a) \(PT\Leftrightarrow3x-2x=2-3\Leftrightarrow x=-1\)

Vậy: \(S=\left\{-1\right\}\)

b) \(PT\Leftrightarrow-2x+3x=-7+22\Leftrightarrow x=15\)

Vậy: \(S=\left\{15\right\}\)

c) \(PT\Leftrightarrow8x-5x=3+12\Leftrightarrow3x=15\Leftrightarrow x=5\)

Vậy: \(S=\left\{5\right\}\)

d) \(PT\Leftrightarrow x+4x-2x=12+25-1\Leftrightarrow3x=36\Leftrightarrow x=12\)

Vậy: \(S=\left\{12\right\}\)

e) \(PT\Leftrightarrow x+2x+3x-3x=19+5\Leftrightarrow3x=24\Leftrightarrow x=8\)

Vậy: \(S=\left\{8\right\}\)

a)3x-2=2x-3

=>x=-1

b)7-2x=22-3x

=>x=15

c)8x-3=5x+12

=>3x=15

=>x=5

d)x-12+4x=25+2x-1

=>3x=12

=>x=4

e)x+2x+3x-19=3x+5

=>3x=24

=>x=8

a)3x-2=2x-3

=>x=-1

b)7-2x=22-3x

=>x=15

c)8x-3=5x+12

=>3x=15

=>x=5

d)x-12+4x=25+2x-1

=>3x=36

=>x=12

e)x+2x+3x-19=3x+5

=>3x=24

=>x=8

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 \(x=\dfrac{5}{2}\)

a) x−12+4x=25+2x−1x−12+4x=25+2x−1

⇔5x – 12 = 2x + 24

⇔5x – 2x = 24 + 12

⇔3x = 36

⇔x = 12

Vậy phương trình có nghiệm x = 12.

b) x+2x+3x−19=3x+5x+2x+3x−19=3x+5

⇔6x – 19 = 5x +3x

⇔3x= 24

⇔x= 8

Vậy phương trình có nghiệm x = 8.

a) x−12+4x=25+2x−1x−12+4x=25+2x−1

⇔5x – 12 = 2x + 24

⇔5x – 2x = 24 + 12

⇔3x = 36

⇔x = 12

Vậy x=12 là nghiệm của phương trình

b) x+2x+3x−19=3x+5x+2x+3x−19=3x+5

⇔6x – 19 = 5x +3x

⇔3x= 24

⇔x= 8

Vậy x=8 là nghiệm của phương trình

a)x−12+4x=25+2x−1x−12+4x=25+2x−1

⇔5x – 12 = 2x + 24

⇔5x – 2x = 24 + 12

⇔3x = 36

⇔x = 12

Vậy phương trình có nghiệm x = 12.

b) x+2x+3x−19=3x+5x+2x+3x−19=3x+5

⇔6x – 19 = 5x +3x

⇔3x= 24

⇔x= 8

Vậy phương trình có nghiệm x = 8.

Lời giải:
PT \(\Leftrightarrow (\frac{x+1}{11}-1)-(\frac{2x-5}{15}-1)=(\frac{3x-47}{17}+1)-(\frac{4x-59}{19}+1)\)

\(\Leftrightarrow \frac{x-10}{11}-\frac{2(x-10)}{15}=\frac{3(x-10)}{17}-\frac{4(x-10)}{19}\)

\(\Leftrightarrow (x-10)(\frac{1}{11}+\frac{4}{19}-\frac{2}{15}-\frac{3}{17})=0\)

\(\Leftrightarrow x-10=0\Leftrightarrow x=10\)

\(a,2x\left(x-5\right)+4\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\2x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{5;-2\right\}\)

\(b,3x-15=2x\left(x-5\right)\\ \Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(-2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\-2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{5;\dfrac{3}{2}\right\}\)

\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-1\\2x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{1}{2};3\right\}\)

Câu d xem lại đề

`3x+7=0`

`<=>3x=-7`

`<=>x=-7/3`

Vậy `S={-7/3}`

______________________

`2x(x-2)+2x(5-3x)=0`

`<=>2x(x-2+5-3x)=0`

`<=>2x(3-2x)=0`

`@TH1:2x=0<=>x=0`

`@TH2: 3-2x=0<=>2x=3<=>x=3/2`

Vậy `S={0;3/2}`

3x+7=0

\(\Leftrightarrow3x=-7\Leftrightarrow x=-\dfrac{7}{3}\)

2x(x-2)+2x(5-3x)=0

\(\Leftrightarrow2x\left(x-2+5-3x\right)=0\)

\(\Leftrightarrow2x\left(-2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\-2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-3}{-2}=\dfrac{3}{2}\end{matrix}\right.\)

3x + 7 = 0

\(\Leftrightarrow\) 3x = -7

\(\Leftrightarrow\) x = -7/3

2x(x - 2) + 2x(5 - 3x) = 0

\(\Leftrightarrow\) 2x(x - 2 + 5 - 3x) = 0

\(\Leftrightarrow\) 2x(-2x + 3) = 0

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\-2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)