Tim x,y va z biet
\(\dfrac{x}{3}=\dfrac{y}{y};\dfrac{y}{6}=\dfrac{z}{8}\)va \(3x-2y-z=13\)
Ai lam dung mik tick cho
tim x, y, z biet :
a, \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\) va 2x + 3y - z = 186
b, \(\dfrac{x}{3}=\dfrac{y}{4}\) va \(\dfrac{y}{5}=\dfrac{z}{7}\) va 2x + 3y - z = 327
c, 2x = 3y = 5z va x + y - z = 95
d, \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) va xyz = 810
a)Vì \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)nên \(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{x}{28}\).
Áp dụng t/c dãy tỉ số = nhau, ta có :
\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)
⇒2x = 3.30 = 90 ⇒ x = 45
3y = 3.60 = 180 ⇒ y = 60
z = 3.28 = 84
Ý b) có gì đó sai sai ?
c)Ta có :
\(2x=3y=5z\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)
Áp dụng t/c dãy tỉ số = nhau, ta có :
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)
⇒x = 5.15 = 75
y = 5.10 = 50
z = 5.6 = 30
d)Ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\left(k\in Z\right)\)
⇒ x = 2k ; y = 3k ; z = 5k
⇒ xyz = 2k.3k.5k = 30k3 = 810
⇒ k = 3 Vậy x = 3.2 = 6; y = 3.3 = 9; z = 3.5 = 15Tim x,y va z biet
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)va \(2x+3y+5z=86\)
Áp dụng tinshh chất dãy tỉ số bằng nhau ; ta được :
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{5z}{25}=\dfrac{2x+3y+5z}{6+12+25}=\dfrac{86}{43}=2\)
Do đó :
\(\dfrac{x}{3}=2\Rightarrow x=2.3=6\)
\(\dfrac{y}{4}=2\Rightarrow y=2.4=8\)
\(\dfrac{z}{5}=2\Rightarrow z=2.5=10\)
Vậy x = 6 ; y = 8 ; z = 10
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau , ta có:
\(\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{5z}{25}=\dfrac{2x+3y+5z}{6+12+25}=\dfrac{86}{43}=2\) \
\(\Rightarrow x=2.3=6\)
\(y=2.4=8\)
\(z=2.5=10\)
Tim cac so x,y,z, biet:\(\dfrac{x}{2}\)=\(\dfrac{y}{4}\)=\(\dfrac{z}{6}\) va x-y+z=8
Áp dụng t/c của dãy tỉ số = nhau có:
\(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}=\dfrac{x-y+z}{2-4+6}=\dfrac{8}{4}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot2=4\\y=2\cdot4=8\\z=2\cdot6=12\end{matrix}\right.\)
Tim x, y, z biet
\(\dfrac{12x-15y}{2017}=\dfrac{20z-12x}{2018}=\dfrac{15y-20z}{2019}\) va x+y+z=48
Ta có:
\(\dfrac{12x-15y}{2017}=\dfrac{20z-12x}{2018}=\dfrac{15y-20z}{2019}\)
\(=\dfrac{12x-15y+20z-12x+15y-20z}{2017+2018+2019}\)
\(=\dfrac{0}{2017+2018+2019}=0\)
\(\Rightarrow\left\{{}\begin{matrix}12x-15y=0\\20z-12x=0\\15y-20z=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}12x=15y\\20z=12x\\15y=20z\end{matrix}\right.\)\(\Rightarrow12x=15y=20z\)
\(\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}\)
Áp dụng tích chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{x+y+z}{5+4+3}=\dfrac{48}{12}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.5=20\\y=4.4=16\\z=4.3=12\end{matrix}\right.\)
Vậy ...
Tim x,y,z biet:
\(\dfrac{2x}{3}\) = \(\dfrac{3y}{4}\) = \(\dfrac{4z}{5}\) va -x-y-z=-49
Ta có: \(\dfrac{2x}{3}\) = \(\dfrac{3y}{4}\) = \(\dfrac{4z}{5}\) suy ra: \(\dfrac{2x}{60}\) = \(\dfrac{3y}{60}\) = \(\dfrac{4z}{60}\)
Suy ra:\(\dfrac{x}{30}\) = \(\dfrac{y}{20}\) = \(\dfrac{z}{15}\)
Theo bài ra, ta có:\(\dfrac{x}{30}\) = \(\dfrac{y}{20}\) = \(\dfrac{z}{15}\)
mà x-y-z = -49
Áp dụng ính chất của dãy ti số bằng nhau, ta có:
\(\dfrac{x}{30}\) = \(\dfrac{y}{20}\) = \(\dfrac{z}{15}\) = \(\dfrac{x-y-z}{30-20-15}\) = \(\dfrac{-49}{-5}\)= 9,8
Suy ra: \(\dfrac{x}{30}\) = 9,8 suy ra: x = 9,8. 30 = 294
\(\dfrac{y}{20}\) = 9,8 suy ra: y = 9,8. 20 = 196
\(\dfrac{z}{15}\) = 9,8 suy ra: z = 9,8. 15=147
vậy x = 294; y = 169 và z = 147
chúc bn hk tốt
(câu trả lời của mk sai thik mong thông cảm nhé)
tim x,y,z biet
\(\dfrac{x}{y}\) =\(\dfrac{3}{2}\) ; \(\dfrac{y}{z}\) = \(\dfrac{5}{7}\) va \(|2x-3y+5z|\) =1
2x-3y+5z=1 hoặc =-1
