tìm x
a, 3,5x -2=1/3x+2/7
b, 2.(x-1/40-3(x-1)=2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
TN
0
15 tháng 3 2021
Bài 1:
c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)
Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)
\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)
Suy ra: \(-12x-3=8x-2-6x-8\)
\(\Leftrightarrow-12x-3-2x+10=0\)
\(\Leftrightarrow-14x+7=0\)
\(\Leftrightarrow-14x=-7\)
\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
a)
\(3,5x-2=\dfrac{1}{3}x+\dfrac{2}{7}\\ \Leftrightarrow\dfrac{7}{2}x-\dfrac{1}{3}x=\dfrac{2}{7}+2\\ \Leftrightarrow x\left(\dfrac{7}{2}-\dfrac{1}{3}\right)=\dfrac{16}{7}\\ \Leftrightarrow\dfrac{19}{6}\cdot x=\dfrac{16}{7}\\ \Leftrightarrow x=\dfrac{16}{7}:\dfrac{19}{6}\\ \Leftrightarrow x=\dfrac{96}{133}\)
b)
\(2\left[\dfrac{x-1}{40}-3\left(x-1\right)\right]=2\\ \Leftrightarrow\dfrac{x-1}{40}-3\left(x-1\right)=1\\ \Leftrightarrow\dfrac{x-1-120\left(x-1\right)}{40}=1\\ \Leftrightarrow x-1-120x+120=40\\ \Leftrightarrow-119x+119=40\\ \Leftrightarrow-119x=40-119=-79\\ \Leftrightarrow x=\dfrac{79}{119}\)
a, \(3,5x-2=\dfrac{1}{3}x+\dfrac{2}{7}\Leftrightarrow\dfrac{19}{6}x=\dfrac{2}{7}+2=\dfrac{16}{7}\Leftrightarrow x=\dfrac{16}{7}:\dfrac{19}{6}=\dfrac{96}{133}\)
b, \(2\left[x-\dfrac{1}{40}-3\left(x-1\right)\right]=2\)
\(\Leftrightarrow x-\dfrac{1}{40}-3x+3=1\Leftrightarrow-2x=1+\dfrac{1}{40}-3=-\dfrac{79}{40}\Leftrightarrow x=\dfrac{79}{80}\)