\(5x+1=\dfrac{6}{7}\)
\(5x=\dfrac{6}{7}-1\)
\(5x=\dfrac{-1}{7}\)
\(x=\dfrac{-1}{7}:5=\dfrac{-1}{7}\times\dfrac{1}{5}\)
\(x=\dfrac{-1}{35}\)
5x + 1 = 6/7
=> 5x = 6/7 - 1
=> 5x = -1/7
=> x = -1/35
Vậy...
Tìm x biết: (5x-1)/3=(7y-6)/5=5x+7y-7/4x.
cho đa thức sau: \(5x^7-7x^6+5x^5-4x^4+7x^6-3x^2+1-5x^7-3x^5\)
tìm bậc của nó
Tim x ; y biet: (5x-1)/3 = (7y-6)/5 = (5x-7y-7)/4x
Tìm x biết :
a, 4.(18 - 5x) - 12.(3x - 7) = 15.(2x - 16) - 6(x + 14)
b, 5.(3x + 5) - 4.(2x - 3) = 5x + 3.(2x + 12) + 1
c, 2.(5x - 8) - 3.(4x - 5) = 4.(3x - 4) + 11
d, (3x + 2)(2x + 9) - (x + 2)(6x + 1) = (x + 1) - (x - 6)
e, (8x - 3)(3x + 2) - (4x + 7)(x + 4)= (2x + 1)(5x - 1) - 33
\(\sqrt{\left(x+1\right)^2}=6;\sqrt{\left(5x+1\right)}^2=\dfrac{6}{7}\)
1/7+(2/3x-1)=(6/5x+1/2)-2/3
x.(1/4+1/5)-(1/7+1/8)=0
(5x-1)(2x-1/3)=0
-15/12x+3/7=6/5x-1/2
5x+5x+2=650
-15/12X+3/7=6/5x-1/2
1/2x+7/6=9/5x-4/3
