\(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}=\frac{y-z-25-9}{16-25} \)
\(<=>\frac{x+16}{9}=\frac{2x^3-34}{-9} \)
<=>\(-x-16=2x^3-34\)
<=>\(2x^3+x-18=0\)
=> x=2
=>\(\frac{2+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}=2\)
=>y=57
=>z=41
Tĩm,y,z biết:
a)\(\dfrac{12x-15y}{7}\)=\(\dfrac{20z-12x}{9}\)=\(\dfrac{15y-20z}{11}\)và x+y+z=48
b)\(\dfrac{x+16}{9}\)=\(\dfrac{y-25}{16}\)=\(\dfrac{z+9}{25}\) và \(^{2x^3}\)-1=15
c)\(\dfrac{x}{y+2+1}\)=\(\dfrac{y}{z+x+1}\)=\(\dfrac{z}{x+y-2}\)=x+y+z
d)\(\dfrac{3}{5}\)x=\(\dfrac{2}{3}\)y và \(^{x^2}\)-\(^{y^2}\)= 38
e)\(\dfrac{y+2+1}{x}\)=\(\dfrac{z+x+2}{y}\)=\(\dfrac{x+y+3}{z}\)=\(\dfrac{1}{x+y+z}\)
f)\(\dfrac{1+2y}{18}\)=\(\dfrac{1+4y}{24}\)=\(\dfrac{1+6y}{6x}\)
Tìm x,y, z biết :
\(\dfrac{x}{y}=\dfrac{9}{7};\dfrac{y}{z}=\dfrac{7}{3}\) và x - y + z = -15
tìm x,y,z biết :
\(\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{z}{3}\) và -x+y+-z = 11
4) \(\dfrac{x}{0,5}=\dfrac{y}{0,3}=\dfrac{z}{0,2}\) và -y+x=1
6) \(\dfrac{x+11}{13}=\dfrac{y+12}{14}=\dfrac{z+13}{15}\)và x+y+z=6
7) 5x=4y và x.y=20
Bài 1: Tìm x,y,z:
a) \(\dfrac{x}{y}\)=\(\dfrac{10}{9}\); \(\dfrac{y}{z}\)=\(\dfrac{3}{4}\); x-y+z =78
b)\(\dfrac{x}{y}=\dfrac{9}{7}\);\(\dfrac{y}{z}\)=\(\dfrac{7}{3}\); x-y+z =-15
c)\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=\(\dfrac{z}{3}\); x2 +y2+z2=200
Câu 1 : Biết\(\dfrac{x}{t}=\dfrac{5}{6};\dfrac{y}{z}=\dfrac{1}{5};\dfrac{z}{x}=\dfrac{7}{3}\) ( x; y; z; t khác 0 ). Hãy tìm tỉ số \(\dfrac{t}{y}\)
A. \(\dfrac{t}{y}=\dfrac{14}{25}\) B. \(\dfrac{t}{y}=\dfrac{7}{8}\) C. \(\dfrac{t}{y}=\dfrac{18}{7}\) D. \(\dfrac{t}{y}=\dfrac{6}{7}\)
Tìm x,y,z, biết :
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}\) và x - y + z = 16
Cho x+16/9 = y-25/16 = z+9/25 và 9-x/7 + 11-x/9=2. Khi đó x+y+z= .....
Cho x+16/9=y-25/16=z+9/25 và 2x^3-1=15.Tính x+y+z
