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a.
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
b.
\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
c.
\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
d.
\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
e.
\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
Bài 3
a) x² + 10x + 25
= x² + 2.x.5 + 5²
= (x + 5)²
b) 8x - 16 - x²
= -(x² - 8x + 16)
= -(x² - 2.x.4 + 4²)
= -(x - 4)²
c) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
d) (x + y)² - 9x²
= (x + y)² - (3x)²
= (x + y - 3x)(x + y + 3x)
= (y - 2x)(4x + y)
e) (x + 5)² - (2x - 1)²
= (x + 5 - 2x + 1)(x + 5 + 2x - 1)
= (6 - x)(3x + 4)
Bài 4
a) x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
b) (x - 4)² - 36 = 0
(x - 4 - 6)(x - 4 + 6) = 0
(x - 10)(x + 2) = 0
x - 10 = 0 hoặc x + 2 = 0
*) x - 10 = 0
x = 10
*) x + 2 = 0
x = -2
Vậy x = -2; x = 10
c) x² - 10x = -25
x² - 10x + 25 = 0
(x - 5)² = 0
x - 5 = 0
x = 5
d) x² + 5x + 6 = 0
x² + 2x + 3x + 6 = 0
(x² + 2x) + (3x + 6) = 0
x(x + 2) + 3(x + 2) = 0
(x + 2)(x + 3) = 0
x + 2 = 0 hoặc x + 3 = 0
*) x + 2 = 0
x = -2
*) x + 3 = 0
x = -3
Vậy x = -3; x = -2
\(a.2x^3+6x=2x\left(x^2+3\right)\)
\(=2x\left(x^2+3\right)-2x\left(x^2+3\right)\)
\(=\left(x^2+3\right)\left(2x-2x\right)\)
\(b.5x\left(x-2\right)-3x^2\left(x-2\right)\)
\(=\left(x-2\right)\left(5x-3x^2\right)\)
\(c.3x\left(x-5y\right)-2y\left(5y-x\right)\)
\(=3x\left(x-5y\right)+2\left(x-5y\right)\)
\(=\left(x-5y\right)\left(3x+2\right)\)
\(d.y^2\left(x^2+y\right)-x^3-xy\)
\(=y^2\left(x^2+y\right)-x\left(x^2+y\right)\)
\(=\left(x^2+y\right)\left(y^2-x\right)\)
e. Cái bài này ghi lại đàng hoàng xíu nha t k hỉu
\(f.3x^2\left(y^2-2x\right)-15x\left(2x-y^2\right)\)
\(=3x^2\left(y^2-2x\right)+15x\left(y^2-2x\right)\)
\(=\left(y^2-2x\right)\left(3x^2+15x\right)\)
Bạn cần viết đề bằng công thức toán để được hỗ trợ tốt hơn.
Bài 12:
1) A = x2 - 6x + 11
= (x2 - 6x + 9) + 2
= (x - 3)2 + 2
Ta có: (x - 3)2 ≥ 0 ∀ x
Dấu ''='' xảy ra khi x - 3 = 0 ⇔ x = 3
Do đó: (x - 3)2 + 2 ≥ 2
Hay A ≥ 2
Dấu ''='' xảy ra khi x = 3
Vậy Min A = 2 tại x = 3
2) B = x2 - 20x + 101
= (x2 - 20x + 100) + 1
= (x - 10)2 + 1
Ta có: (x - 10)2 ≥ 0 ∀ x
Dấu ''='' xảy ra khi x - 10 = 0 ⇔ x = 10
Do đó: (x - 10)2 + 1 ≥ 1
Hay B ≥ 1
Dấu ''='' xảy ra khi x = 10
Vậy Min B = 1 tại x = 10
a: A=y(x-4)-5(x-4)
=(x-4)(y-5)
Khi x=14 và y=5,5 thì A=(14-4)(5,5-5)=0,5*10=5
b: \(B=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\)
Khi x=5,2 và y=4,8 thì B=(5,2+4,8)(5,2-5)
=0,2*10=2
d: Khi x=5,75 và y=4,25 thì
D=5,75^3-5,75^2*4,25+4,25^3
=8087/64
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`
\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)
\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)
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.