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

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

\(=6x-9\)

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

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

a) \(=\dfrac{2}{5}x^3-2x^2+6x\)

b) \(=x^2-2x+9-x^2=9-2x\)

c) \(=x^4y-2x^3+xy^2-2y\)

a) 4x + 12x² = 4x(1+ 3x)

b) x³ - 8 + 3x² - 6 = x³ +3x² -14

c) x² -2x +1 -4y² = (x-1)² -4y²

Câu 2:

a) x² + 5x - 3x - 15 = 0

⇔ x² + 2x +1 - 1 - 15 = 0

⇔ ( x+1)² =16

⇔ x + 1 =4

⇔ x = 3

b) 

1 enjoyable

2 Unluckily

3 communal

4 carelessly

5 relaxation