Tìm x :
a. x - 0,5 = 5/6 + 7/12
giúp mình, mình sẽ tick
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng dãy tỉ số bằng nhau:
b.
\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x-y}{2-\left(-5\right)}=\dfrac{-7}{7}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=-5.\left(-1\right)=5\end{matrix}\right.\)
d.
\(\dfrac{4}{x}=\dfrac{7}{y}\Rightarrow\dfrac{y}{7}=\dfrac{x}{4}=\dfrac{y-x}{7-4}=\dfrac{-12}{3}=-4\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.\left(-4\right)=-16\\y=7.\left(-4\right)=-28\end{matrix}\right.\)
a) \(\dfrac{5}{4}+\dfrac{3}{5}=\dfrac{25}{20}+\dfrac{12}{20}=\dfrac{37}{20}\)
b) \(\dfrac{7}{8}+\dfrac{9}{24}=\dfrac{21}{24}+\dfrac{9}{24}=\dfrac{30}{24}=\dfrac{15}{12}\)
c) \(\dfrac{2}{36}+\dfrac{1}{6}+\dfrac{5}{12}=\dfrac{2}{36}+\dfrac{6}{36}+\dfrac{15}{36}=\dfrac{23}{36}\)
x - 4/5 = 3/7
x = 3/7 + 4/5
x = 43/35
x + 3/7 = 4/5
x = 4/5 - 3/7
x = 13/35
19/20 - x = 8/5 - 3/4
19/20 - x = 17/20
x = 19/20 - 17/20
x = 2/20 = 1/10
4/5 x X = 6/9 - 4/7
4/5 x X = 2/21
x = 2/21 : 4/5
x =5/42
x : 7/9 = 6/8
x = 6/8 x 7/9
x = 7/12
2/3 - x/6 = 6/18
x/6 = 2/3 - 6/18
x/6 = 1/3
x/6 = 2/6
=> x =2
1. x = 43/35
2. x = 13/35
3. x = 1/10
còn lại tự tính nhá
tui bận òi
\(\dfrac{1}{2}=\dfrac{6}{12};\dfrac{2}{3}=\dfrac{8}{12};\dfrac{7}{12}=\dfrac{7}{12}\)
Chắc là quy đồng nhỉ :)
\(\dfrac{1\times6}{2\times6}=\dfrac{6}{12};\dfrac{2\times4}{3\times4}=\dfrac{8}{12};\dfrac{7}{12}\)
câu a+b dùng quy tắc chuyển vế
c, 3.(1/2-x)-5.(x-1/10)=-7/4
=>(3.1/2-3x)-(5x-5.1/10)=-7/4
=>3/2-3x-5x+1/2=-7/4
=>(3/2+1/2)-(3x+5x)=-7/4
=> 2-8x=-7/4
=>8x=15/4
=>x=15/4:8
=>x=15/32
a) 2.(1/4 - 3x) = 1/5 - 4x
=> 1/2 - 6x = 1/5 -4x
=> -6x + 4x = 1/5 - 1/2
=> -2x = -3/10 = 3/20
b) 4.(1/3 - x) + 1/2 = 5/6 +x
=> 4/3 - 4x + 1/2 = 5/6 +x
=> -4x - x = 5/6 - 4/3 - 1/2
=> -5x = -1
=> x= 1/5
c) 3. (1/2 - x) -5. ( x - 1/10) = -7/4
=> 3/2 - 3x - 5x + 1/2 = -7/4
=> -3x - 5x = -7/4 - 3/2 - 1/2
=> -8x = -15/4
=> x = 15/32
\(x-\frac{1}{2}=\frac{17}{12}\)
\(x=\frac{23}{12}\)
Vậy \(x=\frac{23}{12}\)
\(x=\frac{23}{12}\)
\(x=1,91\left(6\right)\)
Vậy ...