Cho : \(Cho\dfrac{1+x}{2}=\dfrac{2-y}{3}=\dfrac{4+z}{5}vàx-2y+z=2\)
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}và2x+3y-z=-14\)
\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}vàx-y-z=28\)
1) Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x}{6}=\dfrac{3y}{15}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right).3=-3\\y=\left(-1\right).5=-5\\z=\left(-1\right).7=-7\end{matrix}\right.\)
2) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{19}.8=-\dfrac{224}{19}\\y=-\dfrac{28}{19}.12=-\dfrac{336}{19}\\z=-\dfrac{28}{19}.15=-\dfrac{420}{19}\end{matrix}\right.\)
a, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{3\cdot2+5\cdot3-7}=\dfrac{-14}{14}=-1\\ \Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)
b, \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\Leftrightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{224}{19}\\y=-\dfrac{336}{19}\\z=-\dfrac{420}{19}\end{matrix}\right.\)
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}\)
⇒\(\dfrac{2x}{10}=\dfrac{3y}{15}=\dfrac{z}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{10}=\dfrac{3y}{15}=\dfrac{z}{7}=\dfrac{2x+3y-z}{10+15-7}=\dfrac{-14}{18}=\dfrac{-7}{9}\)
⇒\(\left\{{}\begin{matrix}x=\dfrac{-7}{9}.3=\dfrac{-7}{3}\\y=\dfrac{-7}{9}.5=\dfrac{-35}{9}\\z=\dfrac{-7}{9}.7=\dfrac{-49}{9}\end{matrix}\right.\)
anh nhanh nhất 5 tik nhé. mk rất gấp.
đề: tìm x, y, z
a)\(\dfrac{y}{2}=\dfrac{z}{3}và4x-3y=2z=36\)
b)\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}vàx-2y+3z=14\)
Coi đề lại câu a
b,
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\\ \dfrac{x-1}{2}=\dfrac{2\left(y-2\right)}{2\cdot3}=\dfrac{3\cdot\left(z-3\right)}{3\cdot4}\\ \dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}=\dfrac{x-1-\left(2y-4\right)+3z-9}{2-6+12}=\dfrac{x-1-2y+4+3z-9}{8}=\dfrac{\left(x-2y+3z\right)+\left(4-1-9\right)}{8}=\dfrac{14+\left(-6\right)}{8}=\dfrac{8}{8}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=1\Rightarrow x-1=2\Rightarrow x=3\\\dfrac{2y-4}{6}=1\Rightarrow2y-4=6\Rightarrow2y=10\Rightarrow y=5\\\dfrac{3z-9}{12}=1\Rightarrow3z-9=12\Rightarrow3z=21\Rightarrow z=7\end{matrix}\right.\)
Vậy x = 3; y = 5; z = 7
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)
\(\Rightarrow\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
\(=\dfrac{x-1-2y+4+3z-9}{2-6+12}\)
\(=\dfrac{14-6}{14-6}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=1\\\dfrac{y-2}{3}=1\\\dfrac{z-3}{4}=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)
Tìm x,y,z biết:
a. \(x=\dfrac{y}{6}=\dfrac{z}{3}và2x-3x-4z=24\)
\(b.6x=10y=15z\) và \(x+y-z=90\)
\(c.\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}và5z-3x-4y=50\)
\(d.\dfrac{x}{4}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{3}vàx-y+100=z\)
a: 2x-3y-4z=24
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y-4z}{2\cdot1-3\cdot6-4\cdot3}=\dfrac{24}{-28}=\dfrac{-6}{7}\)
=>x=-6/7; y=-36/7; z=-18/7
b: 6x=10y=15z
=>x/10=y/6=z/4=k
=>x=10k; y=6k; z=4k
x+y-z=90
=>10k+6k-4k=90
=>12k=90
=>k=7,5
=>x=75; y=45; z=30
d: x/4=y/3
=>x/20=y/15
y/5=z/3
=>y/15=z/9
=>x/20=y/15=z/9
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{9}=\dfrac{x-y-z}{20-15-9}=\dfrac{-100}{-4}=25\)
=>x=500; y=375; z=225
\(\dfrac{x}{2}=\dfrac{y}{3},\dfrac{y}{4}=\dfrac{z}{5}vàx+y-z=10\)
\(\dfrac{x}{2}=\dfrac{y}{3}\) ⇒ \(\dfrac{x}{8}=\dfrac{y}{12}\) (1)
\(\dfrac{y}{4}=\dfrac{z}{5}\) ⇒ \(\dfrac{y}{12}=\dfrac{z}{15}\) (2)
Từ (1) và (2) ⇒ \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)\(=\dfrac{x+y-z}{8+12-15}\) \(=\dfrac{10}{5}=2\)
⇒ \(\left\{{}\begin{matrix}\dfrac{x}{8}=2\\\dfrac{y}{12}=2\\\dfrac{z}{15}=2\end{matrix}\right.\) ⇒\(\left\{{}\begin{matrix}x=16\\y=24\\z=30\end{matrix}\right.\)
