1) `2x(3x-1)-(2x+1)(x-3)`

`=6x^2-2x-2x^2+6x-x+3`

`=4x^2+3x+3`

2) `3(x^2-3x)-(4x+2)(x-1)`

`=3x^2-9x-4x^2+4x-2x+2`

`=-x^2-7x+2`

3) `3x(x-5)-(x-2)^2-(2x+3)(2x-3)`

`=3x^2-15x-(x^2-4x+4)-(4x^2-9)`

`=3x^2-15x-x^2+4x-4-4x^2+9`

`=-2x^2-11x+5`

4) `(2x-3)^2+(2x-1)(x+4)`

`=4x^2-12x+9+2x^2+8x-x-4`

`=6x^2-5x+5`

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

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

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

\(D=-2x-8\)

_______________________

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

\(E=\left(x^2-8x+16\right)-\left(x^2+2x\right)-2x+3\)

\(E=\left(x^2-x^2\right)-\left(8x+2x+2x\right)+\left(16+3\right)\)

\(E=-12x+19\)

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

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

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

\(D=4x^2+4x+1-4x^2+12x-9+6x\)

\(D=22x-8\)

___________

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

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

\(E=x^2-8x+16-x^2-2x-2x+3\)

\(E=-12x+19\)

a) Rút gọn biểu thức D = (2x + 1)2 - (2x - 3)2 + 6x:

Bắt đầu bằng việc mở ngoặc:
D = (4x^2 + 4x + 1) - (4x^2 - 12x + 9) + 6x

Tiếp theo, kết hợp các thành phần tương tự:
D = 4x^2 + 4x + 1 - 4x^2 + 12x - 9 + 6x

Tiếp tục đơn giản hóa:
D = 4x^2 - 4x^2 + 4x + 12x + 6x + 1 - 9

Kết quả cuối cùng:
D = 22x - 8

b) Rút gọn biểu thức E = (x - 4)2 - x(x + 2) - 2x + 3:

Bắt đầu bằng việc mở ngoặc:
E = (x^2 - 8x + 16) - (x^2 + 2x) - 2x + 3

Tiếp theo, kết hợp các thành phần tương tự:
E = x^2 - 8x + 16 - x^2 - 2x - 2x + 3

Tiếp tục đơn giản hóa:
E = x^2 - x^2 - 8x - 2x - 2x + 16 + 3

Kết quả cuối cùng:
E = -12x + 19

\(a,=x^2-6x+9-x^2+6x=9\\ b,=4x^2+4x+1-4x^2+9-4x-8=2\\ c,=\left(2x^2-2x-x+1\right):\left(x-1\right)\\ =\left(x-1\right)\left(2x-1\right):\left(x-1\right)=2x-1\)

`a)(x-3)^2-x(x-6)`

`=x^2-6x+9-x^2+6x=9`

`b)(2x+1)^2-(3+2x)(2x-3)-4(x+2)`

`=4x^2+4x+1-(4x^2-9)-4x-8`

`=2`

`c)(2x^2-3x+1):(x-1)`

`=(2x^2-2x-x+1):(x-1)`

`=[2x(x-1)-(x-1)]:(x-1)`

`=2x-1`

a) \(\left(x-3\right)^2-x\left(x-6\right)=x^2-6x+9-x^2+6x=9\)

b) \(\left(2x+1\right)^2-\left(3+2x\right)\left(2x-3\right)-4\left(x+2\right)=4x^2+4x+1-4x^2+9-4x-8=2\)

c) \(\left(2x^2-3x+1\right):\left(x-1\right)=\left[2x\left(x-1\right)-\left(x-1\right)\right]:\left(x-1\right)=\left[\left(x-1\right)\left(2x-1\right)\right]:\left(x-1\right)=2x-1\)

Answer:

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

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

\(=4x^2+4x+1+4x^2-4x+1-8x^2+2\)

