`@` `\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`

b: B=(x+1)^2-2(2x-1)(x+1)+4x^2-4x+1

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

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

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

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

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

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

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

b) Ta có: \(5\left(x+2\right)\left(x-2\right)-\dfrac{1}{2}\left(6-8x\right)^2+17\)

\(=5\left(x^2-4\right)-\dfrac{1}{2}\left(64x^2-96x+36\right)+17\)

\(=5x^2-20-32x^2+48x-16+17\)

\(=-27x^2+48x-19\)

\(A=2\sqrt{27}-\sqrt{\left(1-\sqrt{3}\right)^2}+\dfrac{1}{2-\sqrt{3}}\\ =2.3\sqrt{3}-\left|1-\sqrt{3}\right|+\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\\ =6\sqrt{3}-\left(-1+\sqrt{3}\right)+\dfrac{2+\sqrt{3}}{2^2-\sqrt{3^2}}\\ =6\sqrt{3}+1-\sqrt{3}+2+\sqrt{3}\\ =6\sqrt{3}+3\)

\(=6\sqrt{3}-\sqrt{3}+1+2+\sqrt{3}=6\sqrt{3}+3\)

\(a,=5\sqrt{2}-3\sqrt{2}+\sqrt{2}=3\sqrt{2}\\ b,=\dfrac{\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}=\dfrac{2\sqrt{3}}{3-2}=2\sqrt{3}\)

a) đã rút gọn
b) (x-3)(x+3)-(x-3)(x+1)
= (x-3)(x+3-x-1)
= (x-3)2

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

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

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

\(=0-11x+24\)

\(=-11x+24\)

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

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

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

\(=0+0+5\)

\(=5\)

\(A=\left(\dfrac{-\left(\sqrt{2}-1\right)}{\sqrt{2}+1}+\dfrac{\sqrt{2}+1}{\sqrt{2}-1}\right)\cdot\dfrac{1}{6\sqrt{2}}\)

\(=\dfrac{-\left(3-2\sqrt{2}\right)+3+2\sqrt{2}}{1}\cdot\dfrac{1}{6\sqrt{2}}\)

\(=\dfrac{-3+2\sqrt{2}+3+2\sqrt{2}}{6\sqrt{2}}=\dfrac{2}{3}\)

\(B=\left(\dfrac{3-2\sqrt{2}-3-2\sqrt{2}}{-1}\right):6\sqrt{2}=\dfrac{4\sqrt{2}}{6\sqrt{2}}=\dfrac{2}{3}\)

a) \(A=\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\right)\left(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}-\sqrt{x}\right)\)

\(A=\left[\dfrac{\left(\sqrt{x}\right)^3-1^3}{\sqrt{x}-1}+\sqrt{x}\right]\left[\dfrac{\left(\sqrt{x}\right)^3+1^3}{\sqrt{x}+1}-\sqrt{x}\right]\)

\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}-1}+\sqrt{x}\right]\left[\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}-\sqrt{x}\right]\)

\(A=\left(x+\sqrt{x}+1+\sqrt{x}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\)

\(A=\left(x+2\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)\)

\(A=\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)^2\)

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

\(A=\left(x-1\right)^2\)

\(A=x^2+2x+1\)