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Tự dưng 1x+2x+4x+......+125x + 126x là sao

a) \(x^4-y^4\)

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

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

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

b) \(x^2-3y^2\)

\(=x^2-\left(y\sqrt{3}\right)^2\)

\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)

c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)

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

\(=\left(5x-5y\right)\left(x+y\right)\)

\(=5\left(x-y\right)\left(x+y\right)\)

d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)

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

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

\(=\left(5x-y\right)\left(x-5y\right)\)

e) \(\left(4x^2-4x+1\right)-\left(x+1\right)^2\)

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

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

\(=3x\left(x-2\right)\)

f) \(x^3+27\)

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

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

g) \(27x^3-0,001\)

\(=\left(3x\right)^3-\left(0,1\right)^3\)

\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)

h) \(125x^3-1\)

\(=\left(5x\right)^3-1^3\)

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

a: Đặt 2x+1=a; 3x-1=b

Phương trình trở thành \(\left(a+b\right)^3-a^3-b^3=0\)

\(\Leftrightarrow3ab\left(a+b\right)=0\)

=>5x(2x+1)(3x-1)=0

hay \(x\in\left\{0;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)

c: Đặt x-3=a; x+1=b

Theo đề, ta có phương trình \(a^3+b^3=\left(a+b\right)^3\)

=>3ab(a+b)=0

=>(x-3)(x+1)(2x-2)=0

hay \(x\in\left\{3;1;-1\right\}\)

`4x^2-1`

`=(2x)^2-1`

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

`125x^3+1`

`=(5x)^3+1`

`=(5x+1)(25x^2-5x+1)`

`x^4+2x^2+1`

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

`4x^2-12xy+y^2`

`=4x^2-12xy+9y^2-8y^2`

`=(2x-3y)^2-(2sqrt2y)^2`

`=(2x-2sqrt2y-3y)(2x+2sqrt2y-3y)`

a) \(x^3+x^2-2x-8\)

\(=\left(x^3-2x^2\right)+\left(3x^2-6x\right)+\left(4x-8\right)\)

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

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

b) \(125x^3-10x^2+2x-1\)

\(=\left(125x^3-25x^2\right)+\left(15x^2-3x\right)+\left(5x-1\right)\)

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

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

c) \(x^3-4x^2+12x-27\)

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

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

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

d) Đề sai sai, nghiệm ra khá xấu nên bạn xem lại nhé

e) \(x^3-3x^2-3x+1\)

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

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

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

Đề k sai nên chắc phải hỏi lại cô thôi. Dù sao cx camon bn nhiều

a) x⁴ - y⁴

= (x²)² - (y²)²

= (x² - y²)(x² + y²)

= (x - y)(x + y)(x² + y²)

a.\(x^4-y^4=\left(x^2\right)^2-\left(y^2\right)^2=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)g,\(27x^3-0.001=27x^3-0=27x^3\)

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

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

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

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

\(=3x\left(x-2\right)\)

g: \(=\left(3x-\dfrac{1}{10}\right)\left(9x^2+\dfrac{3}{10}x+\dfrac{1}{100}\right)\)

k: \(=\left(5x-1\right)\left(25x^2+5x+1\right)\)