Question

[KIEU TRANG DOAN THI]

1 : phân tích các đa thức sau thành nhân tử

a, \(x^2-2xy+y^2-z^2\)

b, \(x^2-6x+9-9y\)

c, \(\left(x+y\right).\left(x+z\right).\left(z+y\right)+xyz\)

2 : tìm x biết

a, \(x.\left(x-2\right)+x-2=0\)

b, \(5x.\left(x-3-x+3\right)=0\)

Answer 1

[Vi Lê Bình Phương]

a. \(x^2-2xy+y^2-z^2=\\ \left(x^2-2xy+y^2\right)-z^2\\ =\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)

b. \(x^2-6x+9-9y\\ =\left(x^2-6x\right)+\left(9-9y\right)\\ =x\left(x-6\right)+9\left(1-y\right)\)

Answer 2

[Nguyễn Xuân Tiến 24]

1,

a, \(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2\)

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

b, hình như sai đề phải ko bn?

Phải là: \(x^2-6x+9-9y^2\)=\(\left(x-3\right)^2-\left(3y\right)^2\)\(=\left(x-3y-3\right)(x+3y-3)\)

c,\(\left(x+y\right)\left(y+z\right)\left(x+z\right)+xyz\)

=\((x^2+xy+yz+xz)\left(z+y\right)+xyz\)

=\((x^2z+xyz+xz^2)+(x^2y+xy^2+xyz)+\)\(\left(yz^2+y^2z+xyz\right)\)

= xz(x + y + z) + xy(x + y + z) + yz(x + y +z)

=(x+y+z)(xz+xy+yz)

2,a,\(\left(x-2\right)\left(x+1\right)=0\)\(\Rightarrow\)\(\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

b,Đề phải thế này nha:

\(5x\left(x-3\right)-x+3=0\)\(\Rightarrow\)(x - 3)(5x - 1)=0

\(\Rightarrow\)\(\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\)\(\Rightarrow\)\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

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