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a. \(y+30\%y=-1,3\)
\(=>y.\left(1+30\%\right)=-1,3\)
\(=>y.\left(1+\dfrac{3}{10}\right)=-1,3\)
\(=>y.\dfrac{13}{10}=\dfrac{-13}{10}\)
\(=>y=-1\)
b.\(y-25\%y=\dfrac{1}{2}\)
\(=>y.\left(1-25\%\right)=\dfrac{1}{2}\)
\(=>y.\left(1-\dfrac{1}{4}\right)=\dfrac{1}{2}\)
\(=>y.\dfrac{3}{4}=\dfrac{1}{2}\)
\(=>y=\dfrac{1}{2}.\dfrac{4}{3}\)
\(=>y=\dfrac{2}{3}\)
c.\(3\dfrac{1}{3}y+16\dfrac{3}{4}=-13,25\)
\(=>\dfrac{10}{3}y+\dfrac{67}{4}=\dfrac{-53}{4}\)
\(=>\dfrac{10}{3}y=\dfrac{-53-67}{4}\)
\(=>\dfrac{10}{3}y=\dfrac{-120}{4}\)
\(=>\dfrac{10}{3}y=-30\)
\(=>y=-30.\dfrac{3}{10}\)
\(=>y=-9\)
tick cho mk nha
k hiểu chỗ nào thì hỏi mk
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)\(y+30\%y=-1,3\Rightarrow y+\frac{3}{10}y=-1,3\Rightarrow y\left(1+\frac{3}{10}\right)=-1,3\Rightarrow y\times1,3\)\(=-1,3\Rightarrow y=-1\)
b)\(y-25\%y=\frac{1}{2}\Rightarrow y-\frac{1}{4}y=\frac{1}{2}\Rightarrow y\left(1-\frac{1}{4}\right)=\frac{1}{2}\Rightarrow y\times\frac{3}{4}=\frac{1}{2}\Rightarrow y=\frac{1}{2}:\frac{3}{4}\Rightarrow y=\frac{2}{3}\)
c)\(3\frac{1}{3}y+16\frac{3}{4}=13,25\Rightarrow\frac{10}{3}y+\frac{67}{4}=\frac{53}{4}\Rightarrow\frac{10}{3}y=\frac{53}{4}-\frac{67}{4}\Rightarrow\frac{10}{3}y=\frac{-7}{2}\Rightarrow y=\frac{-21}{20}\)
a) ta có : y + 30/100y = -1,3
hay y(1+ 3/10) = - 13/10
y.13/10 = -13/10
y= -13/10:13/10
y= -1
b) y - 25/100y=1/2
hay y(1-1/4)=1/2
y.3/4 = 1/2
y=2/3
c) 10/3y + 16,75 = -13,25
10/3y= -13,25 - 16,75
10/3y= -30
y= - 30:10/3
y= -9
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a, \(\dfrac{3}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)
ta có: \(\dfrac{3}{x}+\dfrac{y}{3}=\dfrac{5}{6}=>\dfrac{3}{x}=\dfrac{5}{6}-\dfrac{y}{3}=\dfrac{5-2y}{6}\)
=>\(\dfrac{3}{x}=\dfrac{5-2y}{6}=>x.\left(5-2y\right)=3.6=18\)
=> x và 5-2y thuộc Ư của 18={1,-1,2,-2,3,-3,6,-6}
vì 5-2y là số lẻ=> 5-2y= +-1 hoặc 5-2y=+-3
xét bảng
5-2y | 1 | -1 | 3 | -3 |
y | 2 | 3 | 1 | 4 |
x | 18 | -18 | 6 | -6 |
vậy giá trị x,y cần tìm là: {x=18.y=2}
{x=-18.y=3}
{x=6, y=1}Ư
{x=-6,y=4}
a) \(y+30\%y=-1.3\)
\(y+\dfrac{3}{10}y=\dfrac{-13}{10}\)
\(y\left(1+\dfrac{3}{10}\right)=\dfrac{-13}{10}\)
\(y\left(\dfrac{10}{10}+\dfrac{3}{10}\right)=\dfrac{-13}{10}\)
\(y\cdot\dfrac{13}{10}=\dfrac{-13}{10}\)
\(y=\dfrac{-13}{10}:\dfrac{13}{10}\)
\(y=\dfrac{-13}{10}\cdot\dfrac{10}{13}\)
\(y=-1\)
Vậy \(y=-1\).
b) \(y-25\%y=\dfrac{1}{2}\)
\(y-\dfrac{1}{4}y=\dfrac{1}{2}\)
\(y\left(1-\dfrac{1}{4}\right)=\dfrac{1}{2}\)
\(y\left(\dfrac{4}{4}-\dfrac{1}{4}\right)=\dfrac{1}{2}\)
\(y\cdot\dfrac{3}{4}=\dfrac{1}{2}\)
\(y=\dfrac{1}{2}:\dfrac{3}{4}\)
\(y=\dfrac{1}{2}\cdot\dfrac{4}{3}\)
\(y=\dfrac{2}{3}\)
Vậy \(y=\dfrac{2}{3}\).
c) \(3\dfrac{1}{3}y+16\dfrac{3}{4}=-13.25\)
\(\dfrac{10}{3}y+\dfrac{67}{4}=\dfrac{-53}{4}\)
\(\dfrac{10}{3}y=\dfrac{-53}{4}-\dfrac{67}{4}\)
\(\dfrac{10}{3}y=-30\)
\(y=-30:\dfrac{10}{3}\)
\(y=-30\cdot\dfrac{3}{10}\)
\(y=-9\)
Vậy \(y=-9\).
a) y+30%y=−1.3y+30%y=−1.3
y+310y=−1310y+310y=−1310
y(1+310)=−1310y(1+310)=−1310
y(1010+310)=−1310y(1010+310)=−1310
y⋅1310=−1310y⋅1310=−1310
y=−1310:1310y=−1310:1310
y=−1310⋅1013y=−1310⋅1013
y=−1y=−1
Vậy y=−1y=−1.
b) y−25%y=12y−25%y=12
y−14y=12y−14y=12
y(1−14)=12y(1−14)=12
y(44−14)=12y(44−14)=12
y⋅34=12y⋅34=12
y=12:34y=12:34
y=12⋅43y=12⋅43
y=23y=23
Vậy y=23y=23.
c) 313y+1634=−13.25313y+1634=−13.25
103y+674=−534103y+674=−534
103y=−534−674103y=−534−674
103y=−30103y=−30
y=−30:103y=−30:103
y=−30⋅310y=−30⋅310
y=−9y=−9
Vậy y=−9y=−9.