
ЗА РЕШЕНИЕ ДАЮ 40 БАЛЛОВ! помогите решить уравнения за 5 класс. 1) 2 1/2х-1 5/8=2
3/4; 2) 1/4+х:3 5/9=7/16; 3) 4 2/5:х-2 3/5=7/10; 4) 3 1/3:(2 1/8-х)=2 2/95) (5/12+1/2х):2 1/2=11/126) (6 3/14-х)*2 1/3=9 5/67) (2 1/10-х):8+1 2/158) 4 1/6:(1/4х+1 4/15)-1 5/6=2/39) 1 1/2х+1/2х=1 2/310) 3 4/5х-1 7/10х=3 1/2 11)2/3х+7/12+1/4х+5/6=5 1/123/5+4/5х+7/10х+1 1/2=5 3/5

Ответы на вопрос






- 2 1/2x - 1 5/8 = 2 3/4
To solve this equation, we need to isolate the variable x. Let's start by adding 1 5/8 to both sides of the equation:
2 1/2x - 1 5/8 + 1 5/8 = 2 3/4 + 1 5/8
2 1/2x = 2 3/4 + 1 5/8
Next, let's find a common denominator for the fractions on the right side of the equation. The common denominator for 4 and 8 is 8, so we can rewrite the fractions as:
2 1/2x = 2 6/8 + 1 10/8
2 1/2x = 2 6/8 + 1 5/8
Now, let's add the fractions:
2 1/2x = 3 11/8
To add the whole numbers and fractions, we need to find a common denominator. The common denominator for 2 and 8 is 8:
2 1/2x = 3 11/8
2 1/2x = 3 22/16
Next, let's convert the mixed numbers to improper fractions:
2 1/2 = (2*2 + 1)/2 = 5/2
3 22/16 = (3*16 + 22)/16 = 70/16
Now, the equation becomes:
5/2x = 70/16
To isolate x, we need to multiply both sides of the equation by the reciprocal of 5/2, which is 2/5:
(2/5) * (5/2x) = (2/5) * (70/16)
x = (2/5) * (70/16)
x = 140/80
Simplifying the fraction, we get:
x = 7/4
Therefore, the solution to the equation is x = 7/4.
- 1/4 + x:3 5/9 = 7/16
To solve this equation, we need to isolate the variable x. Let's start by subtracting 1/4 from both sides of the equation:
1/4 + x:3 5/9 - 1/4 = 7/16 - 1/4
x:3 5/9 = 7/16 - 4/16
x:3 5/9 = 3/16
To simplify the equation, let's convert the mixed number to an improper fraction:
3 5/9 = (3*9 + 5)/9 = 32/9
Now, the equation becomes:
x:32/9 = 3/16
To isolate x, we need to multiply both sides of the equation by the reciprocal of 32/9, which is 9/32:
(9/32) * (x:32/9) = (9/32) * (3/16)
x = (9/32) * (3/16)
x = 27/512
Therefore, the solution to the equation is x = 27/512.
Please let me know if you need help with the remaining equations.


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