1. Nhân
a. (-3x + 2\(\frac{y}{x}\))(2yx-2xy)
b.(-3x - 2)(9\(x^2\)-6x + 4)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 10 2023
Bài 1:
a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)
= \(x^2\) - 16 - 5\(x\) - 5 + \(x^2\) + \(x\)
= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)
= 2\(x^2\) - 4\(x\) - 21
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 10 2023
b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)
= 3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7
= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)
= - 3\(x^2\) + 3\(xy\) - 3
K
0
TL
20 tháng 8 2021
a) \(3xy-6xy^2=3xy\left(1-2y\right)\)
b) \(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)
c) \(x^3-x^2+2\)
d) \(x^2+4x+4-y^2=\left(x^2+4x+4\right)-y^2=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\)
e) \(x^3+4x^2+4x=x\left(x^2+4x+4\right)=x\left(x+2\right)^2\)
f) \(x^2+2x+1-9y^2=\left(x+1\right)^2-\left(3y\right)^2=\left(x-3y+1\right)\left(x+3y+1\right)\)
g) \(6x^2-12x=6x\left(x-2\right)\)
h) \(x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
i) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
12 tháng 1 2023
a: =12x^3y^2-12x^3y^3+6x^2y^2
b: =\(\left(-3x+2\right)\left(5x^2-\dfrac{1}{3}x+4\right)\)
=-15x^3+x^2-12x+10x^2-2/3x+8
=-15x^3+11x^2-38/3x+8
c: =x^2-x-2+3x-x^2
=2x-2
18 tháng 6 2017
bài 1 :
a) 6(x+1)2 - (x-3)(x2 + 3x +9) + (x-2)2
= 6( x2 + 2x + 1 ) - (x3 + 3x2 + 9x - 3x2 - 9x - 27 ) + x2 - 4x + 4
= 6x2 + 12x + 6x - x3 - 3x2 - 9x + 3x2 + 9x + 27 + x2 - 4x + 4
= -x3 + 7x2 + 14x + 31 (1)
Thay x = 2 vào biểu thức (1) ta được :
\(\left(-2\right)^3+7.2^2+14.2+31\) = 79
Vậy với x = 2 giá trị của biểu thức (1) là 79
b) \(\left(2x-1\right)\left(3x+1\right)+\left(3x-4\right)\left(3-2x\right)\)
= 6x2 + 2x - 3x - 1 + 9x - 6x2 - 12 + x
= 9x - 13 (2)
Thay x= \(\dfrac{9}{8}\) Vào biểu thức (2) ta được :
9.\(\dfrac{9}{8}\) - 13 = \(-\dfrac{23}{8}\)
Vậy với x = 9/8 giá trị của biểu thức (2) là -\(\dfrac{23}{8}\)
18 tháng 11 2019
a)có khả năng sai đề bài
b)Liệu có sai đề bài không
c)\(=\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)(phân số cuối có âm vì (1-x)=-(x-1)
\(=\frac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)(Hơi tắt)
\(=\frac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{1}{x^2+x+1}\)
d)\(=\frac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{x^2+2xy+x^2-2xy+4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x^2+4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x}{x-2y}\)
22 tháng 10 2023
2:
a: \(x^2-12x+20\)
\(=x^2-2x-10x+20\)
=x(x-2)-10(x-2)
=(x-2)(x-10)
b: \(2x^2-x-15\)
=2x^2-6x+5x-15
=2x(x-3)+5(x-3)
=(x-3)(2x+5)
c: \(x^3-x^2+x-1\)
=x^2(x-1)+(x-1)
=(x-1)(x^2+1)
d: \(2x^3-5x-6\)
\(=2x^3-4x^2+4x^2-8x+3x-6\)
\(=2x^2\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x-2\right)\left(2x^2+4x+3\right)\)
e: \(4y^4+1\)
\(=4y^4+4y^2+1-4y^2\)
\(=\left(2y^2+1\right)^2-\left(2y\right)^2\)
\(=\left(2y^2+1-2y\right)\left(2y^2+1+2y\right)\)
f; \(x^7+x^5+x^3\)
\(=x^3\left(x^4+x^2+1\right)\)
\(=x^3\left(x^4+2x^2+1-x^2\right)\)
\(=x^3\left[\left(x^2+1\right)^2-x^2\right]\)
\(=x^3\left(x^2-x+1\right)\left(x^2+x+1\right)\)
g: \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-3\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)\left(x^2+x-2\right)-3\left(x^2+x-2\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x-3\right)\)
\(=\left(x^2+x-3\right)\left(x+2\right)\left(x-1\right)\)
h: \(\left(x^2+2x\right)^2-2\left(x+1\right)^2-1\)
\(=\left(x^2+2x+1-1\right)^2-2\left(x+1\right)^2-1\)
\(=\left[\left(x+1\right)^2-1\right]^2-2\left(x+1\right)^2-1\)
\(=\left(x+1\right)^4-2\left(x+1\right)^2+1-2\left(x+1\right)^2-1\)
\(=\left(x+1\right)^4-4\left(x+1\right)^2\)
\(=\left(x+1\right)^2\left[\left(x+1\right)^2-4\right]\)
\(=\left(x+1\right)^2\left(x+1+2\right)\left(x+1-2\right)\)
\(=\left(x+1\right)^2\cdot\left(x+3\right)\left(x-1\right)\)
i: \(x^2+4xy+4y^2-4\left(x+2y\right)+3\)
\(=\left(x+2y\right)^2-4\left(x+2y\right)+3\)
\(=\left(x+2y\right)^2-\left(x+2y\right)-3\left(x+2y\right)+3\)
\(=\left(x+2y\right)\left(x+2y-1\right)-3\left(x+2y-1\right)\)
\(=\left(x+2y-1\right)\left(x+2y-3\right)\)
j: \(x\cdot\left(x+1\right)\left(x+2\right)\left(x+3\right)-3\)
\(=\left(x^2-3x\right)\left(x^2-3x+2\right)-3\)
\(=\left(x^2-3x\right)^2+2\left(x^2-3x\right)-3\)
\(=\left(x^2-3x+3\right)\left(x^2-3x-1\right)\)
a, \(\left(-3x+\frac{2y}{x}\right)\left(2yx-2xy\right)=\left(-3x+\frac{2y}{x}\right).0=0\)
b, \(\left(-3x-2\right)\left(9x^2-6x+4\right)=\left(-3x-2\right)\left[\left(3x\right)^2-2.3x+1+3\right]\)
\(=\left(-3x-2\right)\left[\left(3x-1\right)^2+3\right]=\left(-3x-2\right)\left(3x-1\right)^2+3\left(-3x-2\right)\)
a,\(\left(-3x+2\frac{y}{x}\right)\left(2yx-2xy\right)\)
\(=-3x\left(2yx-2xy\right)+2\frac{y}{x}\left(2yz-2xy\right)\)
\(=\left(-6x^2y+6x^2y\right)+\left(\frac{2y^2x}{x}-\frac{2y^2x}{x}\right)\)
\(=0\)
b,\(\left(-3x-2\right)\left(9x^2-6x+4\right)\)
\(=-3x\left(9x^2-6x+4\right)-2\left(9x^2-6x+4\right)\)
\(=-27x^3+18x^2-12x-18x^2+12x-8\)
\(=-27x^3-8\)