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

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

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

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

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

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

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

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

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

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

\(a,\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=4xy\\ b,\left(x+y\right)^2+\left(x-y\right)^2-2\left(x+y\right)\left(x-y\right)=\left(x+y-x+y\right)^2=4y^2\\ c,\left(x^2-1\right)\left(x^2-x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\\ =\left(x-1\right)\left(x^3+1\right)\\ =x^4-x^3+x-1\)

a. (x + y)² - (x - y)²

= (x + y - x + y)(x + y + x - y)

= 2y . 2x

= 4xy

b. (x + y)² + (x - y)² - 2(x + y)(x - y)

= (x² + 2xy + y²) + (x² - 2xy + y²) - 2(x² - y²)

= x² + 2xy + y² + x² - 2xy + y² - 2x² + 2y²

= x² + x² - 2x² + 2xy - 2xy + y² + y² + 2y²

= 4y²

c. (x² - 1)(x² - x + 1)

= x⁴ - x³ + x² - x² + x - 1

= x⁴ - x³ + x - 1

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

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

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

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

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

Lời giải:

ĐK: $x\geq 0; x\neq 1$

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

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

gúp mình với

A = x ( x + y ) - y ( x + y )

A = ( x + y ) ( x - y )

A = x\(^2\) - y\(^2\)

Tại x = \(\dfrac{-1}{2}\) và y = -2 ta có 

\(\left(\dfrac{-1}{2}\right)^2-\left(-2\right)^2\) \(=\) \(\dfrac{-15}{4}\)

\(A=x\left(x+y\right)-y\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y\right)\)

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

Thay \(x=-\dfrac{1}{2}\) và \(y=-2\) vào biểu thức \(A\), ta có:

\(A=\left(-\dfrac{1}{2}\right)^2-\left(-2\right)^2\)

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

\(=-\dfrac{15}{4}\)

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

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

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

\(=0-11x+24\)

\(=-11x+24\)

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

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

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

\(=0+0+5\)

\(=5\)

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

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

\(=\dfrac{-3x+7\sqrt{x}-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\) (ĐK: \(x\ne4;x\ne9;x\ge0\))

\(A=\dfrac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

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

\(A=\dfrac{2\sqrt{x}-9-\left(x-3\sqrt{x}+\sqrt{x}-3\right)-\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(A=\dfrac{2\sqrt{x}-9-x+2\sqrt{x}+3-2x+3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(A=\dfrac{-3x+7\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

`@` `\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