Question

[Uyên Phạm]

Bài 1 : Phân tích đa thức thành nhân tử

a) x²-6x-y²+9

b) 25-4x²-4xy -y²

c) x²+2xy+y²- xz-yz

d) x²-4xy+4y²-z²+4tz-4t²

Bài 2 : Phân tích đa thức thành nhân tử

a) ax²+cx²-ay+ay²-cy+cy²

b) ax^2+ay^2-bx^2-by^2+b-a

c) ac^2-ad-bc^2+cd+bd-c^3

Bài 3 : Tìm x

a) x(x-5)-4x+20=0

b) x(x+6)-7x-42=0

c) x^3-5x^2+x-5=0

d) x^4-2x^3+10x²-20x=0

Answer 1

[T.Thùy Ninh]

Bài 1:

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

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

\(b,25-4x^2-4xy-y^2=25-\left(2x+y\right)^2\)

\(=\left(5-2x-y\right)\left(5+2x+y\right)\)

\(c,x^2+2xy+y^2-xz-yz\)

\(=\left(x+y\right)^2-z\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\) \(d,x^2-4xy+4y^2-z^2+4tz-4t^2\)

\(=\left(x-2y\right)^2-\left(x-2t\right)^2=\left(x-2y-x+2t\right)\left(x-2y+x-2t\right)\)Bài 3,

\(a,x\left(x-5\right)-4x+20=0\)

\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(b,x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\Rightarrow\left[{}\begin{matrix}x+6=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)\(c,x^3-5x^2+x-5=0\)

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

\(\Leftrightarrow\left(x^2+1\right)\left(x-5\right)=0\)

Ta có: \(x^2+1\ge1\Rightarrow x-5=0\Rightarrow x=5\)

\(d,x^4-2x^2+10x^3-20=0\)

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

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

ta có:

\(x^2+1\ge1\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)