Question

Câu 14. (1,0 điểm) Phân tích các đa thức sau thành nhân tử:

a) \(x^3 + 10x^2 + 25x\);

b) \(x^2 - 9y^2 - 8x + 16\).

Câu 15. (1,0 điểm) Tìm \(x\) biết:

a) \(2x(x-3) - x + 3 = 0\);

b) \((3x-1)(2x+1) - (x+1)^2 = 5x^2\).

Answer 1

Câu 14:

a: \(x^3+10x^2+25x\)

\(=x\left(x^2+10x+25\right)\)

\(=x\left(x+5\right)^2\)

b: \(x^2-9y^2-8x+16\)

\(=\left(x^2-8x+16\right)-9y^2\)

\(=\left(x-4\right)^2-9y^2\)

\(=\left(x-4-3y\right)\left(x-4+3y\right)\)

Câu 15:

a: \(2x\left(x-3\right)-x+3=0\)

=>2x(x-3)-(x-3)=0

=>(x-3)(2x-1)=0

=>\(\left[{}\begin{matrix}x-3=0\\2x-1=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

b: \(\left(3x-1\right)\left(2x+1\right)-\left(x+1\right)^2=5x^2\)

=>\(6x^2+3x-2x-1-x^2-2x-1=5x^2\)

=>\(5x^2-x-2=5x^2\)

=>-x-2=0

=>x+2=0

=>x=-2

Answer 2

Câu 14

a)x³+10x²+25x

=x(x²+10x+25)

=x(x+5)²

b)x²-9y²-8x+16

=(x²-8x+16)-9y²

=(x-4)²-9y²

=[(x-4)-3y][(x-4)+3y]

=(x-4-3y)(x-4+3y)

Answer 3

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

   2x(x-3)-1(x-3)=0

   (x-3)(2x-1)=0

TH1:x-3=0            TH2:2x-1=0

      x=3                        2x=1

                                    x=1/2

Vậy x=3 hoặc x=1/2