a. 9x² + 6x + 1 - 9x² + 3x = 9x + 1

b. x³ - 2x² + 4x + 2x² - 4x + 8 - x³ + 3x = 3x + 8

a, `(x-3)(x^2+3x+9)-(x^2-1)(9x+27)`

`=x^3-3^3-(9x^3+27x^2-9x-27)`

`=x^3-3^3-9x^3-27x^2+9x+27`

`=-8x^3-27x^2+9x`

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

`=x^3-2^3-x(x^2-9)`

`=x^3-8-x^3+9x`

`=9x-8`

a) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x^2-1\right)\left(9x+27\right)\)

\(=x^3-27-\left(9x^3+27x^2-9x-27\right)\)

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

\(=-8x^3-27x^2+9x\)

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

\(=x^3-8-x\left(x^2-9\right)\)

\(=x^3-8-x^3+9x\)

\(=9x-8\)

a: \(A=\left(a+b\right)^3+\left(a-b\right)^3-2a^3\)

\(=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-2a^3\)

\(=6ab^2\)

b: \(B=\left(2x-1\right)\left(4x^2+2x+1\right)-8\left(x+2\right)\left(x^2-2x+4\right)\)

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

\(=8x^3-1-8x^3-64=-65\)

c: \(C=\left(x+2y\right)\left(x^2-2xy+4y^2\right)+\left(2x-y\right)\left(4x^2+2xy+y^2\right)-9x^3\)

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

\(=x^3+8y^3+8x^3-y^3-9x^3=7y^3\)

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

b, \(\left(3x-1\right)^2-9x\left(x+1\right)=9x^2-6x+1-9x^2-9x=1-15x\)

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

Câu 1:

a: \(2x^2\left(x^2+3x+\frac12\right)\)

\(=2x^2\cdot x^2+2x^2\cdot3x+2x^2\cdot\frac12\)

\(=2x^4+6x^3+x^2\)

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

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

\(=x^2-x-2-x^2-4x-4=-5x-6\)

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

\(=9x^2+6x+1-9x^2-27x\)

=-21x+1

Câu 2:

a: \(\left(x+2\right)^2-x\left(x+4\right)+10\)

\(=x^2+4x+4-x^2-4x+10\)

=4+10

=14

=>Biểu thức này không phụ thuộc vào biến

b: \(\left(x+3\right)\left(4x-1\right)-\left(2x+1\right)^2-7x+3\)

\(=4x^2-x+12x-3-4x^2-4x-1-7x+3\)

=12x-x-7x-4x-3-1+3

=x(12-1-7-4)-1

=-1

=>Biểu thức này không phụ thuộc vào biến

Câu 3:

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

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

=>5x+4=2

=>5x=2-4=-2

=>\(x=-\frac25\)

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

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

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

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

=>3x+4=-3

=>3x=-7

=>\(x=-\frac73\)

Câu 5

Xét ΔADC và ΔBCD có

AD=BC

DC chung

AC=BD

Do đó: ΔADC=ΔBCD

=>\(\hat{ACD}=\hat{BDC}\)

=>\(\hat{OCD}=\hat{ODC}\)

=>ΔOCD cân tại O

=>OC=OD

Ta có: OC+OA=AC

OD+OB=BD

mà OC=OD và AC=BD

nên OA=OB

Lời giải:

\(B=\frac{3}{x-1}\sqrt{\frac{(x-1)^2}{(3x)^2}}=\frac{3}{x-1}|\frac{x-1}{3x}|\)

\(=\frac{3}{x-1}.\frac{1-x}{3x}=\frac{-1}{x}\)

\(B=\dfrac{3}{x-1}.\sqrt{\dfrac{x^2-2x+1}{9x^2}}=\dfrac{3}{x-1}.\sqrt{\left(\dfrac{x-1}{3x}\right)^2}\)

\(=\dfrac{3}{x-1}.\left|\dfrac{x-1}{3x}\right|=\dfrac{3}{x-1}.\dfrac{1-x}{3x}=-\dfrac{1}{x}\)

...

`Answer:`

`a)`

`A=5(x+1)^2-3(x-3)^2-4(x^2-4)`

`=>A=5(x^2+2x+1)-3(x^2-6x+9)-4x^2+16`

`=>A=5x^2+10x+5-3x^2+18x-27-4x^2+16`

`=>A=(5x^2-3x^2-4x^2)+(10x+18x)+(5-27+16)`

`=>A=-2x^2+28x-6`

`b)`

`B=5(x+1)^2-3(x-3)^2-4(x+2)(x-2)`

`=2x(3x+5)-3(3x+5)-2x(x^2-4x+4)-[(2x)^2-3^2]`

`=6x^2+10x-9x-15-2x^3+8x^2-8x-4x^2+9`

`=(6x^2-4x^2+8x^2)-2x^3+(10x-9x-8x)+(-15+9)`

Thay `x=-7` vào ta được:

`B=10(-7)^2-2(-7)^3-7(-7)-6`

`=>B=10.49-2(-343)+49-6`

`=>B=490+686+49-6`

`=>B=1219`

a: Ta có: \(3x\left(2x+1\right)+\left(2x-3\right)\left(x+1\right)\)

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

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

\(a,3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

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

\(=\left(3x^2+5x^2-8x^2\right)+\left(-6x-5x\right)+24\)

\(=0-11x+24\)

\(=-11x+24\)

\(b,\left(7x-3\right)\left(2x+1\right)-\left(5x-2\right)\left(x+4\right)-9x^2+17x\)

\(=14x^2+7x-6x-3-5x^2-20x+2x+8-9x^2+17x\)

\(=\left(14x^2-5x^2-9x^2\right)+\left(7x-6x-20x+2x+17x\right)+\left(-3+8\right)\)

\(=0+0+5\)

\(=5\)