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8 tháng 4 2020

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

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

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

\(\Leftrightarrow-4x=2x+6\)

\(\Leftrightarrow-6x=6\)

\(\Leftrightarrow x=-1\)
2) \(\left(2x-1\right)^2-\left(2x+1\right)^2=4\left(x-3\right)\)

\(\Leftrightarrow\left(2x-1+2x+1\right)\left(2x-1-2x-1\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow4x\left(-2\right)-4x+12=0\)

\(\Leftrightarrow-12x=-12\)

\(\Leftrightarrow x=1\)

3)\(\left(2x+3\right)^2-\left(2x+3\right)\left(2x-4\right)+\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(2x+3\right)\left(2x+3-2x+4\right)+\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow7\left(2x+3\right)+x^2-4x+4=0\)

\(\Leftrightarrow x^2+10x+25=0\)

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

\(\Leftrightarrow x=-5\)

4) \(8x^3-\left(x+1\right)^3=3x-3\)

\(\Leftrightarrow8x^3-\left(x^3+3x+3x^2+1\right)-3x+3=0\)

\(\Leftrightarrow7x^3-3x^2-6x+2=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x^2+4x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-2+3\sqrt{2}}{7}\\x=\frac{-2-3\sqrt{2}}{7}\end{matrix}\right.\)

5)\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)

\(\Leftrightarrow\left(3x\right)^3-2^3-\left(\left(3x\right)^3-1^3\right)=x-4\)

\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)=x-4\)

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

\(\Leftrightarrow x=-3\)

21 tháng 7 2021

a/ 2x\(^{^{ }3}\)-3\(^{^{ }3}\)-2x\(^3\)-1\(^{^{ }3}\)=-28

b/x\(^{^{ }3}\)+2\(^{^{ }3}\)-x\(^3\)+2=10

c/3x\(^3\)+5\(^3\)-3x(3x\(^2\)-1)=3x\(^3\)+5\(^3\)-3x\(^3\)+3x=125+3x

d/ x\(^6\)-(x\(^3\)+1)(x\(^2\)-x+1)= x\(^6\)-(x\(^6\)-x\(^4\)+x\(^3\)+x\(^2\)-x+1)=x\(^4\)-x\(^3\)-x\(^2\)+x-1

31 tháng 10 2021
(3x-2)(2x-4)=1-12x²
1 tháng 7 2016

1.\(x^2-2x-4y^2-4y=\left(x+2y\right)\left(x-2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)

2.\(x^4+2x^3-4x-4=\left(x^2+2\right)\left(x^2-2\right)+2x\left(x^2-2\right)=\left(x^2-2\right)\left(x^2+2x-2\right)\)

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

4.\(x^3-4x^2-9x+36=x^2\left(x-4\right)-9\left(x-4\right)=\left(x-3\right)\left(x+3\right)\left(x-4\right)\)

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

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

7.\(\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)+\left(x-5\right)\left(2x+1\right)=\left(x-5\right)\left(x-5+x+5+2x+1\right)\)\(=\left(x-5\right)\left(4x+1\right)\)

8.\(\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)=\left(3x-2\right)\left(3x-6\right)=3\left(3x-2\right)\left(x-2\right)\)

a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

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

3 tháng 8 2020

Bài 1:

a) \(4x\left(3x-1\right)-2\left(3x+1\right)-\left(x+3\right)\)

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

\(=12x^2-11x-5\)

b) \(=\left(-2x^2-1xy+2y^2\right)\left(-1x^2y\right)\)

\(=\left[\left(-1x^2y\right)\left(-2x^2\right)\right]-\left[\left(-1x^2y\right).1xy\right]+\left[\left(-1x^2y\right).2y^2\right]\)

\(=\left(2x^4y\right)-\left(-1x^3y^2\right)+\left(-2x^2y^3\right)\)

\(=2x^4y+1x^3y^2-2x^2y^3\)

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

\(=\left(4x.3x^2\right)-\left(4x.x\right)-\left[\left(2x\right)^2+2.2x.3+3^2\right]\left(6x^2-3x+1\right)\)

\(=12x^3-4x^2-\left(4x^2+12x+9\right)\left(6x^2-3x+1\right)\)

\(=12x^3-4x^2-\left[4x^2\left(6x^2-3x+1\right)+12x\left(6x^2-3x+1\right)+9\left(6x^2-3x+1\right)\right]\)

\(=12x^3-4x^2-\left[\left(24x^4-12x^3+4x^2\right)+\left(72x^3-36x^2+12x\right)+\left(36x^2-27x+9\right)\right]\)

\(=12x^3-4x^2-24x^4+12x^3-4x^2-72x^3+36x^2-12x-36x^2+27x-9\)

\(=-48x^3-8x^2-24x^4+15x-9\)

3 tháng 8 2020

Bài 2 ạ