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5 tháng 9 2017

(6x+1)(2x-5)=12x2-30x+2x-5=12x2-28x-5

(2x+5)2-2x(2x+8)=4x2+20x+25-4x2-16x=4x+25

(3x-5)(2x-1)-(2x+3)(3x+7)+30x=6x2-3x-10x+5=6x2-13x+5

(X-1)2-(x+1)(x-1)=x2-2x+1-x2+1=-2x+2

(3x+2)(9x2-6x+4)-(3+x)(x-3)=27x3+8+9-x2=27x3-x2+17

14 tháng 8 2021

Bài 1

A= (x-2)(2x-1)-2x(x+3)=2x2-x-4x+2-2x2-6x=-11x+2

14 tháng 8 2021

Bài 1:

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

\(A=2x^2-x-4x+2-2x^2-6x\)

\(A=-11x+2\)

b) \(B=\left(3x-2\right)\left(2x+1\right)-\left(6x-1\right)\left(x+2\right)\)

\(B=6x^2+3x-4x-2-6x^2-12x+x+2\)

\(B=-12x\)

c) \(C=6x\left(2x+3\right)-\left(4x-1\right)\left(3x-2\right)\)

\(C=12x^2+18x-12x^2+8x+3x-2\)

\(C=29x-2\)

d) \(D=\left(2x+3\right)\left(5x-2\right)+\left(x+4\right)\left(2x-1\right)-6x\left(2x-3\right)\)

\(D=10x^2-4x+15x-6+2x^2-x+8x-4-12x^2+18x\)

\(D=36x-10\)

18 tháng 12 2022

a: =18x^3y^2-12x^3y^3+6x^2y^2

b: (-3x+2)(5x^2-1/3x+4)

=-12x^3+x^2-12x+10x^2-2/3x+8

=-12x^3+11x^2-38/3x+8

c: =x^2-x-2+3x-x^2

=2x-2

d: =4x^2+12x+9-4x^2+25-(x-1)(x^2+12)

=12x+34-x^3-12x+x^2+12

=-x^3+x^2+46

15 tháng 10 2023

2:

a: \(9x^2-1=\left(3x\right)^2-1=\left(3x-1\right)\left(3x+1\right)\)

b: \(2\left(x-1\right)+x^2-x\)

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

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

c: \(3x^2+14x-5\)

\(=3x^2+15x-x-5\)

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

3: 

a: \(2x\left(x-1\right)-2x^2=4\)

=>\(2x^2-2x-2x^2=4\)

=>-2x=4

=>x=-2

b: \(x\left(x-3\right)-\left(x+2\right)\left(x-1\right)=5\)

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

=>\(x^2-3x-x^2-x+2=5\)

=>-4x=3

=>x=-3/4

c: \(4x^2-25+\left(2x+5\right)^2=0\)

=>\(\left(2x-5\right)\left(2x+5\right)+\left(2x+5\right)^2=0\)

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

=>4x(2x+5)=0

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

29 tháng 11 2022

1: \(=3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)

2: \(=8x^2+12x-10x-15-4\left(2x^2-x+4x-2\right)+10x+7\)

\(=8x^2+12x-8-8x^2+4x-16x+8\)

\(=0\)

3: \(=\left(6x+1-6x+1\right)^2=4\)

5: \(=x^3+3x^2+3x+1+x^3-3x^2+3x-1+x^3-3x\left(x^2-1\right)\)

\(=3x^3+6x-3x^3+3x=9x\)

12 tháng 7 2019

\(a,\left(6x+1\right)\left(x+2\right)-2x\left(3x-5\right)\)

\(=6x^2+12x+x+2-6x^2+10x\)

\(=23x+2\)

12 tháng 7 2019

a) (6x + 1)(x + 2) - 2x(3x - 5)

= 6x2 + 12x + x + 2 - 6x2 + 10x

= (6x2 - 6x2) + (12x + x + 10x) + 2

= 23x + 2

b) (2x - 1)2 - (2x - 3)(2x + 3)

