\(a,=\left(x^3+3x^2-x^2-3x+x+3\right):\left(x+3\right)\\ =\left(x+3\right)\left(x^2-x+1\right):\left(x+3\right)\\ =x^2-x+1\\ b,=\left(x^3+2x^2-x^2-2x+3x+6\right):\left(x+2\right)\\ =\left(x+2\right)\left(x^2-x+3\right):\left(x+2\right)\\ =x^2-x+3\)

\(\left(\dfrac{1}{x}+x-2\right):\left(\dfrac{1}{x^2-x}+1-\dfrac{3}{x-1}\right)\)

\(=\dfrac{x^2-2x+1}{x}:\dfrac{1+x^2-x-3x}{x\left(x-1\right)}\)

\(=\dfrac{\left(x-1\right)^2}{x}\cdot\dfrac{x\left(x-1\right)}{x^2-4x+1}=\dfrac{\left(x-1\right)^3}{x^2-4x+1}\)

a:\(2x^2y\left(3x^2-5xy+4y^2\right)-6x^4y-10x^3y^2+8x^2y^3\)

b: \(\left(x^2-9\right)^2-\left(x+3\right)\left(x-3\right)\left(x^2+9\right)\)

\(=x^4-18x^2+81-\left(x^4-81\right)\)

\(=-18x^2+162\)

\(a,\) \(2x^2y\left(3x^2-5xy+4y^2\right)\)

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

\(b,\) \(\left(x^2-9\right)^2-\left(x+3\right)\left(x-3\right).\left(x^2+9\right)\)

\(=\left(x^2-9\right)^2-\left(x^2-3^2\right)\left(x^2+9\right)\\ =\left(x^2-9\right)^2-\left(x^4-9^2\right)\\ =x^4-18x+81-x^4+81\\ =-18x+162\)

a) = x^2 - 9 - (x^2 + 3x - 10)

= -3x + 1

b) = 3x + 1 - 3x + 19

= 20

a: \(\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\)

\(=x^2-9-x^2-3x+10\)

\(=-3x+1\)

b: \(\dfrac{27x^3+1}{9x^2-3x+1}-\left(3x-19\right)\)

\(=3x+1-3x+19\)

=20

a) \(2xy\left(3x^2-5xy+4y^2\right)=6x^3y-10x^2y^2+8xy^3\)

b) \(\left(x-3\right)^2+\left(x+5\right)\left(5-x\right)=x^2-6x+9+25-x^2=34-6x\)

a: \(2xy\left(3x^2-5xy+4y^2\right)=6x^3y-10x^2y^2+8xy^3\)

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

\(=x^2-6x+9+25-x^2\)

=-6x+34

a: =>x>=0 và x^2+x=x^2

=>x=0

b: =>x>=2 và x^2-4x-3=x^2-4x+4

=>-3=4(loại)

\(a)ĐK:x\ge0\)

\(pt\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2+x=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x=0\left(tm\right)\end{matrix}\right.\)

Vậy, pt có nghiệm duy nhất là x=0

\(b)ĐK:x\ge2+\sqrt{7}\)

\(pt\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\x^2-4x-3=(x-2)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-4x-3=x^2-4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\-3=4\end{matrix}\right.\)(vô lý)

Vậy pt vô nghiệm

Tham khảo

a)

-7x²(3x - 4y)

= -7x².3x + 7x^2ư.4y

= -21x² + 28x²y

b)

(x - 3)(5x - 4)

= x.5x - x.4 - 3.5x + 3.4

= 5x² - 4x - 15x + 12

= 5x² - 19x + 12

c)

(2x - 1)² = 4x² - 4x + 1

d)

(x + 3)(x - 3) = x² - 3² = x² - 9

a, = -21x³ +28x²y

b, = 5x² -4x -15x +12

= 5x² -19x +12

c, =x² -9

a)-21x³+28x²y

b)5x²-19x+12

c)4x²-4x+1

d)x²-9

a) \(\left(5x-4y^3\right)\left(2x^2-1+y\right)=10x^3-5x+5xy-8x^2y^3+4y^3-4y^4=-4y^4+10x^3+4y^3-8x^2y^3+5xy-5x\)

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

\(a,=10^3-5x+5xy-8y^3x^2+4y^3-4y^4\\ b,=3\)

tick mik nha

a: \(\left(5x-4y^3\right)\left(2x^2+y-1\right)\)

\(=10x^2+5xy-5x-8x^2y^3-4y^4+4y^3\)

b: \(\left(x-3\right)\left(2x-1\right)-x\left(2x-7\right)\)

\(=2x^2-x-6x+3-2x^2+7x\)

=3

a: \(=x^2-4-x^2+x+8=x+4\)

a) (x-2)(x+2)-x(x-1)+8
= x²-4-x²+x+8
= (x²-x²)+(-4+8)+x
= 4+x
b) bn viết lại đề đi:v
đọc khó quá.

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

b) \(x\left(3x-18\right)-3\left(x-4\right)\left(x-2\right)+8=3x^2-18x-3x^2+18x-24+8=-16\)