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Tra loi: Vi x+3/y+5 = 3/5
=>(x+3).5 = (y+5) .3
=>5x+15=3y+15
=>5x=3y
=>x=3k;y=5k( k thuoc Z; k khac 0)
Ma x+y=16=> 3k+5k=16
=> k=2
=> x= 3 .2 =6 ;y=2.5=10
Vay x=6 va y+10
\(\frac{x}{3}=\frac{7}{y}\)
\(\Rightarrow x.y=7.3\)
\(\Rightarrow x.y=21\)
\(\Rightarrow x.y=21=7.3=3.7=-7.\left(-3\right)=-3.\left(-7\right)\)
\(\Leftrightarrow x\in\left\{3;7;-3;-7\right\}\)
\(\Leftrightarrow y\in\left\{3;7;-3;-7\right\}\)
sửa vì bài này cần thêm điều kiện x,y thuôc z
\(\frac{x}{3}=\frac{7}{y}\)
\(xy=21\)
do x,y thuộc z
\(x;y\in\left\{\left(1;21\right);\left(21;1\right);\left(-1;-21\right);\left(-21;-1\right);\left(-3;-7\right)\right\}\)còn nữa (3;7);(7;3);(-7;-3)
Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!
a> x=13/12:2/3=13/8. vậy x=13/8
b>x=5/4 nhân 5/4=25/16. vậy...
c>x=7/4+1/2=9/4. vậy x=+-9/4
d>x=x=1-1/4=3/4. vậy...
a) \(0,5-\frac{2}{3}x=\frac{-7}{12}\)
\(\frac{1}{2}-\frac{2}{3}x=\frac{-7}{12}\)
\(\frac{2}{3}x=\frac{1}{2}-\frac{-7}{12}\)
\(\frac{2}{3}x=\frac{13}{12}\)
\(x=\frac{13}{12}:\frac{2}{3}\)
\(x=\frac{13}{8}\)
b) \(\frac{3}{4}-x:\frac{5}{4}=\frac{-1}{2}\)
\(x:\frac{5}{4}=\frac{3}{4}-\frac{-1}{2}\)
\(x:\frac{5}{4}=\frac{5}{4}\)
\(x=\frac{5}{4}.\frac{5}{4}\)
\(x=\frac{25}{16}\)
c) \(\left|x-\frac{1}{2}\right|-\frac{1}{4}=\frac{3}{2}\)
\(\left|x-\frac{1}{2}\right|=\frac{3}{2}+\frac{1}{4}\)
\(\left|x-\frac{1}{2}\right|=\frac{7}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{7}{4}\\x-\frac{1}{2}=\frac{-7}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{-5}{4}\end{cases}}}\)
Vậy x = 9/4 hoặc x = -5/4
d) \(\left|1-x\right|=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}1-x=\frac{1}{4}\\1-x=\frac{-1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{5}{4}\end{cases}}}\)
Vậy x = 3/4 hoặc 5/4
a) Ta có: \(\frac{3+x}{5+y}=\frac{3}{5}\)
=> (3 + x).5 = 3(5 + y)
=> 15 + 5x = 15 + 3y
=> 5x = 3y
=> x = 3/5y
Mà x + y = 16
hay 3/5y + y = 16
=> (3/5 + 1).y = 16
=> 8/5.y = 16
=> y = 16 : 8/5
=> y = 10
=> x = 16 - 10 = 6
Vậy x = 6; y = 10
b) Ta có: \(\frac{x-7}{y-6}=\frac{7}{6}\)
=> (x - 7).6 = 7.(y - 6)
=> 6x - 42 = 7y - 42
=> 6x = 7y
=> x = 7/6y
Mà x - y = -4
hay 7/6y - y = -4
=> 1/6y = -4
=> y = -4 : 1/6
=> y = -24
=> x = -4 - 24 = -28
Vậy x = -28; y = -24
\(\Rightarrow36^{15}<6^x<36^{16}\Rightarrow6^{30}<6^x<6^{32}\Rightarrow30
TA CÓ: \(\frac{3+x}{5+y}=\frac{3}{5}\)và \(x+y=16\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{3+x}{5+y}=\frac{3}{5}\Leftrightarrow\frac{3+x}{3}=\frac{5+y}{5}=\frac{3+x+y+5}{3+5}=\frac{24}{8}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{3+x}{3}=3\Leftrightarrow3+x=9\Leftrightarrow x=6\\\frac{5+y}{5}=3\Leftrightarrow5+y=15\Leftrightarrow y=10\end{cases}}\)
Vậy \(x=6\)và \(y=10\)