`@` `\text {Ans}`

`\downarrow`

`A= (2x - 3)^2 - (2x + 3)^2`

`= [(2x - 3) - (2x + 3)]*[(2x - 3) + (2x + 3)]`

`= (2x - 3 - 2x - 3) * (2x - 3 + 2x + 3)`

`= -6 * 4x`

`= -24x`

`A=(2x-3)^2-(2x+3)^2`

`A=(2x-3-2x-3)(2x-3+2x+3)`

`A=-6.4x=-24x`

a) \(=x^2-5-x^2+49=44\)

b) Nhân tử cuối cùng bạn ghi gì vậy?

(2x+3y)²+(3x-2y)²-2(2x+3y)(2x+3y) (3x-2y) đó bạn

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

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

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

\(=0-11x+24\)

\(=-11x+24\)

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

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

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

\(=0+0+5\)

\(=5\)

a: \(=\dfrac{\left(x+1\right)\left[\left(3x-2\right)-\left(2x+5\right)\left(x-1\right)\right]}{x+1}\)

=3x-2-2x^2+2x-5x+5

=-2x^2+3

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

c: =x^3-3x^2+3x-1-x^3-1+9x^2-1

=6x^2+3x-3

\(a,\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x^2-1\right)\right]:\left(x+1\right)\)

\(=\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x-1\right)\left(x+1\right)\right]:\left(x+1\right)\)

\(=\left[\left(x+1\right)\left(3x-2-\left(2x+5\left(x-1\right)\right)\right)\right]:\left(x+1\right)\)

\(=\left[\left(x+1\right)\left(3x-2-2x^2+2x-5x+5\right)\right]:\left(x+1\right)\)
\(=\left[\left(x+1\right)\left(-2x^2+3\right)\right].\dfrac{1}{x+1}\)

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

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

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

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

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

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

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

\(=-3x^2+3x-2-3x+9x^2-1+3x\)

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

a: \(=2x^2-6x+x-3-20x+8x^2\)

\(=10x^2-25x-3\)

b: \(=x^2+4x+4-2\left(x^2-9\right)+10\)

\(=x^2+4x+14-2x^2+18\)

\(=-x^2+4x+32\)

Lời giải:

ĐK: $x\geq 0; x\neq 1$

\(A=\left[\frac{(\sqrt{x}-1)(x+2\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{\sqrt{x}+2}{(\sqrt{x}+1)(\sqrt{x}+2)}\right].\frac{\sqrt{x}-1}{(\sqrt{x}-1)(2\sqrt{x}+3)}\)

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

a: \(N=\left(2x-3y\right)\left(2x+3y\right)=\left(2x\right)^2-\left(3y\right)^2\)

\(=4x^2-9y^2\)

Thay x=1/2 và y=1/3 vào N, ta được:

\(N=4\cdot\left(\dfrac{1}{2}\right)^2-9\left(\dfrac{1}{3}\right)^2\)

\(=4\cdot\dfrac{1}{4}-9\cdot\dfrac{1}{9}\)

=1-1

=0

b: \(N=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=\left(2x-y\right)\left[\left(2x\right)^2+2x\cdot y+y^2\right]\)

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

Khi x=1 và y=3 thì \(N=8\cdot1^3-3^3=8-27=-19\)

a: \(A=x^2-2xy+y^2+x^2+2xy+y^2-2x^2-x\)

=-x

=-2

a, `(8x^3-4x^2): 4x -(4x^2-5x) : 2x + (2x)^2`

`=4x (2x^2-x) : 4x - 2x(2x-5/2 ) :2x + 4x^2`

`=2x^2-x-2x+5/2+4x^2`

`=6x^2-3x+5/2`

b, `(3x^3-x^2y) :x^2 -(xy^2+x^2y) :xy + 2x(x+1)`

`=x^2 (3x-y) :x^2 -xy(y+x) + (2x^2+2x)`

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

`=2x^2+4x-2y`