Ta có \(\dfrac{x}{2}=\dfrac{y}{3}\) => \(\dfrac{1}{4}\cdot\dfrac{x}{2}=\dfrac{1}{4}\cdot\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\left(1\right)\)
\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{1}{3}\cdot\dfrac{y}{4}=\dfrac{1}{3}\cdot\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\left(2\right)\)
Từ ( 1 ) và ( 2 ) ta có
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\) và x+y-z=10
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)
\(\Rightarrow\dfrac{x}{8}=2\Rightarrow x=2\cdot8=16\)
\(\dfrac{y}{12}=2\Rightarrow=2\cdot12=24\)
\(\dfrac{z}{15}=2\Rightarrow z=2\cdot15=30\)
vậy x = 16; y = 24; z = 30
Chúc bn học tốt
\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}vàx+y-z=10\)
Ta có :
\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{y}{4}=\dfrac{z}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{y}{12}\\\dfrac{y}{12}=\dfrac{z}{5}\end{matrix}\right.\) => \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\) và \(x+y-z=10\)
Từ : \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}\)
=> \(\dfrac{10}{5}=2\)
Với : \(\dfrac{x}{8}=2\Rightarrow x=16\)
Với : \(\dfrac{y}{12}=2\Rightarrow y=24\)
Với: \(\dfrac{z}{15}=2\Rightarrow z=30\)
Vậy x,y,z là : 16,24,30
\(\dfrac{x}{3}=\dfrac{2y}{5}=\dfrac{3z}{8}vàx-y+z=95\)
\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{\dfrac{5}{2}}=\dfrac{z}{\dfrac{8}{3}}=\dfrac{x-y+z}{3-\dfrac{5}{2}+\dfrac{8}{3}}=\dfrac{95}{\dfrac{19}{6}}=30\\ \Rightarrow\left\{{}\begin{matrix}x=90\\y=30\cdot\dfrac{5}{2}=75\\z=30\cdot\dfrac{8}{3}=80\end{matrix}\right.\)
Tìm x,y,z biết:
a) \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{4}\) và x-y+z=-21
b)\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) và \(x^2-2y^2+z^2=44\)
\(\dfrac{x}{2}=\dfrac{y}{3}\text{⇒}\dfrac{x}{10}=\dfrac{y}{15}\)
\(\dfrac{y}{5}=\dfrac{z}{4}\text{⇒}\dfrac{y}{15}=\dfrac{z}{12}\)
⇒\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-21}{-3}=7\)
⇒x=70;y=105;z=84
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)⇒\(\dfrac{x^2}{4}=\dfrac{2y^2}{18}=\dfrac{z^2}{25}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x^2}{4}=\dfrac{2y^2}{18}=\dfrac{z^2}{25}=\dfrac{x^2-2y^2+z^2}{4-18+25}=\dfrac{44}{11}=4\)
⇒x=8;y=12;z=20
Tìm x,y,z biết:
a, x : y : z = 10 : 3 : 4 và x + 2y - 3z = -20
b, \(\dfrac{x}{2}\) = \(\dfrac{y}{3}\) và \(\dfrac{y}{5}\) = \(\dfrac{z}{4}\) và x - y + z = -49
c, \(\dfrac{x}{2}\)= \(\dfrac{y}{3}\) =\(\dfrac{z}{4}\) và xy + \(z^2\)= 88
d, \(\dfrac{x}{5}\)= \(\dfrac{y}{7}\) = \(\dfrac{z}{3}\) và \(x^2\) + \(y^2\) + \(z^2\) = 415
Giải hộ mk nha
Tìm x,y,z biết:
a) 3x=2y, 7y=5z và x-y+z=32
b) \(\dfrac{x}{2}\)=\(\dfrac{y}{3}\) và x.y=24
c)\(\dfrac{x-1}{2}\)=\(\dfrac{y-2}{3}\)=\(\dfrac{z-3}{4}\) và 2x+3y-z=50
d)\(\dfrac{x}{2}\)=\(\dfrac{y}{3}\)=\(\dfrac{z}{5}\) và x.y.z=810
Cho đơn thức A=\(\dfrac{6}{7}x^2y^2.\left(-3\dfrac{1}{2}x^2y\right)\)
a. Thu gọn đơn thức A
b. Tính giá trị đơn thức A biết \(\dfrac{x}{y}=\dfrac{-2}{3}vàx-y=5\)
a: \(A=\dfrac{6}{7}x^2y^2\cdot\dfrac{-7}{2}x^2y=-3x^4y^3\)
b: Áp dụng tính chất của dãy tỉ số bằng nhau,ta được:
\(\dfrac{x}{-2}=\dfrac{y}{3}=\dfrac{x-y}{-2-3}=\dfrac{5}{-5}=-1\)
Do đó: x=2; y=-3
\(A=-3x^4y^3=-3\cdot2^4\cdot\left(-3\right)^3=3\cdot27\cdot16=81\cdot16=1296\)
\(A=\dfrac{6}{7}x^2y^2.\left(-3\dfrac{1}{2}x^2y\right)\)
\(=\dfrac{6}{7}x^2y^2.\left(-\dfrac{7}{2}\right)x^2y\)
\(=-3x^4y^3\)
b)Có: \(\dfrac{x}{y}=-\dfrac{2}{3}\Leftrightarrow\dfrac{x}{2}=\dfrac{-y}{3}=\dfrac{x-y}{2+3}=\dfrac{5}{5}=1\)
\(\Rightarrow x=2;y=-3\)
Tại \(x=2;y=-3\) , giá trị của biểu thức là:
\(-3.2^4.\left(-3\right)^3=-3.16.\left(-27\right)=1296\)