrightarrow x=2;y=3;z=5\)
a) Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\)
\(\Leftrightarrow\dfrac{a}{8}=\dfrac{b}{12}\)(1)
Ta có: \(\dfrac{b}{4}=\dfrac{c}{5}\)
nên \(\dfrac{b}{12}=\dfrac{c}{15}\)(2)
Từ (1) và (2) suy ra \(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}\)
mà a+b+c=2
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}=\dfrac{a+b+c}{8+12+15}=\dfrac{2}{35}\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{a}{8}=\dfrac{2}{35}\\\dfrac{b}{12}=\dfrac{2}{35}\\\dfrac{c}{15}=\dfrac{2}{35}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{16}{35}\\b=\dfrac{24}{35}\\c=\dfrac{30}{35}=\dfrac{6}{7}\end{matrix}\right.\)
Vậy: \(a=\dfrac{16}{35}\); \(b=\dfrac{24}{35}\); \(c=\dfrac{6}{7}\)
b) Ta có: 2a=3b=5c
nên \(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}\)
mà a+b-c=3
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}=\dfrac{a+b-c}{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}=\dfrac{3}{\dfrac{19}{30}}=\dfrac{90}{19}\)
Do đó:
\(\left\{{}\begin{matrix}2a=\dfrac{90}{19}\\3b=\dfrac{90}{19}\\5c=\dfrac{90}{19}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{45}{19}\\b=\dfrac{30}{19}\\c=\dfrac{18}{19}\end{matrix}\right.\)
Vậy: \(a=\dfrac{45}{19}\); \(b=\dfrac{30}{19}\); \(c=\dfrac{18}{19}\)
3a) A=\(\dfrac{5}{x+xy+xyz}+\dfrac{5}{y+yz+1}+\dfrac{5xyz}{z+xz+xyz}\)
=\(\dfrac{5}{x\left(1+y+yz\right)}+\dfrac{5}{y+yz+1}+\dfrac{5xy}{1+x+xy}\)
=\(\dfrac{5}{x\left(1+y+zy\right)}+\dfrac{5x}{x\left(1+zy+y\right)}+\dfrac{5xy}{x\left(1+y+zy\right)}\)
=\(\dfrac{5+5x+5xy}{x\left(1+yz+y\right)}\)
=\(\dfrac{5x\left(yz+1+y\right)}{x\left(1+yz+y\right)}=5\)
\(\left\{{}\begin{matrix}3\left(x-1\right)=2\left(y-2\right)\\5\left(y-2\right)=4\left(z-3\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3\left(x-1\right)}{6}=\dfrac{2\left(y-2\right)}{6}\\\dfrac{5\left(y-2\right)}{20}=\dfrac{4\left(z-3\right)}{20}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=\dfrac{y-2}{3}\\\dfrac{y-2}{4}=\dfrac{z-3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1}{8}=\dfrac{y-2}{12}\\\dfrac{y-2}{12}=\dfrac{z-3}{15}\end{matrix}\right.\Leftrightarrow\dfrac{x-1}{8}=\dfrac{y-2}{12}=\dfrac{z-3}{15}\Leftrightarrow\dfrac{2x-2}{16}=\dfrac{3y-6}{36}=\dfrac{z-3}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x-2}{16}=\dfrac{3y-6}{36}=\dfrac{z-3}{15}=\dfrac{2x-2+3y-6-z+3}{16+36-15}=\dfrac{\left(2x+3y-z\right)+\left(3-2-6\right)}{37}=\dfrac{79-5}{37}=\dfrac{74}{37}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.8+1=17\\y=2.12+2=26\\z=2.15+3=33\end{matrix}\right.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{3a+b+c}{a}=\dfrac{a+3b+c}{b}=\dfrac{a+b+3c}{c}=\dfrac{3a+b+c+a+3b+c+a+b+3c}{a+b+c}=\dfrac{5a+5b+5c}{a+b+c}=\dfrac{5\left(a+b+c\right)}{a+b+c}=5\)\(\Rightarrow\left\{{}\begin{matrix}3a+b+c=5a\\a+3b+c=5b\\a+b+3c=5c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b+c=2a\\a+c=2b\\a+b=2c\end{matrix}\right.\)
\(M=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}=\dfrac{2c}{c}+\dfrac{2a}{a}+\dfrac{2b}{b}=2+2+2=6\)