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PT
2
16 tháng 10 2020
a) 2x - x3 + 4y - 8y3
= ( 2x + 4y ) - ( x3 + 8y3 )
= 2( x + 2y ) - ( x + 2y )( x2 - 2xy + 4y2 )
= ( x + 2y )( 2 - x2 + 2xy - 4y2 )
b) -3x2 + 11x + 14
= -3x2 - 3x + 14x + 14
= -3x( x + 1 ) + 14( x + 1 )
= ( x + 1 )( 14 - 3x )
16 tháng 10 2020
a) 2x - x3 + 4y - 8y3
= (2x + 4y) - (x3 + 8y3)
= 2 (x + y) - [x3 + (2y)3]
= 2 (x + y) - (x + y)(x2 - 2xy + 4y2)
= (x + y)( 2 - x2 + 2xy - 4y2) (Thật sự là câu này mình vẫn chưa chắc chắn lắm =)))
b) -3x2 + 11x + 14
= -3x2 - 3x + 14x + 14
= (-3x2 - 3x) + (14x + 14)
= -3x(x + 1) + 14(x + 1)
= (-3x + 14)(x + 1)
=))
9 tháng 10 2021
a) \(x^2-2x-4y^2-4y=\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x-1-2y-1\right)\left(x-1+2y+1\right)\)
\(=\left(x-2y-3\right)\left(x+2y\right)\)
b) \(x^2-4x^2y^2+y^2+2xy=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-4x^2y^2=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
c) \(x^6-x^4+2x^3+2x^2=\left(x^6+2x^3+1\right)-\left(x^4-2x^2+1\right)\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2=\left(x^3+1-x^2+1\right)\left(x^3+1+x^2-1\right)=x^2\left(x^3-x^2+2\right)\left(x+1\right)\)
d) \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-8y^3=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)
HN
14 tháng 12 2017
a) \(\dfrac{4y^2}{11x^4}:\left(-\dfrac{8y}{33x^2}\right)\)
\(=\dfrac{4y^2}{11x^4}.\left(-\dfrac{33x^2}{8y}\right)\)
\(=-\dfrac{4y^2.33x^2}{11x^4.8y}\)
\(=-\dfrac{3y}{2x^2}\)
b) \(\dfrac{x^2-4}{3x+12}.\dfrac{x+4}{2x-4}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}.\dfrac{x+4}{2\left(x-2\right)}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)\left(x+4\right)}{3\left(x+4\right).2\left(x-2\right)}\)
\(=\dfrac{x+2}{6}\).
ET
4 tháng 12 2018
a) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}=\dfrac{15x.2y^2}{7y^3.x^2}=\dfrac{30}{7xy}\)
b) \(\dfrac{4y^2}{11x^4}.\left(-\dfrac{3x^2}{8y}\right)=\dfrac{-4y^2.3x^2}{11x^4.8y}=\dfrac{-3y}{22x^2}\)
c) \(\dfrac{x^3-8}{5x+20}.\dfrac{x^2+4x}{x^2+2x+4}\\ =\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{5\left(x+4\right)}.\dfrac{x\left(x+4\right)}{x^2+2x+4}\\ =\dfrac{x^2-2x}{5}\)
KV
23 tháng 2 2023
A) 3x² - x(3x - 5) = 9
3x² - 3x² + 5x = 9
5x = 9
x = 9/5
--------------------
B) 5x² + 9x - 2 = 0
5x² + 10x - x - 2 = 0
(5x² + 10x) - (x + 2) = 0
5x(x + 2) - (x + 2) = 0
(x + 2)(5x - 1) = 0
x + 2 = 0 hoặc 5x - 1 = 0
*) x + 2 = 0
x = -2
*) 5x - 1 = 0
5x = 1
x = 1/5
Vậy x = -2; x = 1/5
---------------------
D) 4(5 - 3x) = 5x - 5
20 - 12x = 5x - 5
-12x - 5x = -5 - 20
-17x = -25
x = 25/17
--------------------
E) 2x² - 11x + 14 = 0
2x² - 4x - 7x + 14 = 0
(2x² - 4x) - (7x - 14) = 0
2x(x - 2) - 7(x - 2) = 0
(x - 2)(2x - 7) = 0
x - 2 = 0 hoặc 2x - 7 = 0
*) x - 2 = 0
x = 2
*) 2x - 7 = 0
2x = 7
x = 7/2
Vậy x = 2; x = 7/2
4 tháng 10 2021
1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)
\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)
\(=x^2+4x+3-\dfrac{12}{2x+3}\)
6 tháng 8 2021
a, \(x-2y+x^2-4y^2=\left(x-2y\right)+\left(x-2y\right)\left(x+2y\right)=\left(x-2y\right)\left(1+x+2y\right)\)
b, \(x^2-4x^2y^2+y^2+2xy=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
c, \(x^6-x^4+2x^3+2x^2=x^6+2x^3+1-x^4+2x^2-1\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2=\left(x^3-x^2+2\right)\left(x^3+x^2\right)\)
\(=x^2\left(x+1\right)\left(x^3-x^2+2\right)\)
d, \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-\left(2y\right)^3=\left(x+1-2y\right)\left(x+1+2y\right)\)
6 tháng 8 2021
a) Ta có: \(x-2y+x^2-4y^2\)
\(=\left(x-2y\right)+\left(x-2y\right)\left(x+2y\right)\)
\(=\left(x-2y\right)\left(1+x+2y\right)\)
b: Ta có: \(x^2-4x^2y^2+y^2+2xy\)
\(=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
a) 2x - x3 + 4y - 8y3
= ( 2x + 4y ) - ( x3 + 8y3 )
= 2( x + 2y ) - ( x + 2y )( x2 - 2xy + 4y2 )
= ( x + 2y )( 2 - x2 + 2xy - 4y2 )
b) -3x2 + 11x + 14
= -3x2 + 14x - 3x + 14
= -x( 3x - 14 ) - ( 3x - 14 )
= ( 3x - 14 )( -x - 1 )