Tìm y :
a) y−3 = 4
y : 3 = 4
b) y−4 = 5
y : 4 = 5
c) y−2 = 3
y : 2 = 3
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a) y : 4/11 x 3/5 = 27/4
y : 4/11 = 27/4 : 3/5
y : 4/11 = 45/4
y = 45/4 x 4/11
y = 45/11
b) y + 7/2 - 5/9 = 14/3
y + 7/2 = 14/3 + 5/9
y + 7/2 = 47/9
y = 47/9 - 7/2
y = 31/18
a) y : 4/11 x 3/5 = 27/4
y : 4/11 = 27/4 : 3/5
y : 4/11 = 45/4
y = 45/4 x 4/11
y = 45/11
b) y + 7/2 - 5/9 = 14/3
y + 7/2 = 14/3 + 5/9
y + 7/2 = 47/9
y = 47/9 - 7/2
y = 31/18
bài 2 : a) \(\dfrac{a-1}{2}=\dfrac{b+3}{4}=\dfrac{c-5}{6}\)
áp dụng dảy tỉ số bằng nhau
ta có : \(\dfrac{5\left(a-1\right)-3\left(b+3\right)-4\left(c-5\right)}{5.2-3.4-4.6}\)
\(=\dfrac{5a-5-3b-9-4c+20}{10-12-24}=\dfrac{\left(5a-3b-4c\right)-5-9+20}{-26}\)
\(=\dfrac{46+6}{-26}=\dfrac{52}{-26}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a-1}{2}=-2\\\dfrac{b+3}{4}=-2\\\dfrac{c-5}{6}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a-1=-4\\b+3=-8\\c-5=-12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=-3\\b=-11\\c=-7\end{matrix}\right.\)
vậy \(a=-3;b=-11;c=-7\)
b) ta có : \(3a=2b\Leftrightarrow6a=4b=5c\Leftrightarrow\dfrac{6a}{2}=\dfrac{4b}{2}=\dfrac{5c}{2}\)
áp dụng dảy tỉ số bằng nhau
ta có \(\dfrac{-60a-60b+60c}{-10.2-15.2+12.2}=\dfrac{60\left(-a-b+c\right)}{-20-30+24}\)
\(=\dfrac{60\left(-52\right)}{-26}=\dfrac{-3120}{-26}=120\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{6a}{2}=120\\\dfrac{4b}{2}=120\\\dfrac{5c}{2}=120\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6a=240\\4b=240\\5c=240\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=40\\b=60\\c=48\end{matrix}\right.\)
vậy \(a=40;b=60;c=48\)
a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)
y = \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)
y = \(\dfrac{4}{3}\)
b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)
y - 0,5 + 0,5 = \(\dfrac{3}{4}\)
y = \(\dfrac{3}{4}\)
c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2
0,8 - 0,4y = 0,2
0,4y = 0,8 - 0,2
0,4y = 0,6
y = 1,5
d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{14}{9}\)
y = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)
y = \(\dfrac{29}{36}\)
e, y : \(\dfrac{5}{4}\) = \(\dfrac{9}{5}\) + \(\dfrac{1}{2}\)
y : \(\dfrac{5}{4}\) = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{8}\)
f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y = \(\dfrac{4}{5}\)
y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\)) = \(\dfrac{4}{5}\)
2y = \(\dfrac{4}{5}\)
y = \(\dfrac{2}{5}\)
bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 => x-1/3=y-2/4=z-3/5
áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1
do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự
Bài 1:
a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )
a) y × 2/5 + y × 3/5 = 5/7
y x ( 2/5 + 3/5) = 5/7
y x 5/5 = 5/7
y x 1 = 5/7
y = 5/7 : 1
y = 5/7
b) y × 7/2 - y × 3/2 = 3/4
y x ( 7/2 - 3/2) = 3/4
y x 4/2 = 3/4
y x 2 = 3/4
y = 3/4 : 2
y = 3/4 x 1/2
y = 3/8
`a,yxx(2/5+3/5)=5/7`
`y xx1=5/7`
`y=5/7:1`
`y=5/7`
`b,y xx(7/2-3/2)=3/4`
`y xx2=3/4`
`y =3/4:2`
`y=3/2`
Phương pháp giải:
- Muốn tìm số bị trừ ta lấy hiệu cộng số trừ.
- Muốn tìm số bị chia ta lấy thương nhân với số chia.
Lời giải chi tiết:
a)
● y−3 = 4
y = 4+3
y = 7
● y : 3 = 4
y = 4 × 3
y = 12
b)
● y−4 = 5
y = 5+4
y = 9
● y : 4 = 5
y = 5 × 4
y = 20
c)
● y−2 = 3
y = 3+2
y = 5
● y : 2 = 3
y = 3 × 2
y = 6