tìm y nếu y+y : 12/5 = 9/4
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Vì x tỉ lệ thuận với y theo hệ số tỉ lệ k
nên x=ky
hay k=3
Vậy: x=3y
Thay \(y=-\dfrac{5}{3}\) vào x=3y, ta được:
\(x=3\cdot\dfrac{-5}{3}=-5\)
Thay \(y=\dfrac{4}{9}\) vào x=3y, ta được:
\(x=3\cdot\dfrac{4}{9}=\dfrac{4}{3}\)
ta có \(\dfrac{x}{y}=\dfrac{12}{4}=3\)
khi \(y=\dfrac{-5}{3}\) thì \(x=-\dfrac{5}{3}.3=-5\)
khi \(y=\dfrac{4}{9}\) thì \(x=\dfrac{4}{9}.3=\dfrac{4}{3}\)
\(\frac{7}{y}=\frac{1}{12}\)
\(=>\frac{7}{y}=\frac{7}{84}\)
\(=>y=84\)
\(\frac{9}{10}-y.\frac{3}{4}=\frac{2}{5}\)
\(y.\frac{3}{4}=\frac{9}{10}-\frac{2}{5}=\frac{1}{2}\)
\(y=\frac{1}{2}:\frac{3}{4}\)
\(y=\frac{1}{2}.\frac{4}{3}\)
\(y=\frac{2}{3}\)
\(\frac{5}{6}:\left[y+\frac{7}{9}\right]=\frac{3}{4}\)
\(y+\frac{7}{9}=\frac{5}{6}:\frac{3}{4}\)
\(y+\frac{7}{9}=\frac{5}{6}.\frac{4}{3}\)
\(y+\frac{7}{9}=\frac{10}{9}\)
\(y=\frac{10}{9}-\frac{7}{9}\)
\(y=\frac{3}{9}\)
\(y=\frac{1}{3}\)
1, 7/y=1/12
=>7.12=y
=>y=84
2, 9/10-y.3/4=2/5
=>y.3/4=9/10-2/5
=>y.3/4=1/2
=>y=1/2:3/4=2/3
3, 5/6[y+7/9]=3/4
=>y+7/9=5/6:3/4
=>y+7/9=10/9
=>y=10/9-7/9
=>y=1/3
\(a.\)\(y-\frac{11}{3}=\frac{12}{5}\)
\(y=\frac{12}{5}+\frac{11}{3}\)
\(y=\frac{91}{15}\)
\(b.2-y=\frac{15}{6}\)
\(y=2-\frac{15}{6}\)
\(y=-\frac{1}{2}\)
\(c.y\times2\frac{2}{3}=\frac{4}{9}\)
\(y=\frac{4}{9}:2\frac{2}{3}\)
\(y=\frac{1}{6}\)
\(d.y:1\frac{2}{5}=\frac{21}{10}\)
\(y=\frac{21}{10}\times1\frac{2}{5}\)
\(y=\frac{147}{5}\)
a: Áp dụng tính chất của DTSBN, ta được:
x/5=y/2=(x-y)/(5-2)=9/3=3
=>x=15; y=6
b: =>(x-3)/12=3/(x-3)
=>(x-3)^2=36
=>(x-9)(x+3)=0
=>x=9 hoặc x=-3
c; x/2=y/3
=>x/10=y/15
y/5=z/4
=>y/15=z/12
=>x/10=y/15=z/12=(x-y-z)/(10-15-12)=-49/-17=49/17
=>x=490/17; y=735/17; z=588/17
Lời giải:
$z=(x+y+z)-(x+y)=21-4=17$
$y=z-5=17-5=12$
$2k=z+x=(x+y+z)-y=21-12=9$
$k=\frac{9}{2}$
Không đáp án nào đúng.
Bài 9:
Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)
Vậy: (x,y,z,t)=(-10;6;34;-18)
Bài 11:
Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)
\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)
Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)
\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)
Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)
\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)
Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)
\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)
Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)
\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)
Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)
\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)
Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)
a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)
b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)
c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)
\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)
\(y+y:\frac{12}{5}=\frac{9}{4}\)
\(y+y\times\frac{5}{12}=\frac{9}{4}\)
\(y\times\left(1+\frac{5}{12}\right)=\frac{9}{4}\)
\(y\times\frac{17}{12}=\frac{9}{4}\)
\(y=\frac{9}{4}:\frac{17}{12}\)
\(y=\frac{27}{17}\)