a)\(60x^2+35x-60x^2+15x=100\)

  35x+15x=100

  50x=100   =>x=2

b)\(10x^2-35x+16x-10x^2=5\)

  -35x+16x=5

   -19x=5    =>x=-5/19

a)

  35x+15x=100

  50x=100   

=>x=2

b)

  -35x+16x=5

   -19x=5   

=>x=-5/19

Câu 1:

\(\frac{2x-1}{3}=\frac{3x+1}{4}\)

\(\frac{8x-4}{12}=\frac{9x+3}{12}\)

\(\Rightarrow8x-4=9x+3\)

\(8x-9x=3+4\)

\(-x=7\Rightarrow x=-7\)

Câu 2:

\(\frac{3x}{5x+4}=\frac{12}{28}\)

\(\Rightarrow\left(3x\right).28=12.\left(5x+4\right)\)

\(84x=60x+48\)

\(84x-60x=48\)

\(24x=48\)

\(\Leftrightarrow x=2\)

Câu 3 :

\(\frac{x}{2}=\frac{8}{x}\)

\(\Rightarrow x.x=8.2\)

\(x^2=16\)

\(x^2=\left(\pm4\right)^2\)

\(\Rightarrow x\left\{-4;4\right\}\)

học tốt !!!

Câu 1: \(\frac{2x-1}{3}=\frac{3x+1}{4}\)

\(\Leftrightarrow2\left(2x-1\right)=3\left(3x+1\right)\)

\(\Leftrightarrow4x-2=9x+3\)

<=> -9x+4x=3+2

<=> -5x=5

<=> x=-1

Câu 3: \(\frac{x}{2}=\frac{8}{x}\left(x\ne0\right)\)

<=> x²=16

<=> \(\orbr{\begin{cases}x=4\\x=-4\end{cases}\left(tmđk\right)}\)

Câu 2: Bạn cũng nhân chéo lên nhưng nhớ đi điều kiện: \(x\ne\frac{-4}{5}\)

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

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

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: \(x=-2\)

\(5x-\frac{1}{3}=3x+\frac{2}{7}=8-\frac{5x}{2}\)

\(\Leftrightarrow5x-3x=\frac{2}{7}+\frac{1}{3}\)

\(\Leftrightarrow2x=\frac{13}{21}\)

\(\Leftrightarrow x=\frac{13}{42}\)

Thử lại:

\(5x-\frac{1}{3}=5\cdot\frac{13}{42}-\frac{1}{3}=\frac{17}{14}\)

\(3x+\frac{2}{7}=3\cdot\frac{13}{42}+\frac{2}{7}=\frac{17}{14}\)

\(8-\frac{5x}{2}=8-5\cdot\frac{13}{42}\div2=\frac{607}{84}\)( vô lý)

Vậy không có giá trị nào của x thoả mãn

a)x+15=-7

x=(-7)-15

x=-22

b)\(\left(5x-2\right).8^3=8^4\)

(5x-2).512=4096

(5x-2)=4096:512

(5x-2)=8

5x=8+2

5x=10

x=10:5

x=2

c)|x-1|=2

=>x-1=2      hoac   x-1=-2

x=2+1                    x=-2+1

x=3                          x=-1

a/ => x = (-7) - 15 = -22

Vậy x = -22

b/ (5x - 2) . 8³ = 8⁴

=> 5x - 2 = 8

=> 5x = 10

=> x = 2

Vậy x = 2

c/ => x - 1 = 2 => x = 3

hoặc x - 1 = -2 => x = -1

Vậy x = 3 ; x = -1

Mấy bài này căng vậy?

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

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

<=> -80x = -480

<=> x = 6

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

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

<=> 15x + 25 - 8x + 12 - 5x - 6x - 36 - 1 = 0

<=> -4x = 0

<=> x = 0

c) 2(5x-8)-3(4x-5)=4(3x-4)+11

= 10x - 16 - 12x + 15 = 12x - 16 + 11

= -14x = -4

= x =\(\frac{2}{7}\)

d) 5x-3{4x-2[4x-3(5x-2)]}=182

= 5x - 3 . [4x - 2(4x - 15x + 6)]

= 5x - 3 . (4x - 8x + 30x - 12)

= 5x - 12x + 24x - 90x + 36

= -73x + 36 = 182

=> -73x = 182 - 36 = 146

=> x = 146 : (-73) = -2

~Hok tốt~

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

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

<=> -80x = -480

<=> x = 6

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

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

<=> 15x + 25 - 8x + 12 - 5x - 6x - 36 - 1 = 0

<=> -4x = 0

<=> x = 0

c) 2(5x-8)-3(4x-5)=4(3x-4)+11

= 10x - 16 - 12x + 15 = 12x - 16 + 11

= -14x = -4

= x = 2/7

d) 5x-3{4x-2[4x-3(5x-2)]}=182

= 5x - 3 . [4x - 2(4x - 15x + 6)]

= 5x - 3 . (4x - 8x + 30x - 12)

= 5x - 12x + 24x - 90x + 36

= -73x + 36 = 182

=> -73x = 182 - 36 = 146

=> x = 146 : (-73) = -2