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

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

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: \(x=-2\)

Bài 2: 

a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

=>-13x=26

hay x=-2

b: \(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)

hay \(x\in\left\{1;\dfrac{1}{5}\right\}\)

c: \(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)

hay \(x\in\left\{-5;2\right\}\)

`a)4x(x+1)+(3-2x)(3+2x)=15`

`<=>4x^2+4x+9-4x^2=15`

`<=>4x=6`

`<=>x=3/2`

Vậy `S={3/2}`

`b)3x(x-20012)-x+20012=0`

`<=>3x(x-20012)-(x-20012)=0`

`<=>(x-20012)(3x-1)=0`

`<=>` $\left[\begin{matrix} x=20012\\ x=\dfrac{1}{3}\end{matrix}\right.$

Vậy `S={1/3;20012}`

a) 4x(x + 1) + (3 – 2x)(3 + 2x) = 15

⇔4x² + 4x + (9 – 4x²) = 15

⇔ 4x² + 4x + 9 – 4x² = 15

⇔4x = 15 – 9

⇔x=1,5

b)3x(x – 2001²) – x + 2001² = 0

⇔3x(x – 2001²) – (x – 2001²) = 0

⇔(x – 2001²)(3x – 1) = 0

⇔x – 2001² = 0 hay 3x – 1 = 0

⇔x = 2001² hoặc x = \(\dfrac{1}{2}\)

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 

3,26 + 4/5 =?

làm nhanh lên giúp mình nhé

3,26+4/5=3,26+0,8=3,34

K cho mình cái

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

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

\(\Leftrightarrow-2x-1=12x+27\Leftrightarrow-14x-28=0\Leftrightarrow x=-2\)

\(a,\Leftrightarrow4x^2-20x-4x^2+7x-3=23\\ \Leftrightarrow-13x=-26\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^2+4x+4+4x^2-12x+9=5x^2+35x\\ \Leftrightarrow-43x=-13\\ \Leftrightarrow x=\dfrac{13}{43}\)

a) \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=23\)

\(\Leftrightarrow4x^2-20x-4x^2+7x-3=23\)

\(\Leftrightarrow13x=-26\Leftrightarrow x=-2\)

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

\(\Leftrightarrow x^2+4x+4+4x^2-12x+9=5x^2+35x\)

\(\Leftrightarrow43x=13\Leftrightarrow x=\dfrac{13}{43}\)

a) 3/35 - (3/5 + x) = 2/7

=> 3/5 + x= 3/35- 2/7

=> 3/5 +x = -1/5

=> x = -1/5 -3/5

=> x = -4/5

b) 3/7 +1/7 : x = 3/14

=> 1/7 : x= 3/14 -3/7

=> 1/7 : x = -3/14

=> x = 1/7 : -3/14 

=> x = -2/3

c) (5x-1).(2x-1/3)=0

=> \(\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}5x=0+1=1\\2x=0+\dfrac{1}{3}=\dfrac{1}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{3}:2=\dfrac{1}{6}\end{matrix}\right.\)

Học tốt :D

a)x=-4/5

b)x=-2/3

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

Vậy.........

mik lười mong bn thông cảm

a) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\\ \Rightarrow\dfrac{3}{5}+x=-\dfrac{1}{5}\\ \Rightarrow x=-\dfrac{4}{5}\)

b) \(\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\\ \Rightarrow\dfrac{1}{7}:x=-\dfrac{3}{14}\\ \Rightarrow x=-\dfrac{2}{3}\)

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

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

\(\Leftrightarrow72-20x-36x+84=30x-240-6x-84\)

\(\Leftrightarrow156-56x=24x-324\)

\(\Leftrightarrow-80x+480=0\Leftrightarrow x=-6\)

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

\(\Leftrightarrow15x+25-8x+12=5x+6x-36+1\)

\(\Leftrightarrow7x+37=11x-35\)

\(\Leftrightarrow-4x+72=0\Leftrightarrow x=18\)

c, \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

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

\(\Leftrightarrow-2x-1=12x-5\)

\(\Leftrightarrow-14x+4=0\Leftrightarrow x=\frac{2}{7}\)

d, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)

\(\Leftrightarrow5x-3\left[4x-15x+6\right]=182\)

\(\Leftrightarrow5x-3\left(-11x+6\right)=182\)

\(\Leftrightarrow5x+33x-18-182=0\)

\(\Leftrightarrow38x-200=0\Leftrightarrow x=\frac{100}{19}\)

eeeeeeeeeeeeeeeeeee

e

ê

e

e

ê

e

e

e

e

e

e

eeeEE

ê

e

e

e

a) \(\Rightarrow72-20x-36x+84=30x-240-6x-84\)

\(\Rightarrow80x=480\Rightarrow x=6\)

b) \(\Rightarrow15x+25-8x+12=5x+6x+36+1\)

\(\Rightarrow4x=0\Rightarrow x=0\)

c) \(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)