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a) 25x² - 16
= (5x)² - 4²
= (5x - 4)(5x + 4)
b) 16a² - 9b²
= (4a)² - (3b)²
= (4a - 3b)(4a + 3b)
c) 8x³ + 1
= (2x)³ + 1³
= (2x + 1)(4x² - 2x + 1)
d) 125x³ + 27y³
= (5x)³ + (3y)³
= (5x + 3y)(25x² - 15xy + 9y²)
e) 8x³ - 125
= (2x)³ - 5³
= (2x - 5)(4x² + 10x + 25)
g) 27x³ - y³
= (3x)³ - y³
= (3x - y)(9x² + 3xy + y²)
a) \(25x^2-16=\left(5x-4\right)\left(5x+4\right)\)
b) \(16a^2-9b^2=\left(4a-3b\right)\left(4a+3b\right)\)
c) \(8x^3+1=\left(2x+1\right)\left(4x^2-2x+1\right)\)
d) \(125x^3+27y^3=\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)
e) \(8x^3-125=\left(2x-5\right)\left(4x^2-10x+25\right)\)
g) \(27x^3-y^3=\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
a: \(\left(4x^2+12xy+9y^2\right):\left(2x+3y\right)=\left(2x+3y\right)^2:\left(2x+3y\right)=2x+3y\)
d: \(\left(x^2+6xy+9y^2\right):\left(x+3y\right)=\left(x+3y\right)^2:\left(x+3y\right)=x+3y\)
e: \(\dfrac{64y^3-27}{4y-3}=\dfrac{\left(4y-3\right)\left(16y^2+12y+9\right)}{4y-3}=16y^2+12y+9\)
a, \(4x^2+12xy+9y^2=\left(2x+3y\right)^2\)
\(\Rightarrow\left(4x^2+12xy+9y^2\right):\left(2x+3y\right)\)
\(=\left(2x+3y\right)^2:\left(2x+3y\right)\\ =2x+3y\)
b,\(x^2+6xy+9y^2=\left(x+3y\right)^2\)
\(\Rightarrow\left(x^2+6xy+9y^2\right):\left(x+3y\right)\\ =\left(x+3y\right)^2:\left(x+3y\right)\\ =x+3y\)
c, \(64y^3-27=\left(4y-3\right)\left(16y^2+12y+9\right)\)
\(\Rightarrow\left(64x^3-27\right):\left(4y-3\right)\\ =\left[\left(4y-3\right)\left(16x^2+12x+9\right)\right]:\left(4y-3\right)\\ =16x^2+12x+9\)
1: \(=-\left(x^2+2x+2\right)=-\left(x^2+2x+1+1\right)=-\left(x+1\right)^2-1< =-1\)
Dấu '=' xảy ra khi x=-1
2: \(=-\left(4x^2-12x-10\right)\)
\(=-\left(4x^2-12x+9-19\right)\)
\(=-\left(2x-3\right)^2+19< =19\)
Dấu '=' xảy ra khi x=3/2
3: \(=-\left(x^2+4x+4-4\right)=-\left(x+2\right)^2+4< =4\)
Dấu '=' xảy ra khi x=-2
h, \(27x^3-8=\left(3x-2\right)\left(9x^2+6x+4\right)\)
\(\Rightarrow\left(27x^3-8\right):\left(3x-2\right)\\ =\left(3x-2\right)\left(9x^2+6x+4\right):\left(3x-2\right)\\ =9x^2+6x+4\)
g, \(x^4-2x^2+1=\left(x^2-1\right)^2\)
\(\Rightarrow\left(x^4-2x^2+1\right):\left(1-x^2\right)\\ =\left(x^2-1\right)^2:\left(1-x^2\right)\\ =x^2-1\)
Giải:
1) \(\left(x^2-y\right)^3\)
\(=x^6-3x^4y+4x^2y^2-y^3\)
Vậy ...
2) \(\left(x-2+y\right)^3\)
\(=\left(x-2\right)^3+3\left(x-2\right)^2y+3\left(x-2\right)y^2+y^3\)
\(=x^3-3x^2+16x-2^3+3\left(x^2-4x-4\right)y+3\left(x-2\right)y^2+y^3\)
\(=x^3-3x^2+16x-2^3+3x^2-12x-12y+3\left(xy^2-2y^2\right)+y^3\)
\(=x^3-3x^2+16x-2^3+3x^2-12x-12y+3xy^2-6y^2+y^3\)
\(=x^3+4x-8-12y+3xy^2-6y^2+y^3\)
Vậy ...
3) \(\left(z+y^2\right)^3\)
\(=z^3+3z^2y^2+3zy^4+y^6\)
Vậy ...
4) \(\left(x-y+z\right)^3\)
\(=\left(x-y\right)^3+3\left(x-y\right)^2z+3\left(x-y\right)z^2+z^3\)
\(=x^3-3x^2y+3xy^2-y^3+3\left(x^2-2xy+y^2\right)z+3\left(xz^2-yz^2\right)+z^3\)
\(=x^3-3x^2y+3xy^2-y^3+3x^2-6xy+3y^2z+3xz^2-3yz^2+z^3\)
\(=-3x^2y+3xy^2-y^3+4x^2-6xy+3y^2z+3xz^2-3yz^2+z^3\)
Vậy ...
a)\(x^3+8\)
= \(x^3+2^3\)
= (x + 2)(\(x^2-2x+4\))
b) \(27+m^3\)
= \(3^3+m^3\)
= (3 + m)(9 - 3m + \(m^2\))
c)\(x^3-8\)
= \(x^3-2^3\)
= (x - 2)(\(x^2+2x+4\))
d) \(27-y^3\)
= \(3^3-y^3\)
= (3 - y)(9 + 3y + \(y^2\))
a. \(x^3+8=x^3+2^3=x^3+6x^2+12x+8\)
b. \(27+m^3=3^3+m^3=9+27m+9m^2+m^3\)
c. \(x^3-8=x^3-2^3=x^3-6x^2+12x-8\)
d. \(27-y^3=3^3+y^3=9-27y+9y^2-y^3\)