= 4x2 - 4x + 1 - 4x2 + 9

= (4x2 - 4x2) - 4x + (1 + 9)

= -4x + 10

c) (2x - 3)3  - (3x  + 1)(5 - 4x) - 16x2

= 8x3 - 36x2 + 54x - 15x + 12x2 - 5 + 4x - 16x2

= 8x3 - (36x2 - 12x2 + 16x2) + (54x - 15x + 4x) - 5

= 8x3 - 40x2 + 43x - 5

d) (3x + 2) - (x - 5) - x(3x - 13)

= 3x  + 2 - x + 5 - 3x2 + 13x

= (3x - x + 13x) + (2 + 5) - 3x2

= 15x + 7 - 3x2

8 tháng 9 2021

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

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

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

12 tháng 12 2020

Bạn chú ý đăng lẻ câu hỏi! 1/

a/ \(=x^3-2x^5\)

b/\(=5x^2+5-x^3-x\)

c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)

d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)

e/ \(=x^4-x^2+2x^3-2x\)

f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)

12 tháng 12 2020

cảm ơn bạn đã nhắc

 

8 tháng 6 2019

Tìm x:

1. 3x (2x + 3) - (2x + 5).(3x - 2) = 8

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=0 \)

\(\Leftrightarrow-2x+10=0\Leftrightarrow x=5\)

Vậy x = 5

2. 4x (x -1) - 3(x2 - 5) -x2 = (x - 3) - (x + 4)

\(\Leftrightarrow4x^2-4x-3x^2+15-x^2=x-3-x-4\)

\(\Leftrightarrow-4x+15=-7\)

\(\Leftrightarrow-4x=-22\Leftrightarrow x=\frac{11}{2}\)

Vậy x = \(\frac{11}{2}\)

3. 2 (3x -1) (2x +5) - 6 (2x - 1) (x + 2) = -6

\(\Leftrightarrow2\left(6x^2+15x-2x-5\right)-6\left(2x^2+4x-x-2\right)=-6\)

\(\Leftrightarrow12x^2+30x-4x-10-12x^2-24x+6x+12=-6\)

\(\Leftrightarrow8x=-8\Leftrightarrow x=-1\)

Vậy x = -1

4. 3 ( 2x - 1) (3x - 1) - (2x - 3) (9x - 1) - 3 = -3

\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-18x^2+2x+27x-3-3=-3\)

\(\Leftrightarrow18x^2-6x-9x+3-18x^2+2x+27x-6=-3\)

\(\Leftrightarrow14x=0\Leftrightarrow x=0\)

Vậy x = 0

5. (3x - 1) (2x + 7) - ( x + 1) (6x - 5) = (x + 2) - (x - 5)

\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5=7\)

\(\Leftrightarrow18x=9\Leftrightarrow x=\frac{1}{2}\)

Vậy x = \(\frac{1}{2}\)

6. 3xy (x + y) - (x + y) (x2 + y2 + 2xy) + y3 = 27

\(\Leftrightarrow3x^2y+3xy^2-\left(x+y\right)^3+y^3=27\)

\(\Leftrightarrow3x^2y+3xy^2-x^3-y^3-3x^2y-3xy^2+y^3=27\)

\(\Leftrightarrow-x^3=27\)

\(\Leftrightarrow x=-3\)

Vậy x = -3

7. 3x (8x - 4) - 6x (4x - 3) = 30

\(\Leftrightarrow24x^2-12x-24x^2+12x=30\)

\(\Leftrightarrow0=30\) ( vô lý)

Vậy pt vô nghiệm

8. 3x (5 - 2x) + 2x (3x - 5) = 20

\(\Leftrightarrow15x-6x^2+6x^2-10x=20\)

\(\Leftrightarrow5x=20\Leftrightarrow x=4\)

Vậy x = 4