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+ \(2x=3y\Rightarrow x=\frac{3y}{2}\left(1\right)\)
+ \(5y=7z\Rightarrow z=\frac{5y}{7}\left(2\right)\)
Thay (1) và (2) vào 3x - 7y + 5z = - 30
Ta có \(3.\frac{3y}{2}-7y+5.\frac{5y}{7}=-30\Rightarrow y=-28\)
Thay y = - 28 vào (1) => x = - 42
Thay y = - 28 vào (2) => x = -20
\(\frac{x}{3}=\frac{y}{2};\frac{y}{7}=\frac{z}{5}\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{x+y+z}{21+14+10}=\frac{3x-7y+5z}{3.21-7.14+5.10}=-\frac{30}{15}=-2\)
\(\Rightarrow\frac{x+y+z}{45}=-2\Rightarrow x+y+z=-90\)
2x=3y;5y=7z
=>x/3=y/2;y/7=z/5
=>x/21=x/14;y/14=z/10
=>x/21=y/14=z/10
=>3x/63=7y/98=5z/50
áp dụng tính chất của dãy tỉ số bằng nhau ta có:
3x/63=7y/98=5z/50=3x-7y+5z/63-98+50=30/15=2
suy ra : 3x/63=2 =>3x=126 =>x=126:3=42
7y/98=2 =>7y =196 =>y=196:7=28
5z/50=2 =>5z = 100 => z=100:5=20
\(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{21}=\frac{y}{14}\left(1\right)\)
\(5y=7z\Rightarrow\frac{y}{7}=\frac{z}{5}\Rightarrow\frac{y}{14}=\frac{z}{10}\left(2\right)\)
Từ 1 và 2
=> \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)
Áp dụng tính chất dãy tỉ số = nhau ta có
\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\Rightarrow\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2\)
\(\frac{x}{21}=2\Rightarrow x=42\)
\(\frac{y}{14}=2\Rightarrow y=28\)
\(\frac{z}{10}=2\Rightarrow z=20\)
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
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:
\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)
2x=3y=>x/y=3/2=>x/3=y/2=>x/21=y/14
5y=7z=>y/z=7/5=>y/7=z/5=>y/14=z/10
=>x/21=y/14=z/10
áp dụng....ta có:
\(\frac{x}{21}=\frac{y}{14}=\frac{x}{10}=>\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{4x-7y+5z}{63-98+50}=-\frac{30}{15}=-2\)
3x/63=-2=>3x=-126=>x=-42
7y/98=-2=>7y=-196=>y=-28
5z/50=-2=>5z=-100=>z=-20
vậy....
Thank You Very Much!!!!:v