Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu

a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14) 

=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84) 

=> 156 -  56x = 24x - 324 

=>  24x + 56x = 324 + 156 

=> 80x = 480 

=> x = 480 : 80 =  6 

Vậy x = 6 

b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1 

=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1 

=> (15x - 8x) + (25 + 12) = 11x + 37 

=> 7x + 37 = 11x + 37 

=> 11x - 7x = 0 

=>  x = 0 

Bài 1:

a) Ta có: 7x+12=0

\(\Leftrightarrow7x=-12\)

hay \(x=-\frac{12}{7}\)

Vậy: \(x=-\frac{12}{7}\)

b) Ta có: 5x-2=0

\(\Leftrightarrow5x=2\)

hay \(x=\frac{2}{5}\)

Vậy: \(x=\frac{2}{5}\)

c) Ta có: 12-6x=0

\(\Leftrightarrow6x=12\)

hay x=2

Vậy: x=2

d) Ta có: -2x+14=0

⇔-2x=-14

hay x=7

Vậy: x=7

Bài 2:

a) Ta có: 3x+1=7x-11

⇔3x+1-7x+11=0

⇔-4x+12=0

⇔-4x=-12

hay x=3

Vậy: x=3

b) Ta có: 2x+x+12=0

⇔3x+12=0

⇔3x=-12

hay x=-4

Vậy: x=-4

c) Ta có: x-5=3-x

⇔x-5-3+x=0

⇔2x-8=0

⇔2x=8

hay x=4

Vậy: x=4

d) Ta có: 7-3x=9-x

⇔7-3x-9+x=0

⇔-2x-2=0

⇔-2x=2

hay x=-1

Vậy: x=-1

e) Ta có: 5-3x=6x+7

⇔5-3x-6x-7=0

⇔-9x-2=0

⇔-9x=2

hay \(x=\frac{-2}{9}\)

Vậy: \(x=\frac{-2}{9}\)

f) Ta có: 11-2x=x-1

⇔11-2x-x+1=0

⇔12-3x=0

⇔3x=12

hay x=4

Vậy: x=4

g) Ta có: 15-8x=9-5

⇔15-8x=4

⇔8x=11

hay \(x=\frac{11}{8}\)

Vậy: \(x=\frac{11}{8}\)

Bài 3:

a) Ta có: 0,25x+1,5=0

⇔0,25x=-1,5

hay x=-6

Vậy: x=-6

b) Ta có: 6,36-5,2x=0

⇔5,2x=6,36

hay \(x=\frac{159}{130}\)

Vậy: \(x=\frac{159}{130}\)

15-8x=9-5x 

<=> 5x - 8x = 9 - 15 

<=> -3x = -6

<=> x = 2 

nhấn vào đúng 0 sẽ có câu trả lời

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

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

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) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)

\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)

b) \(\dfrac{39}{7}:x=13\)

\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)

c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)

\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)

\(\dfrac{14}{5}x=34+50=84\)

\(x=\dfrac{84}{\dfrac{14}{5}}=30\)

d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)

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

e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)

\(\dfrac{1}{6}x=\dfrac{5}{12}\)

\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)

g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)

\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)

\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)

\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)

\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)

h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)

\(x=1\)

Mỏi tay woa bn làm nốt nha!!

a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)

=>-38x=7

hay x=-7/38

b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)

=>1/2x=0

hay x=0

c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)

=>-29x=15

hay x=-15/29

d: \(\Leftrightarrow x^2+2x-x-3=5\)

\(\Leftrightarrow x^2+x-8=0\)

\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)

e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)

\(\Leftrightarrow-25x^2=4\)

\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)

a) \(3x-2=2x-3\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

b) \(3-4y+24+6y=y+27+3y\)

\(\Leftrightarrow-2y=0\Leftrightarrow y=0\)

c) \(7-2x=22-3x\)

\(\Leftrightarrow x-15=0\)

\(\Leftrightarrow x=15\)

d) \(8x-3=5x+12\)

\(\Leftrightarrow3x-15=0\Leftrightarrow x=5\)

e) \(x-12+4x=25+2x-1\)

\(\Leftrightarrow3x-36=0\)

\(\Leftrightarrow x=12\)

f) \(x+2x+3x-19=3x+5\)

\(\Leftrightarrow3x-24=0\)

\(\Leftrightarrow x=8\)

\(A,5x-25=0\)

\(\Leftrightarrow5x-5^2=0\)

\(\Leftrightarrow5\left(x-1\right)=0\)

\(\Leftrightarrow x-1=0\)

\(\Rightarrow x=1\)

Chúc bạn học tốt !

* 4x - 1 = 3x - 2

⇔ 4x - 3x = -2 + 1

⇔ x = -1

Vậy tập nghiệm của pt là S = {-1}

* \(\frac{3}{4}-3x=0\)

⇔ \(\frac{3}{4}-\frac{3x.4}{4}=0\)

⇒ 3 - 12x = 0

⇔ 12x = 3

⇔ x = \(\frac{3}{12}=\frac{1}{4}\)

Vậy tập nghiệm của pt là S = \(\left\{\frac{1}{4}\right\}\)

* 3x - 2 = 2x + 3

⇔ 3x - 2x = 3 + 2

⇔ x = 5

Vậy tập nghiệm của pt là S = {5}

* 2(x - 3) = 5(x + 4)

⇔ 2x - 6 = 5x + 20

⇔ 2x - 5x = 20 + 6

⇔ -3x = 26

⇔ x = \(\frac{-26}{3}\)

Vậy tập nghiệm của pt là S = \(\left\{\frac{-26}{3}\right\}\)