câu ở dưới mình ghi sai đề

x⁴+2002x²+2001x+2002

mk đang cần gấp lắm.mọi người giúp mk nha.ai nhanh tay nhất mk k cho

       \(x^4+2002x^2-2001x+2002\)

\(=x^4+2002x^2+x-2002x+2002\)

\(=\left(x^4+x\right)+\left(2002x^2-2002x+2002\right)\)

\(=x\left(x^3+1\right)+2002\left(x^2-x+1\right)\)

\(=x\left(x+1\right)\left(x^2-x+1\right)+2002\left(x^2-x+1\right)\)

\(=\left(x^2-x+1\right)\left[x\left(x+1\right)+2002\right]\)

\(=\left(x^2-x+1\right)\left(x^2+x+2002\right)\)

1A:

a: \(x^3+2x=x\left(x^2+2\right)\)

b: \(3x-6y=3\left(x-2y\right)\)

c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)

\(=5\left(x+3y\right)\left(1-3x\right)\)

d: \(3\left(x-y\right)-5x\left(y-x\right)\)

\(=3\left(x-y\right)+5x\left(x-y\right)\)

\(=\left(x-y\right)\left(5x+3\right)\)

1A. a. x(x²+2) 

b. 3(x-2y)

c. 5(x+3y)(1-3x) 

d. (x-y) (3-5x)

1B. a. 2x(2x-3)

b.xy(x²-2xy+5)

c. 2x(x+1)(x+2)

d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)

1B:

a: \(4x^2-6x=2x\left(2x-3\right)\)

b: \(x^3y-2x^2y^2+5xy\)

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

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

x2 - x = x.x - x.1 = x(x - 1)

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

x²-x

= x(x-1)

=4x²-4=4(x²-1)=4(x-1)(x+1)

4(x²-1)

đíu bt làm nka

\(x^2+x+\dfrac{1}{4}-\dfrac{1}{4}+4=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\)(vô lí)

Vậy pt vô nghiệm 

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

Giải thích: Sử dụng hằng đẳng thức 1 và 3 

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