`@` `\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/ \(A=\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)+3\left(x-1\right)\left(x^2+x+1\right)\)

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

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

Thay x = 2 vào A được:

\(=-3.2^2+7.2-4=-2\)

Vậy: Giá trị của A khi x = 2 là -2

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b/ \(B=126y^3+\left(x-5y\right)\left(x^2+25y^2+5xy\right)\)

\(=126y^3+x^3-125y^3\)

Thay x = -5 và y = -3 vào B được: 

\(126.\left(-3\right)^3+\left(-5\right)^3-125.\left(-3\right)^3=-152\)

Vậy: Giá trị của B tại x = -5 và y = -3 là -152

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c/ \(C=a^3+b^3-\left(a^2-2ab+b^2\right)\left(a-b\right)\)

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

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

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

Thay a = -4 và b = 4 vào C được:

\(2.4^3+3.\left(-4\right)^2.4-3.\left(-4\right).4^2=512\)

Vậy: Giá trị của C tại a = -4 vào b = 4 là 512

a:Ta có: \(A=\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)+3\left(x-1\right)\left(x^2+x+1\right)\)

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

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

\(=-3\cdot2^2+7\cdot2-4\)

\(=-12-4+14=-2\)

c: Ta có: \(C=a^3+b^3-\left(a-b\right)\left(a^2-2ab+b^2\right)\)

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

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

\(=2\cdot4^3+3\cdot\left(-4\right)^2\cdot4-3\cdot\left(-4\right)\cdot4^2\)

\(=128+192+192=512\)

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

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

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

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

a: \(=2x^2-6x+x-3-20x+8x^2\)

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

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

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

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

1) \(\Rightarrow16x^2+24x+9+9x^2-24x+16+4-25x^2=x\)

\(\Rightarrow x=29\)

2)

a) \(=x^2-9-x^2+6x-9=6x-18\)

b) \(=\left(3x-1+2x+1\right)^2=\left(5x\right)^2=25x^2\)

\(A=\left(x+2\right)^3-\left(x-2\right)^3-12x^2=x^3+6x^2+12x+8-x^3+6x^2-12x+8-12x^2=16\)

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

=-x

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

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

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

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