TH1: \(\dfrac{x}{y}\)=\(\dfrac{3}{2}\)=>\(\dfrac{x}{3}\)=\(\dfrac{y}{2}\)=>\(\dfrac{x}{15}\)=\(\dfrac{y}{10}\)
\(\dfrac{y}{z}\)=\(\dfrac{5}{7}\)=>\(\dfrac{y}{5}\)=\(\dfrac{z}{7}\)=>\(\dfrac{y}{10}\)=\(\dfrac{z}{14}\)
\(\Rightarrow\)\(\dfrac{x}{15}\)=\(\dfrac{y}{10}\)=\(\dfrac{z}{14}\)=>\(\dfrac{2x}{30}\)=\(\dfrac{3y}{30}\)=\(\dfrac{5z}{70}\)
Áp dụng tính chát dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x-3y+5z}{30-30+70}\)=\(\dfrac{1}{70}\)
=>x=1.15:7=\(\dfrac{3}{14}\)
y=\(\dfrac{1}{7}\)
z=\(\dfrac{1}{5}\)
TH2:............=-1 tự tính nhé làm tương tựmình còn phải ôn bài
a,\(\dfrac{y+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
b, 10x = 6y va 2x2 - y2 = -28
Tim x,y,z(cau a)
tim x,y ( cau b)
\(a)\dfrac{y+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{y+z+x+x+z+2+x+y-3}{x+y+z}\)
\(=\dfrac{\left(x+y+z\right)+\left(x+y+z\right)+\left(1+2-3\right)}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
Lại có: \(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
\(\Rightarrow2=\dfrac{1}{x+y+z}\Rightarrow2\left(x+y+z\right)=1\Rightarrow x+y+z=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y+z+1}{x}=2\\\dfrac{x+z+2}{y}=2\\\dfrac{x+y-3}{z}=2\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y+z+x+1=3x\\x+y+z+2=3y\\x+y+z-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}+1=3x\\\dfrac{1}{2}+2=3y\\\dfrac{1}{2}-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1+\dfrac{1}{2}}{3}\\y=\dfrac{\dfrac{1}{2}+2}{3}\\z=\dfrac{\dfrac{1}{2}-3}{3}\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=\dfrac{-5}{6}\end{matrix}\right.\)
Chúc bạn học tốt!
Tim x,y va z
\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{6}=\dfrac{z}{8}\)va \(3x-2y-z=13\)
Từ \(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}\)
Và \(\dfrac{y}{6}=\dfrac{z}{8}\Rightarrow\)\(\dfrac{y}{12}=\dfrac{z}{16}\)
Suy ra \(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\)\(\Rightarrow\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}=\dfrac{3x-2y-z}{27-24-16}=\dfrac{13}{-13}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=-1\Rightarrow x=-1\cdot9=-9\\\dfrac{y}{12}=-1\Rightarrow y=-1\cdot12=-12\\\dfrac{z}{16}=-1\Rightarrow z=-1\cdot16=-16\end{matrix}\right.\)
Ta có :
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{x}{9}=\dfrac{y}{12}\)(1)
\(\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{y}{12}=\dfrac{z}{16}\)(2)
Từ (1) và (2) , suy ra \(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ; ta được :
\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}=\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}=\dfrac{3x-2y-z}{27-24-16}=\dfrac{13}{-13}=-1\)
Do đó :
\(\dfrac{x}{9}=-1\Rightarrow x=-1.9=-9\)
\(\dfrac{y}{12}=-1\Rightarrow y=-1.12=-12\)
\(\dfrac{z}{16}=-1\Rightarrow z=-1.16=-16\)
Vậy x = -9 ; y = -12 ; z = -16
\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{6}=\dfrac{z}{8}\)
\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12};\dfrac{y}{12}=\dfrac{z}{16}\)
\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\)
\(\Rightarrow\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{3x}{54}=\dfrac{y}{24}=\dfrac{z}{32}\)
\(=\dfrac{3x-2y-z}{27-24-16}\)
\(=\dfrac{13}{-13}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x=9.-1=-9\\y=12.-1=-12\\z=16.-1=-16\end{matrix}\right.\)
tim x, y biet
a ,\(\dfrac{x}{2}=\dfrac{y}{5}\) va x . y = 3,6
b , \(\dfrac{x}{3}=\dfrac{y}{4}\) va x . y =108
giup minh nhe minh dang can gap
a)Ta có :
\(\dfrac{x}{2}=\dfrac{y}{5}=k\)
Mà x.y=3,6 => 2k+5k=3,6=>7k=3,6
Vậy k = \(\dfrac{18}{35}\)
\(x=2k\Rightarrow x=\dfrac{36}{35}\)
\(y=5k\Rightarrow y=\dfrac{18}{7}\)
\(a,\dfrac{x}{2}=\dfrac{y}{5}\)
\(\rightarrow\)\(x.5=y.2\)
\(x.x.5=y.x.2\)
\(x^2.5=3,6.2\)
\(x^2.5=7,2\)
\(x^2=1,44\)
\(\rightarrow x=1,2\) hoặc \(x=-1,2\)
Ý b bạn làm tường tự nha