c: \(\left(2x+3\right)^2+\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\)

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

\(=8x^2+18-4x^2+9=4x^2+27\)

d: \(\left(x-1\right)\cdot\left(x^2+x+1\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\)

\(=\left(x-1\right)\left(x^2+x\cdot1+1^2\right)-\left(2x+3\right)\left[\left(2x\right)^2-2x\cdot3+3^2\right]\)

\(=x^3-1-8x^3-27=-7x^3-28\)

e: \(\left(x+1\right)^3-\left(x-1\right)^3-6x^2\)

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

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

=2

\(=12x^3-10x+18x^2-15\)

\(\left(6x^3-7x^2-x+2\right):\left(2x+1\right)\\ =\left[\left(6x^3-6x^2\right)-\left(x^2-x\right)-\left(2x-2\right)\right]:\left(2x+1\right)\\ =\left[\left(x-1\right)\left(6x^2-x-2\right)\right]:\left(2x+1\right)\\ =\left\{\left(x-1\right)\left[\left(6x^2+3x\right)-\left(4x+2\right)\right]\right\}:\left(2x+1\right)\\ =\left[\left(x-1\right)\left(3x-2\right)\left(2x+1\right)\right]:\left(2x+1\right)\\ =\left(x-1\right)\left(3x-2\right)\)

\(\left(6x^2-5\right)\left(2x+3\right)=2x\left(6x^2-5\right)+3\left(6x^2-5\right)=12x^3-10x+18x^2-15\)

a: (1-2x)^3-(1+2x)^3

\(=1^3-3\cdot1^2\cdot2x+3\cdot1\cdot\left(2x\right)^2-8x^3-8x^3-12x^2-6x-1\)

\(=1-6x+12x^2-8x^3-8x^3-12x^2-6x-1\)

\(=-16x^3-12x\)

b: \(=x^3-6x^2+12x-8-x^3-x^2+8\)

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

c: \(=x^3+8-12x+6x^2-x^3+6x^2+12x\)

\(=12x^2+8\)

Tìm được A = 10 ( x 2 + 1 ) ( x 2 − 1 ) 2

\(6x^2-2x\left(3x+\dfrac{3}{2}\right)=9\)

\(\Rightarrow6x^2-6x^2-3x=9\)

\(\Rightarrow-3x=9\)

\(\Rightarrow x=\dfrac{9}{-3}\)

\(\Rightarrow x=-3\)

\(6x^2-2x\left(3x+\dfrac{3}{2}\right)=9\\ \Leftrightarrow6x^2-6x^2-3x=9\\ \Leftrightarrow3x=9\\ \Leftrightarrow x=3\)

\(6x^2-2x\left(3x+\dfrac{3}{2}\right)=9\\ \Rightarrow6x^2-\left(6x^2+3x\right)=9\\ \Rightarrow6x^2-6x^2-3x=9\\ \Rightarrow-3x=9\\ \Rightarrow x=9:-3\\ \Rightarrow x=-3\)

a,x(2x-1)-(x-1)^2-x^2=0

<=>x(2x-1-x)-(x-1)^2=0

<=>x(x-1)-(x-1)^2=0

<=>(x-x+1)(x-1)=0

<=>x-1=0

<=>x=1

b,(x+2)^3-x^3-6x^2=4

<=>x^3+6x^2+12x+8-x^3-6x^2=4

<=>12x+8=4

<=>x=-1/3

tick mik nha

`a)x(2x-1)-(x-1)^2-x^2=0`

`<=>2x^2-x-x^2+2x-1-x^2=0`

`<=>x-1=0`

`<=>x=1`

Vậy `x=1.`

`b)(x+2)^3-x^3-6x^2=4`

`<=>x^3+6x^2+12x+8-x^3-6x^2=4`

`<=>12x+8=4`

`<=>12x=-4`

`<=>x=-1/3`

Vậy `x=-1/3.`

a: Ta có: \(x\left(2x-1\right)-\left(x-1\right)^2-x^2=0\)

\(\Leftrightarrow2x^2-x-x^2+2x-1-x^2=0\)

\(\Leftrightarrow x=1\)

b: Ta có: \(\left(x+2\right)^3-x^3-6x^2=4\)

\(\Leftrightarrow x^3+6x^2+12x+8-x^3-6x^2=4\)

\(\Leftrightarrow12x=-4\)

hay \(x=-\dfrac{1}{3}\)

câu 1 là (x-y).(6x^2-4y^2+\(\dfrac{1}{2}\)xy) nha 

1. \(\left(x-y\right)\left(6x^2-4y^2+\dfrac{1}{2}xy\right)\)

\(=6x^3-4xy^2+\dfrac{1}{2}x^2y-6x^2y+4y^3-\dfrac{1}{2}xy^2\)

\(=6x^3+4y^3-\dfrac{11}{2}x^2y-\dfrac{9}{2}xy^2\)

===========

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

\(=18x+6x^2-3-x-6x^2-15x\)

\(=2x-3\)

Chúc bạn học tốt!