Bài 1:

a) \(A=x^2-2xy+y^2+x^2+2xy+y^2-2x^2+2y^2-4y^2+4=4\)

b) \(B=\left(x^3-3x^2+3x-1\right)+1020=\left(x-1\right)^3+1020=\left(11-1\right)^3+1020\)

\(=10^3+1020=1000+1020=2020\)

\(=x^3-y^3+2y^3=x^3+y^3=\dfrac{8}{27}+\dfrac{1}{27}=\dfrac{1}{3}\)

\(a,\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x+1-x+3\right)=0\\ \Leftrightarrow4x=0\Leftrightarrow x=0\\ c,\Leftrightarrow2\left(x^2+x+\dfrac{1}{4}\right)=0\\ \Leftrightarrow2\left(x+\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)

1a,5xy²-10xyz+5xz² = 5x(y²-2xy+z²)

                                 = 5x(y-z)²

b,x²-4y²+x+2y = (x-2y)(x+2y)+x+2y

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

2a,x(x-3)-x+3 = 0

<=> x(x-3)-(x-3) = 0

<=> (x-1)(x-3) = 0

Xét x-1 = 0 => x = 1

Xét x-3 = 0 => x = 3

b, Câu b bạn ghi lại đề giúp mình ko có mờ quá

1: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)

\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)

\(\Leftrightarrow6x=6\)

hay x=1

3: Ta có: \(x^2+3x-10=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Câu 1: 

1: Ta có: \(P=\left(x+4\right)^2+\left(x+5\right)\left(x-5\right)-2x\left(x+1\right)\)

\(=x^2+8x+16+x^2-25-2x^2-2x\)

\(=6x-9\)

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

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

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

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

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

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

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

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

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

Bài II

\(A\left(x\right)⋮B\left(x\right)\Leftrightarrow2x^3+x^2-4x+m=\left(2x-1\right)\cdot a\left(x\right)\)

Thay \(x=\dfrac{1}{2}\Leftrightarrow\dfrac{1}{4}+\dfrac{1}{4}-2+m=0\Leftrightarrow m=\dfrac{3}{2}\)

Bài I

\(1,\Rightarrow x^2-4x-x^2+6x-9=0\\ \Rightarrow2x=9\Rightarrow x=\dfrac{9}{2}\\ 2,\Rightarrow x^2-3x-10=0\\ \Rightarrow x^2+2x-5x-10=0\\ \Rightarrow\left(x+2\right)\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)