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bài 3:
a, đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)
A/D tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2-z^2}{25+49-9}=\dfrac{585}{65}=9\)
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
\(\dfrac{3}{2}x=\dfrac{4}{3}y=\dfrac{5}{4}z\Rightarrow\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{4}}=\dfrac{z}{\dfrac{4}{5}}\\ \Rightarrow\dfrac{x}{\dfrac{2}{3}}=\dfrac{2y}{\dfrac{3}{2}}=\dfrac{z}{\dfrac{4}{5}}=\dfrac{x-2y+z}{\dfrac{2}{3}-\dfrac{3}{2}+\dfrac{4}{5}}=-\dfrac{16}{-\dfrac{1}{30}}=480\)
suy ra: \(x=\dfrac{480}{\dfrac{2}{3}}=720\\ 2y=\dfrac{480}{\dfrac{3}{2}}=320\Rightarrow y=160\\ z=\dfrac{480}{\dfrac{4}{5}}=600\)
Ta có:
\(\dfrac{3}{2}x=\dfrac{4}{3}y=\dfrac{5}{4}z\Rightarrow\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{4}}=\dfrac{z}{\dfrac{4}{5}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{4}}=\dfrac{z}{\dfrac{4}{5}}=\dfrac{2y}{\dfrac{3}{2}}=\dfrac{x-2y+z}{\dfrac{2}{3}-\dfrac{3}{2}+\dfrac{4}{5}}=\dfrac{-16}{\dfrac{-1}{30}}=480\)
\(\Rightarrow\left\{{}\begin{matrix}x=320\\y=360\\z=384\end{matrix}\right.\)
Vậy...
\(\frac{27}{4}=\frac{-x}{3}=>x=-\frac{81}{4}\notinℤ\)
\(^{y^2=\frac{4}{9}=\left(\frac{2}{3}\right)^2=>y=\pm\frac{2}{3}\notinℤ}\)
\(\frac{27}{4}=\frac{\left(z+3\right)}{-4}=\left(z+3\right)=-27=\left(-3\right)^3=>z+3=-3=>z=-6\)
\(+)|t|-2=-54=>|t|=-52\)(vô lí)
\(+)|t|-2=54=>|t|=56=>t=\pm56\)
x=\(\dfrac{-4.\left(-10\right)}{8}=5\).
y=\(\dfrac{-10.\left(-7\right)}{5}=14.\)
z=\(\dfrac{-7.\left(-24\right)}{14}=12.\)
2)\(x+y+z=9^2=81\)
Ta có:\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}\left(1\right)\)
\(\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{20}=\dfrac{z}{28}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{x+y+z}{15+20+28}=\dfrac{81}{63}=\dfrac{9}{7}\)
\(\Rightarrow x=\dfrac{135}{7};y=\dfrac{180}{7};z=36\)
Ta có: \(x+y+z=18\)
\(\dfrac{x+1}{3}=\dfrac{y+2}{5}=\dfrac{z+3}{5}\)
\(\Rightarrow\dfrac{x+1}{3}=\dfrac{y+2}{5}=\dfrac{z+3}{5}and=\dfrac{\left(y+z\right)+\left(2+3\right)}{5}+\dfrac{\left(x+1\right)}{3}\)
\(\Leftrightarrow\dfrac{5+\left(y+z\right)}{5}+\dfrac{1+x}{3}\)
\(and\dfrac{5}{5}=1\)
\(\Rightarrow x=1-\dfrac{1}{3}=\dfrac{2}{3}\) vậy \(x=2\)
Ps: tự làm tiếp nha mình mới làm tới đó
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y+z}{2+3+5}=\dfrac{20}{10}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=2\\\dfrac{y}{3}=2\\\dfrac{z}{5}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=6\\z=10\end{matrix}\right.\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(4;6;10\right)\)
Áp dụng t/c của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y+z}{2+3+5}=\dfrac{20}{10}=2\)
Ta có: \(\dfrac{x}{2}=2\) => x=4
\(\dfrac{y}{3}=2\) => y=6
\(\dfrac{z}{5}=2\) => z=10
Vậy x=4; y=6 ; z=10
a) Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}=\dfrac{x-3y+4z}{4-3.3+4.9}=\dfrac{63}{31}=2\)
\(\Rightarrow x=8\)
\(\Rightarrow y=6\)
\(\Rightarrow z=18\)
b. c. Xem lại đề.
Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\) => \(\left(\dfrac{x}{3}\right)^2=\left(\dfrac{y}{4}\right)^2=\left(\dfrac{z}{5}\right)^2\)
=> \(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{25}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{25}=\dfrac{2x^2+y^2-z^2}{2.9+16-25}=\dfrac{9}{18+16-25}=\dfrac{9}{9}=1\)
=> \(\left\{{}\begin{matrix}\dfrac{x^2}{9}=1\Rightarrow\dfrac{x}{3}=1\Rightarrow x=3\\\dfrac{y^2}{16}=1\Rightarrow\dfrac{y}{4}=1\Rightarrow y=4\\\dfrac{z^2}{25}=1\Rightarrow\dfrac{z}{5}=1\Rightarrow z=5\end{matrix}\right.\)
Vậy x = 3, y = 4, z = 5
Từ \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\) và \(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\)
\(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=2\Rightarrow x=2\cdot8=16\\\dfrac{y}{12}=2\Rightarrow y=2\cdot12=24\\\dfrac{z}{15}=2\Rightarrow z=2\cdot15=30\end{matrix}\right.\)
Theo bài ta có :
\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\)
\(x+y-z=10\)
\(\Leftrightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8-12+15}=\dfrac{10}{5}=2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=2\Leftrightarrow x=16\\\dfrac{y}{12}=2\Leftrightarrow y=24\\\dfrac{z}{15}=2\Leftrightarrow z=30\end{matrix}\right.\)
Vậy ....