\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}và2x+y-z=6\)
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
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.\(6x=10y=15zvàx+y-z=90\)
\(1,\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y+4z}{2-18+12}=\dfrac{24}{-4}=-6\\ \Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-36\\z=-18\end{matrix}\right.\\ 2,\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x+3-4y-12+5z-25}{-6-16+30}=\dfrac{50-34}{8}=\dfrac{16}{8}=2\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=4\\y+3=8\\z-5=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=5\\z=17\end{matrix}\right.\)
\(3,6x=10y=15z\Leftrightarrow\dfrac{6x}{30}=\dfrac{10y}{30}=\dfrac{15z}{30}\\ \Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{x+y-z}{5+3-2}=\dfrac{90}{6}=15\\ \Leftrightarrow\left\{{}\begin{matrix}x=75\\y=45\\z=30\end{matrix}\right.\)
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}và2x+3y-z=-14\)
\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}vàx-y-z=28\)
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:
\(\dfrac{2x}{10}=\dfrac{3y}{15}=\dfrac{z}{7}=\dfrac{2x+3y-z}{10+15-7}=\dfrac{-14}{18}=\dfrac{-7}{9}\)
⇒\(\left\{{}\begin{matrix}x=\dfrac{-7}{9}.3=\dfrac{-7}{3}\\y=\dfrac{-7}{9}.5=\dfrac{-35}{9}\\z=\dfrac{-7}{9}.7=\dfrac{-49}{9}\end{matrix}\right.\)
Tìm x,y,z
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}và2x+3y-z=50\)
Bài 1: Tìm x, y biết :
\(a,\dfrac{x}{5}=\dfrac{y}{6};\dfrac{y}{8}=\dfrac{z}{7}vàx+y+z=138\)
\(b,7x=5yvàxy=140\)
\(c,3x=8y=6zvà2x+3y-z=210\)
\(d,\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}và2x+3y-z=50\)
\(a,\dfrac{x}{5}=\dfrac{y}{6};\dfrac{y}{8}=\dfrac{x}{7}\) và \(x+y+z=138\)
\(\dfrac{x}{5}=\dfrac{y}{6}\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{24}\) \(\left(1\right)\)
\(\dfrac{y}{8}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{24}=\dfrac{z}{21}\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=\dfrac{x+y+z}{20+24+21}=\dfrac{138}{65}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{20}=\dfrac{138}{65}\\\dfrac{y}{24}=\dfrac{138}{65}\\\dfrac{z}{21}=\dfrac{138}{65}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{553}{13}\\y=\dfrac{3312}{65}\\z=\dfrac{2898}{65}\end{matrix}\right.\)
Vậy.......
Tìm x,y,z
\(\dfrac{3x-2y}{37}=\dfrac{5y-3z}{15}=\dfrac{2z-5y}{2}và2x+3y-z=-4\)
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}\)và 2x+y-z=81
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{2}\)và 5x-y+3z=124
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)và x.y.z=810
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}\)và\(x^2.y^2.z^2=288^2\)
a.
Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=3k\\z=4k\end{matrix}\right.\)
Thế vào \(2x+y-z=81\)
\(\Rightarrow2.5k+3k-4k=81\)
\(\Rightarrow9k=81\)
\(\Rightarrow k=9\)
\(\Rightarrow\left\{{}\begin{matrix}x=5k=45\\y=3k=27\\z=4k=36\end{matrix}\right.\)
b.
Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{2}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\\z=2k\end{matrix}\right.\)
Thế vào \(5x-y+3z=124\)
\(\Rightarrow5.3k-5k+3.2k=124\)
\(\Rightarrow16k=124\)
\(\Rightarrow k=\dfrac{31}{4}\) \(\Rightarrow\left\{{}\begin{matrix}x=3k=\dfrac{93}{4}\\y=5k=\dfrac{155}{4}\\z=2k=\dfrac{31}{2}\end{matrix}\right.\)
c.
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)
Thế vào \(xyz=810\)
\(\Rightarrow2k.3k.5k=810\)
\(\Rightarrow k^3=27\)
\(\Rightarrow k=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k=6\\y=3k=9\\z=5k=15\end{matrix}\right.\)
d.
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=6k\end{matrix}\right.\)
Thế vào \(x^2y^2z^2=288^2\)
\(\Rightarrow\left(2k\right)^2.\left(3k\right)^2.\left(6k\right)^2=288^2\)
\(\Rightarrow\left(k^2\right)^3=64\)
\(\Rightarrow k^2=4\)
\(\Rightarrow k=\pm2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k=4\\y=3k=6\\z=6k=12\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=2k=-4\\y=3k=-6\\z=6k=-12\end{matrix}\right.\)
Tìm x, y, z
\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\)
Áp dụng tích chất của dãy tỉ số bằng nhau, ta có
\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}\\ =\dfrac{x+y+2+y+z+1+z+x-3}{z+x+y}=\dfrac{2\left(x+y+z\right)+\left(1+2-3\right)}{z+x+y}=2\\ Vì\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\\ =>2=\dfrac{1}{x+y+z}=>2\left(x+y+z\right)=1=>x+y+z=\dfrac{1}{2}\\ =>\dfrac{x+y+2}{z}=2=>x+y+2=2z\\ \dfrac{y+z+1}{x}=2=>y+z+1=2x\\ \dfrac{z+x-3}{y}=2=>z+x-3=2y\\ \dfrac{1}{x+y+z}=2=>x+y+z=\dfrac{1}{2}\)
+) x+y+z = \(\dfrac{1}{2}=>y+z=\dfrac{1}{2}-x=>\dfrac{1}{2}-x+1=2x=>3x=\dfrac{3}{2}=>x=\dfrac{1}{2}\)
+)\(x+y+z=\dfrac{1}{2}=>x+y=\dfrac{1}{2}-z=>\dfrac{1}{2}-z+2=2z=>3z=\dfrac{5}{2}=>z=\dfrac{5}{6}\)
\(=>x+y+z=\dfrac{1}{2}+\dfrac{5}{6}+y=\dfrac{1}{2}=>\dfrac{4}{3}+y=\dfrac{1}{2}=>y=\dfrac{-5}{6}\)
Vậy \(x=\dfrac{1}{2}\\ y=\dfrac{-5}{6}\\ z=\dfrac{5}{6}\)
Ê mấy bọn 7B Nguyễn Lương Bằng ơi bài 2 Toán chiều làm thế này đúng chưa! Góp ý nha!
Tìm các số x, y, z biết:
a) \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{4}\) và x + z - y = -49
b) \(\dfrac{x}{3}=\dfrac{y}{-2};\dfrac{x}{6}=\dfrac{z}{7}\) và 3x - z + 2y = 3
Lm hết nha mọi ngừi ^^
a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-49}{7}=-7\)
Do đó: x=-70; y=-135; z=-84
a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
a) \(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{y}{5}=\dfrac{z}{4}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x+z-y}{10+12-15}=-\dfrac{49}{7}=-7\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-7\right).10=-70\\y=\left(-7\right).15=-105\\z=\left(-7\right).12=-84\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{-2}\\\dfrac{x}{6}=\dfrac{z}{7}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{-4}=\dfrac{z}{7}=\dfrac{3x}{18}=\dfrac{2y}{-8}=\dfrac{3x-z+2y}{18-7-8}=\dfrac{3}{3}=1\)
\(\Rightarrow\left\{{}\begin{matrix}x=1.6=6\\y=1.\left(-4\right)=-4\\z=1.7=7\end{matrix}\right.\)