SORY I'M IN GRADE 6

Thợ Đào Mỏ Panda, mày bị điên à, không biết còn trả lời làm cái quái gì

Đã có đáp án:

2x^4-6x^3+x^2+6x-3

=2x^2-6x^3-3x^2-2x^2-6x-3

=2x^2(x^2-1)-6x(x^2-1)+3(x^2-1)

=(x^2-1)(2x^2-6x+3)

\(=x\left(2x^2-x-6\right)\)

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

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

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

x(2x^2-x-6)

x(2x^2-4x+3x-6)

x[2x(x-2)+3(x-2)]

x(2x+3)(x-2)

a, x²+x³-4x+4=x²(x+1)-4(x+1)=(x+1)(x²-4)=(x+1)(x-2)(x+2)

2x^5-6x^4-2a^2x^3-6ax^3

 =(2x^5-2a^2x^3)-(6x^4+6ax^3)

 =2x^3(x^2-a^2)-6x^3(x+a)

 =2x^3(x-a)(x+a)-6x^3(x+a)

 =(x+a)(2x^4-2x^3a-6x^3)

 =(x+a) 2x^3 (x-a-3)

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

6x³ + 5x² - 7x - 4

= (6x³ + 5x - 7x) - 4

= x (6x² - 5 - 7) - 2²

= x (6x² - 12) - 2²

= x [6 (x² - 2)] - 2²

= x [6 (x² - \(\sqrt{2}^2\))] - 2²

= x [6 (x +\(\sqrt{2}\)) (x -\(\sqrt{2}\))] - 2²

= (x - 2²) [6 (x +\(\sqrt{2}\)) (x -\(\sqrt{2}\))

b) 2x³ - x² + x - 2

= (2x³ - x² - x) - 2

= x (2x² - x - 1) - 2

= (x - 2) (2x² - x - 1)

(mik ko biet dug ko, neu sai mog bn thog cam)

a, Cách 1 : \(x^2+5x+6=x^2+2x+3x+6=\left(x+2\right)\left(x+3\right)\)

Cách 2 : \(x^2+5x+6=x^2+2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}+6\)

\(=\left(x+\frac{5}{2}\right)^2-\frac{1}{4}=\left(x+2\right)\left(x+3\right)\)

b, Cách 1 : \(x^2-x-6=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\)

Cách 2 : \(x^2-x-6=x^2-x+\frac{1}{4}-\frac{1}{4}-6=\left(x-\frac{1}{2}\right)^2-\frac{25}{4}=\left(x-3\right)\left(x+2\right)\)

c, Cách 1 : \(x^2+6x+8=x^2+4x+2x+8=\left(x+2\right)\left(x+4\right)\)

Cách 2 : \(x^2+6x+8=x^2+6x+9-1=\left(x+3\right)^2-1=\left(x+2\right)\left(x+4\right)\)

d, Cách 1 : \(x^2-2x-8=x^2+2x-4x-8=\left(x-4\right)\left(x+2\right)\)

Cách 2 : \(x^2-2x-8=x^2-2x+1-9=\left(x-1\right)^2-9=\left(x-4\right)\left(x+2\right)\)

\(=x^4-x^3+3x^3-3x^2+3x^2-3x+9x-9\\ =\left(x-1\right)\left(x^3+3x^2+3x+9\right)\\ =\left(x-1\right)\left(x+3\right)\left(x^2+3\right)\)

\(x^4+2x^3+6x-9=x^3\left(x-1\right)+3x^2\left(x-1\right)+3x\left(x-1\right)+9\left(x-1\right)\)

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

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

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

\(x^4+2x^3+6x-9=0\)\(\)

\(\Rightarrow x^4+3x^3-x^3+9x-3x-9+3x^2-3x^2=0\)

\(\Rightarrow\left(x^4+3x^3+3x^2+9x\right)-\left(x^3+3x^2+3x+9\right)=0\)

\(\Rightarrow x\left(x^3+3x^2+3x+9\right)-\left(x^3+3x^2+3x+9\right)=0\)

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

\(\Rightarrow\left(x-1\right)\left[x^2\left(x+3\right)+3\left(x+3\right)\right]=0\)

\(\Rightarrow\left(x-1\right)\left(x^2+3\right)\left(x+3\right)=0\)

Mà \(x^2+3>0\) với mọi x\(\in\)R.

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)