k đi rồi giúp cho

k rồi giúp cho

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

b)\(6x^4+7x^2+2=6x^4+4x^2+3x^2+2=\left(3x^2+2\right)\left(2x^2+1\right)\)

c)\(2x^2-3x-2700=2x^2+72x-75x+2700=\left(2x-75\right)\left(x+36\right)\)

a)x³+x²+4

=x³-x²+2x+2x²-2x+4

=x(x²-x+2)+2(x²-x+2)

=(x+2)(x²-x+2)

b)x³-2x-4

=x³+2x²+2x-2x²-4x-4

=x(x²+2x+2)-2(x²+2x+2)

=(x-2)(x²+2x+2)

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

\(x^4+x^2+x\)

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

\(=x^2+2x.\frac{1}{2}+\frac{1}{2}^2-\frac{1}{2}^2+x^4\)

\(=\left(x^2+2x.\frac{1}{2}+\frac{1}{2}^2\right)-\frac{1}{2}^2+x^4\)

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

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

\(=\left(x+\frac{1}{2}\right)^2-\sqrt{\frac{1}{4}}^2+x^4\)

\(=\left(x+\frac{1}{2}-\sqrt{\frac{1}{4}}\right).\left(x+\frac{1}{2}+\sqrt{\frac{1}{4}}\right)+x^4\)

Đến đây dễ rồi .Biến đổi ngoặc bên phải giống ngoặc trái rồi mở ngoặc đặt nhân tử chung là được .

x^4+4=x^4 + 4x^2 +4 - 4x^2=(x^2)^2+ 2.x^2.2+2^2 - (2x)^2 = (x^2+2)-(2x)^2 =(x^2+2-2x)(2^2+2-2x)

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

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

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

a)2x²+3x-5

=2x²+5x-2x-5

=x(2x+5)-(2x+5)

=(x-1)(2x+5)

b)x⁸+x⁴+1

=(x⁴)²+2x⁴+1-x⁴

=(x⁴+1)²-x⁴

=(x⁴+1-x²)(x⁴+x²+1)

=(x⁴-x²+1)(x²-x+1)(x²+x+1)

y^4+64

=(y^2)^2+16y^2+64-16y^2

=(y^2+8-4x)(x^2+8+4x)

x^2+4

=x^2+2x^2+4-2x^2

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

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

x^4+16

=(x^2)^2+4x^2+16-4x^2

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

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

x^4y^4+4

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

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

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

4x^4y^4+1

=4x^4y^4+x^4+1-x^4

=(2x^4y^4+1)^2-(x^2)^2

=(2x^4y^4+1-x^2)(2x^4y^4+1+x^2)

Mình ko bt câu D đúng hay sai nữa. Mà lỡ sai bạn đừng giận mình nha!