\(=(4x^2+4x^2-8x^2)+(4x-4x)+(1+1+2)\)

\(=4\)

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

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

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

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

\(=3x-12\)

C = (2x-3)²-(x+4)(2x-1) -(x+3)²

(Chuyển đổi các hằng đẳng thức)

    = (4x²-12x+9)-(2x²-x+8x-4)-(x²+6x+9)

    =  4x²-12x+9-2x²+x-8x+4-x²-6x-9

(Ta thu gọn các hạng tử đồng dạng với nhau)

    = x²-25x-14 

\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)

Bài 2:

a) \(=x^2-4-x^2-2x-1=-2x-5\)

b) \(=8x^3-1-8x^3-1=-2\)

Bài 3:

a) \(\Rightarrow x^3+8-x^3+2x=15\)

\(\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)

b) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)

\(\Rightarrow7x=14\Rightarrow x=2\)

a) \(\left(x+3\right)\left(x-1\right)-x\left(x-5\right)=x^2+2x-3-x^2+5x=7x-3\)

b) \(\left(2x-3\right)\left(2x+3\right)-4\left(x+2\right)^2=4x^2-9-4x^2-16x-16=-16x-25\)

c) \(=x^3-3x^2+3x-1-x^3-8+3x^2=3x-9\)

1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)

=-27x^3-18x^2+4x+10

2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27

=7x^3+37x^2+46x+33

5:

\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)

\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)

=7x^3-48x^2+8x-35

`1)` Biểu thức xác định `<=>x+1 \ne 0<=>x \ne -1`

`[x^2+2x+1]/[x+1]=[(x+1)^2]/[x+1]=x+1`

`2)` Bth xác định `<=>x(x-3) \ne 0<=>{(x \ne 0),(x \ne 3):}`

`[x^2-6x+9]/[x(x-3)]=[(x-3)^]/[x(x-3)]=[x-3]/x`

`3)` Bth xác định `<=>2x(x+2) \ne 0<=>{(x \ne 0),(x \ne -2):}`

`[x^2-4]/[2x(x+2)]=[(x-2)(x+2)]/[2x(x+2)]=[x-2]/[2x]`

`4)` Bth xác định `<=>5x^2-10x \ne 0<=>5x(x-2) \ne 0<=>{(x \ne 0),(x \ne 2):}`

`[x^2-2x]/[5x^2-10x]=[x(x-2)]/[5x(x-2)]=1/5`

1)

\(ĐKXĐ:x\ne-1\)

\(\dfrac{x^2+2x+1}{x+1}\\ =\dfrac{\left(x+1\right)^2}{x+1}\\ =x+1\)

2)

ĐKXĐ x khác 0 và x khác 3

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

3)

ĐKXĐ: x khác 0 và x khác -2

\(\dfrac{x^2-4}{2x\left(x+2\right)}\\ =\dfrac{\left(x-2\right)\left(x+2\right)}{2x\left(x+2\right)}\\ =\dfrac{x-2}{2x}\)

4)

DKXĐ: x khác 0 và x khác 2

\(\dfrac{x^2-2x}{5x^2-10x}\\ =\dfrac{x\left(x-2\right)}{5x\left(x-2\right)}\\ =\dfrac{1}{5}\)

đk `x≠-1`

`(x^2+2x+1)/(x+1)`

`=((x+1)^2)/(x+1)`

`=x+1`

---------

đk \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne3\end{matrix}\right.\)

`(x^2-6x+9)/(x(x-3))`

`=((x-3)^2)/(x(x-3))`

`=(x-3)/x`

--------

 đk \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-2\end{matrix}\right.\)

`(x^2-4)/(2x(x+2))`

`=((x-2)(x+2))/(2x(x+2))`

`=(x-2)/(2x)`

--------

đk \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)

`(x^2-2x)/(5x^2-10x)`

`=(x(x-2))/(5x(x-2))`

`=x/(5x)`