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bài 3:
a, đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)
A/D tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2-z^2}{25+49-9}=\dfrac{585}{65}=9\)
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
a: \(y:\dfrac{2}{3}=-\dfrac{7}{2}+\dfrac{3}{5}=\dfrac{-35+6}{10}=\dfrac{-29}{10}\)
nên \(y=-\dfrac{29}{10}\cdot\dfrac{2}{3}=\dfrac{-58}{30}=-\dfrac{29}{15}\)
b: \(\dfrac{y}{-7}=-\dfrac{20}{35}\)
nên y/7=20/35
hay y=4
c: \(\dfrac{4}{5}=\dfrac{36}{-y}\)
nên \(y=-36\cdot\dfrac{5}{4}=-45\)
Bài 4: 26 \(\dfrac{1}{4}\)= 26, 25 . Quãng đương là 26,25 . 2,4 = 63 km.
Thời gian đi từ B đến A là : 63 : 30 = 2,1 h
a,\(\dfrac{x}{3}-\dfrac{1}{y}=\dfrac{1}{2}\)
=> \(\dfrac{1}{y}=\dfrac{x}{3}-\dfrac{1}{2}=>\dfrac{1}{y}=\dfrac{2x-3}{6}\)
=> y(2x-3)=6.1=6
=> y và 2x-3 là Ư (6)= {+-1,+-2,+-3,+-6}
| 2x-3 | -1 | 1 | 2 | -2 | 3 | -3 | 6 | -6 |
| x | 1 | 2 | 2,5 | 1/2 | 3 | 0 | 9/2 | -3/2 |
| y | -6 | 6 | 3 | -3 | 2 | -2 | 1 |
-1 |
vậy (x;y)= .......................
b,c làm tương tự
chúc bn học tốt 
a) \(x\)=1 \(y\)= 12
b)\(x\)=4 \(y\)= 14
hoặc \(x\)= 6 \(y \)=21
...
a) \(\dfrac{x}{5}=\dfrac{6}{-10}\)
\(\Rightarrow\) (-10).x=5.6
\(\Leftrightarrow\) (-10).x=30
\(\Leftrightarrow x=30:\left(-10\right)\)
\(\Leftrightarrow\) x=(-3)
Vậy......................
b) \(\dfrac{x}{3}=\dfrac{4}{y}\)
\(\Rightarrow xy=3.4=12\)
Ta có: xy=12=1.12=12.1=2.6=6.2=3.4=4.3=(-1).(-12)=......( bạn tự ghi nốt)
\(\Rightarrow\)(x;y)=(1;12) (12;1) (2;6) (6;2) (3;4) (4;3) (-1;-12) (-12;-1) (-2;-6) (-6;-2) (-3;-4) (-4;-3)
Vậy.....................................
a: x/5=6/-10
=>x/5=-3/5
=>x=-3
b: =>xy=12
=>\(\left(x,y\right)\in\left\{\left(1;12\right);\left(12;1\right);\left(-1;-12\right);\left(-12;-1\right);\left(2;6\right);\left(6;2\right);\left(-2;-6\right);\left(-6;-2\right);\left(3;4\right);\left(4;3\right);\left(-3;-4\right);\left(-4;-3\right)\right\}\)
c: =>x/2=y/7=k
=>x=2k; y=7k
=>\(\left(x,y\right)\in\left\{\left(2k;7k\right);k\in Z\right\}\)
d: 2/x=x/8
=>x^2=16
=>x=4 hoặc x=-4
a) x.21=6.7
x.21=42
x=42:21
x = 2
b) y . 20 = -5.28
y.20 = -140
y = (-140) : 20
y = -7
a)=>x*21=7*6
=>x*21=42
=>x=42/21
x=2
b)=>y*20=(-5)*28
=>y*20=-140
=>y=-140/20
y=-7
b/ Có \(\dfrac{x-7}{y-6}=\dfrac{7}{6}\)
nên \(6.\left(x-7\right)=7.\left(y-6\right)\)
\(\rightarrow\) \(6.x-6.7=7.y-7.6\)
\(\Rightarrow\) \(6x=7y\). Mà \(x-y=-4\) nên \(6x-6y=-24\)
\(\rightarrow\) \(7y-6x=-24\)
\(\rightarrow1y=-24\)
Và \(x-y=-4\) \(\Rightarrow\) \(x=\left(-4\right)+y\) \(=\left(-4\right)+\left(-24\right)\)\(=-28\)
Vậy \(x=-28\) \(;\) \(y=-24\)
a) \(\dfrac{1,28}{3,15}=\dfrac{128}{315}\)
b) \(\dfrac{2}{5}:3\dfrac{1}{4}=\dfrac{2}{5}:\dfrac{13}{4}=\dfrac{2}{5}\cdot\dfrac{4}{13}=\dfrac{8}{65}\)
c) \(\left(-1\dfrac{3}{7}\right):1,24=\dfrac{-10}{7}:\dfrac{31}{25}=\dfrac{-10}{7}\cdot\dfrac{25}{31}=\dfrac{-250}{217}\)
d) \(2\dfrac{1}{5}:3\dfrac{1}{7}=\dfrac{11}{5}:\dfrac{22}{7}=\dfrac{11}{5}\cdot\dfrac{7}{22}=\dfrac{7}{10}\)
\(\dfrac{y-5}{7-y}=\dfrac{2}{-3}\)
\(\Rightarrow\)\(-3\left(y-5\right)=2\left(7-y\right)\)
\(\Rightarrow-3y+15=14-2y\)
\(\Rightarrow-3y+2y=-1\)
\(\Rightarrow-1y=-1\Rightarrow y=1\)
\(\dfrac{y-5}{7-y}=\dfrac{2}{-3}\)
\(\Rightarrow-3\left(y-5\right)=2\left(7-y\right)\)
\(\Rightarrow-3y+15=14-2y\)
\(\Rightarrow-3y+2y=14-15\)
\(\Rightarrow-1y=-1\)
\(\Rightarrow y=\dfrac{-1}{-1}=1\)
Vậy \(y=1\).