\(\dfrac{x}{y}=\dfrac{y}{21}và2x-y=34\)
\(\dfrac{x}{19}=\dfrac{y}{21}và2x-y=34\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{19}=\dfrac{y}{21}=\dfrac{2x-y}{19\cdot2-21}=\dfrac{34}{17}=2\\ \Rightarrow\left\{{}\begin{matrix}x=38\\y=42\end{matrix}\right.\)
\(\dfrac{2x}{19.2}=\dfrac{y}{21}\) và \(\text{ 2x− y = 34}\)
Áp dụng tcdtsbn , ta có:
\(\dfrac{2x}{19.2}=\dfrac{y}{21}=\dfrac{2x-y}{19.2-21}=\dfrac{34}{17}=2\)
\(\Rightarrow x=76\)
\(\Rightarrow y=42\)
Ta có: \(\dfrac{x}{19}=\dfrac{y}{21}=\dfrac{2x}{38}=\dfrac{y}{21}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x+y}{38+21}=\dfrac{34}{59}\)
=> \(\left\{{}\begin{matrix}x=10,94...\\y=12,101...\end{matrix}\right.\)
câu 3 tìm x,y biết
\(\dfrac{x}{y}=\dfrac{17}{3}vàx+y=-60\)
\(\dfrac{x}{19}=\dfrac{y}{21}và2x-y=34\)
\(\dfrac{x}{y}=\dfrac{17}{3}\Rightarrow\dfrac{x}{17}=\dfrac{y}{3}=\dfrac{x+y}{17+3}=\dfrac{-60}{20}=-3\)
x=-3.17=-51
y=-3.3=-9
câu tiếp nha:\(\dfrac{x}{19}=\dfrac{y}{21}=\dfrac{2x}{38}=\dfrac{2x-y}{38-21}=\dfrac{34}{17}=2\)
x=19.2=38
y=21.2=42
Chúc bạn học tốt
\(\dfrac{x}{y}=\dfrac{17}{3}\Rightarrow\dfrac{x}{17}=\dfrac{y}{3}\)và x+y=-60
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{17}=\dfrac{y}{3}=\dfrac{x+y}{17+3}=\dfrac{-60}{20}=-3\)
=>x=-3.17=-51
y=-3.3=-9
b)\(\dfrac{x}{19}=\dfrac{y}{21}\Rightarrow\dfrac{2x}{38}=\dfrac{y}{21}\)và 2x-y=34
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{38}=\dfrac{y}{21}=\dfrac{2x-y}{38-21}=\dfrac{34}{17}=2\)
=>x=2.19=38
y=2.21=42
\(\dfrac{x}{y}=\dfrac{17}{3} \)
\(\dfrac{x}{17}=\dfrac{y}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{17}=\dfrac{y}{3}=\dfrac{x+y}{17+3}=\dfrac{-60}{20}=-3\)
\(\Rightarrow\)\(x=-3\Rightarrow 17 . -3 = -51 \)
\(\Rightarrow y =-3\Rightarrow 3 . -3=-9\)
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}và2x+3y-z=-14\)
\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}vàx-y-z=28\)
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:
\(\dfrac{2x}{10}=\dfrac{3y}{15}=\dfrac{z}{7}=\dfrac{2x+3y-z}{10+15-7}=\dfrac{-14}{18}=\dfrac{-7}{9}\)
⇒\(\left\{{}\begin{matrix}x=\dfrac{-7}{9}.3=\dfrac{-7}{3}\\y=\dfrac{-7}{9}.5=\dfrac{-35}{9}\\z=\dfrac{-7}{9}.7=\dfrac{-49}{9}\end{matrix}\right.\)
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}và2x+y-z=6\)
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) \(\Rightarrow\dfrac{2x}{4}=\dfrac{y}{3}=\dfrac{z}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{4}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2x+y-z}{4+3-4}=\dfrac{6}{3}=2\)
\(\dfrac{x}{2}=2\Rightarrow x=2.2=4\)
\(\dfrac{y}{3}=2\Rightarrow y=2.3=6\)
\(\dfrac{z}{4}=2\Rightarrow z=2.4=8\)
Vậy \(x=4;y=6;z=8\)
Tìm x,y biết: \(\dfrac{x}{19}=\dfrac{y}{21}\)và 2x -y = 34.
A.x = 38; y = 42; B.x = -38; y = -42; C. x = 21; y = 19 D.x = -21; y = -1
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
\(\dfrac{x}{19}=\dfrac{y}{21}\)và\(2x-y=34\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{19}=\dfrac{y}{21}=\dfrac{2x-y}{2\cdot19-21}=\dfrac{34}{17}=2\)
Do đó: x=38;y=42
1.\(x=\dfrac{y}{6}=\dfrac{z}{3}và2x-3y+4z=24\)
2.\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}và5z-3x-4y=50\)
3.\(6x=10y=15zvàx+y-z=90\)
\(1,\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y+4z}{2-18+12}=\dfrac{24}{-4}=-6\\ \Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-36\\z=-18\end{matrix}\right.\\ 2,\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x+3-4y-12+5z-25}{-6-16+30}=\dfrac{50-34}{8}=\dfrac{16}{8}=2\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=4\\y+3=8\\z-5=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=5\\z=17\end{matrix}\right.\)
\(3,6x=10y=15z\Leftrightarrow\dfrac{6x}{30}=\dfrac{10y}{30}=\dfrac{15z}{30}\\ \Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{x+y-z}{5+3-2}=\dfrac{90}{6}=15\\ \Leftrightarrow\left\{{}\begin{matrix}x=75\\y=45\\z=30\end{matrix}\right.\)
Tìm x,y,z
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}và2x+3y-z=50\)