Question

[Nguyễn Đức Tài]

nhờ giải giupws em với a

1. Phân tích các đa thức sau thành nhân tử:

a)     5x² – 10xy

b)    3x(x – y)  –  6(x – y)

c)     2x(x – y) – 4y(y – x)

d)    9x² – 9y²

e)     x² – xy – x + y

f)      xy – xz – y + z

2. Phân tích các đa thức sau thành nhân tử:

 a)a² – 4b²                                        b) x² – y² + 6y - 9                                         

c) (2a + b)² – a²                     d) 16(x – 1)² – 25(x + y)²

e)x² + 10x + 25                f) 25x² – 20xy + 4y²

      g)9x⁴ + 24x² + 16             h) x³ – 125

      i)x⁶ – 1                            k) x³ + 15x² + 75x + 125

3. Tìm x biết :

a) 3x² + 8x = 0              b) 9x² – 25 = 0          c) x³ – 16x = 0     d) x³ + x = 0.

4. Chứng minh rằng với mọi số nguyên a thì: a³ – a chia hết cho 6

Answer 1

[⭐Hannie⭐]

Bài `1`

\(a,5x^2-10xy=5x\left(x-2y\right)\\ b,3x\left(x-y\right)-6\left(x-y\right)=\left(x-y\right)\left(3x-6\right)\\ =3\left(x-y\right)\left(x-2\right)\\ c,2x\left(x-y\right)-4y\left(y-x\right)=2x\left(x-y\right)+4y\left(x-y\right)\\ =\left(x-y\right)\left(2x+4y\right)=2\left(x-y\right)\left(x+2y\right)\\ d,9x^2-9y^2=\left(3x\right)^2-\left(3y\right)^2=\left(3x-3y\right)\left(3x+3y\right)\\ f,xy-xz-y+z=\left(xy-xz\right)-\left(y-z\right)\\ =x\left(y-z\right)-\left(y-z\right)=\left(y-z\right)\left(x-1\right)\)

Bài `3`

\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)

\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)