* Trả lời:

\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)

\(\Leftrightarrow-3+6x-4-12x=-5x+5\)

\(\Leftrightarrow6x-12x+5x=3+4+5\)

\(\Leftrightarrow x=12\)

\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)

\(\Leftrightarrow6x-15-6+24x=-3x+7\)

\(\Leftrightarrow6x+24x+3x=15+6+7\)

\(\Leftrightarrow33x=28\)

\(\Leftrightarrow x=\dfrac{28}{33}\)

\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)

\(\Leftrightarrow1-3x-6x+12=-4x-5\)

\(\Leftrightarrow-3x-6x+4x=-1-12-5\)

\(\Leftrightarrow-5x=-18\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)

\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)

\(\Leftrightarrow-x-5x=-7\)

\(\Leftrightarrow-6x=-7\)

\(\Leftrightarrow x=\dfrac{7}{6}\)

\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)

\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)

\(\Leftrightarrow-15x+3x=4\)

\(\Leftrightarrow-12x=4\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

w

1: \(A=\left(-x+5\right)\left(x-2\right)+\left(x-7\right)\left(x+7\right)\)

\(=-x^2+2x+5x-10+x^2-49=7x-59\)

\(B=\left(3x+1\right)^2-\left(3x-2\right)\left(3x+2\right)\)

\(=9x^2+6x+1-9x^2+4=6x+5\)

=>7x-59=6x+5

=>x=64

2: \(A=\left(5x-1\right)\left(x+1\right)-2\left(x-3\right)^2\)

\(=5x^2+5x-x-1-2x^2+12x-9\)

\(=3x^2+16x-10\)

\(B=\left(x+2\right)\left(3x-1\right)-\left(x+4\right)^2+x^2-x\)

\(=3x^2-x+6x-2-x^2-8x-16+x^2-x\)

\(=3x^2-4x-18\)

=>16x-10=-4x-18

=>20x=-8

hay x=-2/5

dấu [] là giá trị tuyệt đối nha

a) \(=x^2-49-x^2\) \(=-49\)

b) \(=25x^2-1-25x^2-1\) \(=-2\)

c) \(=16x^2-1-16x^2+8x-1\) \(=8x-2\)

d) \(=9x^2-30x+25-9x^2+25\) \(=50-30x\)

vt rõ đề câu a đi lm 1 thể

\(\dfrac{2}{7}\)\(x\) - \(\dfrac{1}{3}\) = \(\dfrac{3}{5}\)\(x\) - 1

\(\dfrac{2}{7}\)\(x\) - \(\dfrac{3}{5}\)\(x\) =  - 1 + \(\dfrac{1}{3}\)

 (\(\dfrac{2}{7}\) - \(\dfrac{3}{5}\))\(x\) =  - \(\dfrac{2}{3}\)

  - \(\dfrac{11}{35}\)\(x\)     = - \(\dfrac{2}{3}\)

         \(x\)     = - \(\dfrac{2}{3}\) : (- \(\dfrac{11}{35}\))

         \(x\)     = \(\dfrac{70}{33}\)

Vậy \(x=\dfrac{70}{33}\) 

|3\(x\) + 7| - 3\(x\) = 7

|3\(x\) + 7| = 7  +3\(x\) Vì 7 + 3\(x\) > 0  nên 3\(x\) >  -7 ⇒ \(x\) >  -\(\dfrac{7}{3}\)

Với \(x\) >  - \(\dfrac{7}{3}\) ta có:

3\(x\) + 7 = 7 + 3\(x\) 

        7  =7 (luôn đúng)

Vậy \(x\) \(\ge\) - \(\dfrac{7}{3}\) 

\(\dfrac{3x^6-4x^3}{x^3}-\dfrac{\left(3x+1\right)^2}{3x+1}-\dfrac{3x^7}{x^5}=0\)

\(\Leftrightarrow3x^3-4-3x-1-3x^2=0\)

\(\Leftrightarrow3x^3-3x^2-3x-5=0\)

\(\Leftrightarrow x\simeq1,9506\)

Bài 1:

a) Ta có: \(82-7\left(3x-4\right)=47\)

\(\Leftrightarrow82-21x+28-47=0\)

\(\Leftrightarrow-21x+63=0\)

\(\Leftrightarrow-21x=-63\)

hay x=3(nhận)

Vậy: x=3

b) Ta có: \(97+4\left(5x-7\right)=129\)

\(\Leftrightarrow97+20x-28-129=0\)

\(\Leftrightarrow20x-60=0\)

\(\Leftrightarrow20x=60\)

hay x=3(nhận)

Vậy: x=3

c) Ta có: \(\left(7x-13\right)\cdot27-12=15\)

\(\Leftrightarrow189x-351-12-15=0\)

\(\Leftrightarrow189x-378=0\)

\(\Leftrightarrow189x=378\)

hay x=2(nhận)

Vậy: x=2

d) Ta có: \(\left(2x+3\right)\cdot13+23=140\)

\(\Leftrightarrow26x+39+23-140=0\)

\(\Leftrightarrow26x-78=0\)

\(\Leftrightarrow26x=78\)

hay x=3(nhận)

Vậy: x=3

đ) Ta có: \(52x+8x-5x=70\)

\(\Leftrightarrow55x=70\)

\(\Leftrightarrow x=\frac{70}{55}\)(loại)

Vậy: x∈∅

e) Ta có: \(19x-3x-x=60\)

\(\Leftrightarrow15x=60\)

hay x=4(nhận)

Vậy: x=4

g) Ta có: \(7\left(3x+1\right)-5\left(3x+1\right)=74\)

\(\Leftrightarrow2\left(3x+1\right)=74\)

\(\Leftrightarrow3x+1=37\)

\(\Leftrightarrow3x=36\)

hay x=12(nhận)

Vậy: x=12

h) Ta có: \(5\left(3x-1\right)+7\left(3x-1\right)=96\)

\(\Leftrightarrow12\left(3x-1\right)=96\)

\(\Leftrightarrow3x-1=8\)

\(\Leftrightarrow3x=9\)

hay x=3(nhận)

Vậy: x=3