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

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

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

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

\(\Leftrightarrow-4x-4=x-4\Leftrightarrow-4x-x=0\Leftrightarrow x=0\)

b, \(\left(x-2\right)\left(x+3\right)-\left(x+4\right)\left(x-7\right)=5-x\)

\(\Leftrightarrow x^2+x-6-x^2-3x+28=5-x\Leftrightarrow-2x+22=5-x\Leftrightarrow x=17\)

c,  thiếu đề 

d, \(3\left(x-7\right)\left(x+7\right)-\left(x-1\right)\left(3x+2\right)=13\)

\(\Leftrightarrow3x^2-147-3x^2+x+2=13\Leftrightarrow x=11+147=158\)

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

\(\Leftrightarrow-5x=0\Leftrightarrow x=0\)

b.\(x^2+x-6-\left(x^2-3x-28\right)=5-x\)

\(\Leftrightarrow5x=-17\Leftrightarrow x=-\frac{17}{5}\)

c.\(5\left(x^2-10x+21\right)-\left(5x^2-9x-2\right)=0\)

\(\Leftrightarrow-41x+107=0\Leftrightarrow x=\frac{107}{41}\)

d.\(3\left(x^2-49\right)-\left(3x^2-x-2\right)=13\Leftrightarrow x=158\)

sửa dòng 2 phần b ;-; 

\(x^2+x-6-x^2+3x+28=5-x\)

\(\Leftrightarrow4x+22=5-x\Leftrightarrow5x=-17\Leftrightarrow x=-\frac{17}{5}\)

\(\dfrac{x+1}{3}=\dfrac{1-3x}{-5}\\ \Rightarrow-5\left(x+1\right)=3\left(1-3x\right)\\ \Rightarrow-5x-5=3-9x\\ \Leftrightarrow-5x+9x=3+5\\ \Rightarrow4x=8\\ \Rightarrow x=\dfrac{8}{4}=2\)

bạn ơi cho mình hỏi là phép tính thứ ba  của bạn là ⇒−5x−5=3−9x thì tại sao bạn tính được như vậy ? Phiền bạn giúp mình nhé 

1/ \(7x-5=13-5x\)

\(\Leftrightarrow12x=18\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy: \(S=\left\{\dfrac{3}{2}\right\}\)

==========

2/ \(19+3x=5-18x\)

\(\Leftrightarrow21x=-14\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

Vậy: \(S=\left\{-\dfrac{2}{3}\right\}\)

==========

3/ \(x^2+2x-4=-12+3x+x^2\)

\(\Leftrightarrow-x=-8\)

\(\Leftrightarrow x=8\)

Vậy: \(S=\left\{8\right\}\)

===========

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

\(\Leftrightarrow-x-5=3x-15\)

\(\Leftrightarrow-4x=-10\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

Vậy: \(S=\left\{\dfrac{5}{2}\right\}\)

==========

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

\(\Leftrightarrow3x+12=-x+4\)

\(\Leftrightarrow4x=-8\)

\(\Leftrightarrow x=-2\)

Vậy: \(S=\left\{-2\right\}\)

[----------]

1. \(7x-5=13-5x\) \(\Leftrightarrow12x=18\Leftrightarrow x=\dfrac{3}{2}\)

2. \(19+3x=5-18x\Leftrightarrow21x=-14\Leftrightarrow x=-\dfrac{2}{3}\)

3. \(x^2+2x-4=-12+3x+x^2\Leftrightarrow-x=-8\Leftrightarrow x=8\)

4. \(-\left(x+5\right)=3\left(x-5\right)\Leftrightarrow-x-5=3x-15\Leftrightarrow4x=10\Leftrightarrow x=\dfrac{5}{2}\)

5. \(3\left(x+4\right)=-x+4\Leftrightarrow3x+12=-x+4\Leftrightarrow4x=-8\Leftrightarrow x=-2\)

1) Ta có: \(7x-5=13-5x\)

\(\Leftrightarrow12x=18\)

hay \(x=\dfrac{3}{2}\)

2) Ta có: \(19+3x=5-18x\)

\(\Leftrightarrow21x=-14\)

hay \(x=-\dfrac{2}{3}\)

3) Ta có: \(x^2+2x-4=x^2+3x-12\)

\(\Leftrightarrow3x-12=2x-4\)

hay x=8

4) Ta có: \(-\left(x+5\right)=3\left(x-5\right)\)

\(\Leftrightarrow-x-5-3x+15=0\)

\(\Leftrightarrow-4x=-10\)

hay \(x=\dfrac{5}{2}\)

b. `|x + 1| + |2x - 3| = |3x - 2|`

Ta có: \(\left|x+1\right|+\left|2x-3\right|\ge\left|x+1+2x-3\right|=\left|3x-2\right|\)

\(\Leftrightarrow\left|3x-2\right|=\left|3x-2\right|\) (luôn đúng với mọi x)

Vậy phương trình có vô số nghiệm.