Question

[Kudo Shinichi]

B1. phân tích đa thức sau thành nhân tử

a. 9x² + 90x + 225 - ( x - 7)²

b. 49( y - 4 )² - 9y² - 36y + 36

B2 tính giá trị biểu thức

a. A = xy - 4y - 5y + 20 với x =14 , y=\(\dfrac{11}{2}\)

b. B= x² + xy - 5x - 5y với x= -5, y= -8

B3. tìm x

a. 4x² - 25 - (2x-5)(2x+7) =0

b. (x³ + 27 ) + (x+3)(x-9) =0

c. (2x³ + 2x²) + (3x² + 3) =0

Answer 1

[Hà Linh]

B1:

a) \(9x^2+90x+225-\left(x-7\right)^2\)

= \(9x^2+90x+225-x^2+14x-49\)

= \(8x^2+104x+176\)

= \(\left(x+2\right)\left(x+11\right)\)

b) \(49\left(y-4\right)^2-9y^2-36y+36\)

= \(49\left(y^2-8y+16\right)-9y^2-36y+36\)

= \(49y^2-392y+784-9y^2-36y+36\)

= \(40y^2-428y+820\)

= \(\left(5y-41\right)\left(8y-20\right)\)

B2:

a) A = \(xy-4y-5y+20=xy-9y+20\)

A = \(y\left(x-9\right)+20\)

Với x = 14, y = \(\dfrac{11}{2}\)

A = \(\dfrac{11}{2}\left(14-9\right)+20=47,5\)

b) B = \(x^2+xy-5x-5y\)

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

Với x = -5, y = -8

B = \(\left(-5-8\right)\left(-5-5\right)=130\)

B3:

a) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\left(2x-5\right)\left(2x+5-2x-7\right)=0\)

\(\left(2x-5\right)\left(-2\right)=0\)

\(x=\dfrac{5}{2}\)

b) \(\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

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

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

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

\(2x^3+5x^2+3=0\)

\(\Rightarrow\) Đề sai rồi, nghiệm khủng bố lắm.