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\(\Rightarrow\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)
\(\Rightarrow\frac{3}{4}x-2x-\frac{1}{4}x=-6+\frac{1}{4}\)
\(\Rightarrow-\frac{3}{2}x=-\frac{23}{4}\)
\(\Rightarrow x=\frac{23}{4}:\frac{3}{2}=\frac{23}{6}\)
\(\Rightarrow\frac{8}{9}x-\frac{1}{3}x=\frac{2}{3}+\frac{11}{3}\)
\(\Rightarrow\frac{5}{9}x=\frac{13}{3}\)
\(\Rightarrow x=\frac{13}{3}:\frac{5}{9}=\frac{39}{5}\)
\(\Rightarrow\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)
\(\Rightarrow\frac{3}{4}x-2x-\frac{1}{4}x=-6+\frac{1}{4}\)
\(\Rightarrow-\frac{3}{2}x=-\frac{23}{4}\)
\(\Rightarrow x=\frac{23}{4}:\frac{3}{2}=\frac{23}{6}\)
\(\Rightarrow\frac{8}{9}x-\frac{1}{3}x=\frac{2}{3}+\frac{11}{3}\)
\(\Rightarrow\frac{5}{9}x=\frac{13}{3}\)
\(\Rightarrow x=\frac{13}{3}:\frac{5}{9}=\frac{39}{5}\)
a) 1/2X + 2/3X- 2/3 = 1/3
X= 1/3+2/3
X=1
b) bạn vt lại đề
c) 2/3X : 1/5 =6
2/3X=6/5
X=18/10=9/5
d) 4X-8=14
4X=22
X=22/4=11/2
Sửa đề : \(\frac{3}{4}x-\frac{1}{4}=2\left(x-3\right)-\frac{1}{4}x\)
\(\frac{3}{4}x-\frac{1}{4}=2x-6-\frac{1}{4}x\)
\(\frac{3}{4}x-\frac{1}{4}=\frac{7}{4}x+2x\)
\(-x-\frac{1}{4}=2x\)
\(-x-\frac{1}{4}-2x=0\)
\(-3x-\frac{1}{4}=0\)
\(-3x=\frac{1}{4}\Leftrightarrow x=-\frac{1}{12}\)
\(\Leftrightarrow\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)
\(\Leftrightarrow\frac{3}{4}x-2x-\frac{1}{4}x=-6+\frac{1}{4}\)
\(\Leftrightarrow\frac{-3}{2}x=\frac{-23}{4}\)
\(\Leftrightarrow x=\frac{23}{6}\)
Ta có:
3/4x-1/4=2(x-3)+1/4x
<=>3/4x-1/4x=2x-6+1/4
<=>x-2x=-23/4
<=>-x=-23/4
<=>x=23/4.
Vậy x=23/4
1) 2(3x + 5) - 6 = 0
2(3x + 5) = 6 + 0
2(3x + 5) = 6
3x + 5 = 6 : 2
3x + 5 = 3
3x = 5 - 3
3x = 2
x = 2 : 3
x = 2/3
2) 5x + 3(4 + 2x) = 25
11x + 12 = 25
11x = 25 - 12
11x = 13
x = 13 : 11
x = 13/11
3) 3(4x + 1) + 2(x - 1) = 105
14x + 1 = 105
14x = 105 - 1
14x = 104
x = 104 : 14
x = 104/14 = 52/7
4) 30 - [2(x - 3) - 2] = 14
[2(x - 3) - 2] = 30 - 14
[2(x - 3) - 2] = 16
2x - 8 = 16
2x = 16 + 8
2x = 24
x = 24 : 2
x = 12
5) 3(x - 8)(4x + 5) - 8(x - 8) = 0
12x2 - 89x - 56 = 0
x = 8 hoặc -7/12
6) 4x = 8
Vì: 41 = 4
42 = 16
=> x thuộc tập hợp rỗng
4x +\(\dfrac{1}{3}\)= 3(x+\(\dfrac{1}{2}\)) 4x +\(\dfrac{1}{3}\)= 3x+\(\dfrac{3}{2}\) 3x+x+\(\dfrac{1}{3}\)=3x+\(\dfrac{3}{2}\) => x+\(\dfrac{1}{3}\)=\(\dfrac{3}{2}\) x= \(\dfrac{3}{2}-\dfrac{1}{3}\) x= \(\dfrac{7}{6}\)