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a)Ta có: \(2x=3y;5y=7z\)và \(x-y-z=-27\)
\(\Rightarrow\frac{x}{3}=\frac{y}{2};\frac{y}{7}=\frac{z}{5}\)và\(x-y-z=-27\)
\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)và \(x-y-z=-27\)
Áp dụng tính chất của dãy tỉ số bằng nhau,ta có:
\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{x-y-z}{21-14-10}=\frac{-27}{-3}=9\)
Ta có:\(\frac{x}{21}=9\Rightarrow x=9.21=189\)
\(\frac{y}{14}=9\Rightarrow y=9.14=126\)
\(\frac{z}{10}=9\Rightarrow z=9.10=90\)
Vậy:\(x=189;y=126\)và\(z=90\)
b) \(\frac{x}{4}=\frac{y}{5}=\frac{z}{6}\)và\(x^2-2y^2+z^2=18\)
\(\Rightarrow\frac{x^2}{16}=\frac{2y^2}{50}=\frac{z^2}{36}\)và\(x^2-2y^2+z^2=18\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x^2}{16}=\frac{2y^2}{50}=\frac{z^2}{36}=\frac{x^2-2y^2+z^2}{16-50+36}=\frac{18}{2}=9\)
Ta có:\(\frac{x^2}{16}=9\Rightarrow x^2=144\Rightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)
\(\frac{2y^2}{50}=9\Rightarrow2y^2=450\Rightarrow y^2=225\Rightarrow\orbr{\begin{cases}y=15\\y=-15\end{cases}}\)
\(\frac{z^2}{36}=9\Rightarrow z^2=324\Rightarrow\orbr{\begin{cases}z=18\\z=-18\end{cases}}\)
Vậy: \(x=12;y=15;z=18\)hoặc \(x=-12;y=-15;z=-18\)
\(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\)
\(7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\)
\(\hept{\begin{cases}\frac{x}{2}=\frac{x}{3}\\\frac{y}{5}=\frac{x}{7}\end{cases}\Rightarrow}\frac{x}{2}=\frac{5y}{15};\frac{3y}{15}=\frac{z}{7}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)
Áp dụng tính chát dãy tỉ số = nhau ta có:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)
\(\Rightarrow\frac{x}{10}=2\Rightarrow x=20\)
\(\frac{y}{15}=2\Rightarrow y=30\)
\(\frac{z}{21}=3\Rightarrow z=63\)
b, Tự làm
c, \(5x=2y\Leftrightarrow\frac{x}{2}=\frac{y}{5}\)
\(2x=3z\Leftrightarrow\frac{x}{3}=\frac{z}{2}\)
\(\Leftrightarrow\frac{x}{2}=\frac{y}{5};\frac{x}{3}=\frac{z}{2}\)
\(\Leftrightarrow\frac{x}{6}=\frac{y}{15}=\frac{x}{6}=\frac{z}{10}\)
\(\Leftrightarrow\frac{x}{6}=\frac{y}{15}=\frac{z}{10}\)
Đặt \(\frac{x}{6}=\frac{y}{15}=\frac{z}{10}=k(k\inℤ)\)
\(\Leftrightarrow\hept{\begin{cases}x=6k\\y=15k\\z=10k\end{cases}}\)
\(\Leftrightarrow x\cdot y=6k\cdot15k=90\)
\(\Leftrightarrow90:k^2=90\Leftrightarrow k^2=1\Leftrightarrow k=\pm1\)
\(\Leftrightarrow\hept{\begin{cases}x=6k\\y=15k\\z=10k\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=15\\z=10\end{cases}}\)hay \(\hept{\begin{cases}x=-6\\y=-15\\z=-10\end{cases}}\)
Vậy \((x,y)\in(6,15);(-6,-15)\)
1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)
2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)
3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)
Áp dụng t/c dtsbn:
\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)
suy ra: x/5 = 45 => x = 225
y/7 = 45 => y = 315
z/9 = 45 => z = 405
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\)
suy ra: \(x=2k;\)\(y=3k;\)\(z=4k\)
Ta có: \(x^2+y^2+z^2=116\)
<=> \(\left(2k\right)^2+\left(3k\right)^2+\left(4k\right)^2=116\)
<=> \(29k^2=116\)
<=> \(k^2=4\)
<=> \(k=\pm2\)
tự làm nốt
minh lam cau b) roi dc co 2/3 thoy ban tham khao nhe phan () la minh giai thich nha dung viet vo bai !!
2x=3y ; 5y = 7z
+) 10x=15y=21z ( Quy dong)
+)10x/210 = 15y/210 = 21z/210 ( BC)
+) x/21 = y/14 = z/10 ( Rut gon)
+) 3x/63 = 7y/98 = 5z/50 = 3x-7y+ 5z / 63 - 98 - 50 = -30/14 = -2
+ x/21 = 2 => ............ phan nay minh chua xong neu xong thi minh pm not cho
Ta có: \(-3x=5y\Rightarrow\dfrac{-3x}{15}=\dfrac{5y}{15}\Rightarrow\dfrac{x}{-5}=\dfrac{y}{3}\Rightarrow\dfrac{x}{35}=\dfrac{y}{-21}\)(1)
\(-2y=7z\Rightarrow\dfrac{-2y}{14}=\dfrac{7z}{14}\Rightarrow\dfrac{y}{-7}=\dfrac{z}{2}\Rightarrow\dfrac{y}{-21}=\dfrac{z}{6}\) (2)
Từ (1) và (2) ta có: \(\dfrac{x}{35}=\dfrac{y}{-21}=\dfrac{z}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{35}=\dfrac{y}{-21}=\dfrac{z}{6}=\dfrac{2x-3y+z}{2\cdot35-3\cdot-21+6}=\dfrac{42}{139}\)
\(\Rightarrow\dfrac{x}{35}=\dfrac{42}{139}\Rightarrow x=\dfrac{1470}{139}\)
\(\Rightarrow\dfrac{y}{-21}=\dfrac{42}{139}\Rightarrow y=-\dfrac{882}{139}\)
\(\Rightarrow\dfrac{z}{6}=\dfrac{42}{139}\Rightarrow z=\dfrac{252}{139}\)