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Ta có:
y=3x(1)
Lấy (1) thay vào 5y=4z<=>5.3x=4z<=>z=15x/4(2)
Lấy (2) thay vào x-z=22<=>x-15x/4=22<=>-11x=88<=>x=-8
=>y=-24 và z=-30
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Có\(\dfrac{3x-5y}{4}=\dfrac{4z+=-3x}{5}=\dfrac{5y-4z}{6}=\dfrac{3x-5y+4z-3x+5y-4z}{4+5+6}=\dfrac{0}{15}=0\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-5y}{4}=0\Rightarrow3x-5y=0\Rightarrow3x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\dfrac{x}{20}=\dfrac{y}{12}\\\dfrac{5y-4z}{6}=0\Rightarrow5y-4z=0\Rightarrow5y=4z\Rightarrow\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y+z}{20+12+15}=\dfrac{16}{47}\)
\(\Rightarrow\dfrac{x}{20}=\dfrac{16}{47}\Rightarrow x=\dfrac{320}{47}\)
\(\Rightarrow\dfrac{y}{12}=\dfrac{16}{47}\Rightarrow y=\dfrac{192}{47}\)
\(\Rightarrow\dfrac{z}{15}=\dfrac{16}{47}\Rightarrow z=\dfrac{240}{47}\)
Vậy \(\left(x;y;z\right)=\left(\dfrac{320}{47};\dfrac{192}{47};\dfrac{240}{47}\right)\)
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a, \(3x=5y=7z=>\dfrac{3x}{105}=\dfrac{5y}{105}=\dfrac{7z}{105}=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}\)
áp dụng tính chất dãy tỉ số = nhau
\(=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}=\dfrac{x+y+z}{35+21+15}=\dfrac{10}{71}\)
\(=>\dfrac{x}{35}=\dfrac{10}{71}=>x=\dfrac{350}{71}\)
\(=>\dfrac{y}{21}=\dfrac{10}{71}=>y=\dfrac{210}{71}\)
\(=>\dfrac{z}{15}=\dfrac{10}{71}=>z=\dfrac{150}{71}\)
b, \(\)\(6x=5y=>\dfrac{x}{5}=\dfrac{y}{6}=>\dfrac{x}{20}=\dfrac{y}{24}\)
có \(7y=8z=>\dfrac{y}{8}=\dfrac{z}{7}=>\dfrac{y}{24}=\dfrac{z}{21}\)
\(=>\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}\)
áp dụng t/c dãy tỉ số = nhau
\(=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}=\dfrac{3x+2y+4z}{60+48+84}=\dfrac{12}{192}=\dfrac{1}{16}\)
\(=>\dfrac{3x}{60}=\dfrac{1}{16}=>x=1,25\)
\(=>\dfrac{2y}{48}=\dfrac{1}{16}=>y=1,5\)
\(=>\dfrac{4z}{84}=\dfrac{1}{16}=>z=1,3125\)
c, \(x:y:z=1:2:3=>\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\)
\(=>x=\dfrac{y}{2},z=\dfrac{3y}{2}\)
thay x,z vào \(x^3+y^3+z^3=36=>\left(\dfrac{y}{2}\right)^3+y^3+\left(\dfrac{3y}{2}\right)^3=36\)
\(=>y=2\)
\(=>x=\dfrac{y}{2}=\dfrac{2}{2}=1,z=\dfrac{3y}{2}=\dfrac{3.2}{2}=3\)
d, \(\dfrac{x}{2}=\dfrac{y}{3}=>x=\dfrac{2y}{3}\)
thay x vào \(3x^3+y^3=51=>3.\left(\dfrac{2y}{3}\right)^3+y^3=51=>y=3\)
\(=>x=\dfrac{2.3}{3}=2\)
c, từ đoạn này á
\(\left(\dfrac{y}{2}\right)^3+y^3+\left(\dfrac{3y}{2}\right)^3=36\)
\(< =>\dfrac{y^3}{8}+\dfrac{8y^3}{8}+\dfrac{27y^3}{8}=36\)
\(=>\dfrac{36y^3}{8}=36=>36y^3=8.36=>y^3=8=>y=2\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
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Ta có : \(3x=y\Rightarrow\frac{x}{1}=\frac{y}{3}\Rightarrow\frac{x}{4}=\frac{y}{12}\left(1\right)\)
\(5y=4z\Rightarrow\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\left(2\right)\)
Từ (1);(2) \(\Rightarrow\frac{x}{4}=\frac{y}{12}=\frac{z}{15}\)
ADTC dãy tỉ số bằng nhau ta có :
\(\frac{x}{4}=\frac{y}{12}=\frac{z}{15}=\frac{x+z}{4+15}=\frac{22}{19}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{4}=\frac{22}{19}\\\frac{y}{12}=\frac{22}{19}\\\frac{z}{15}=\frac{22}{19}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{22}{19}.4=\frac{88}{19}\\y=\frac{22}{19}.12=\frac{264}{19}\\z=\frac{22}{19}.15=\frac{330}{19}\end{cases}}}\)sai thì pk
cho mk xl bài này ko ''sai ''