K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

HQ
Hà Quang Minh
Giáo viên
12 tháng 1

1. 

\({x^3} - 8 = {x^3} - {2^3} = \left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)\)
HQ
Hà Quang Minh
Giáo viên
12 tháng 1

2.

 \(\begin{array}{l}\left( {3x - 2y} \right)\left( {9{x^2} + 6xy + 4{y^2}} \right) + 8{y^3}\\ = \left( {3x - 2y} \right)\left[ {{{\left( {3x} \right)}^2} + 3x.2y + {{\left( {2y} \right)}^2}} \right] + 8{y^3}\\ = {\left( {3x} \right)^3} - {\left( {2y} \right)^3} + 8{y^3}\\ = 27{x^3} - 8{y^3} + 8{y^3}\\ = 27{x^3}\end{array}\)

HQ
Hà Quang Minh
Giáo viên
12 tháng 1
\({x^3} + 27 = {x^3} + {3^3} = \left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right)\)
HQ
Hà Quang Minh
Giáo viên
12 tháng 1
\({x^3} + 8{y^3} - \left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right) = {x^3} + 8{y^3} - \left[ {{x^3} + {{\left( {2y} \right)}^3}} \right] = {x^3} + 8{y^3} - \left( {{x^3} + 8{y^3}} \right) = 0\)
22 tháng 12 2023

Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

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

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

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

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

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

\(=3xy\left(x^2y-2\right)\)

2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)

\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)

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

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

\(=x^2-\left(y^2+6y+9\right)\)

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

\(=\left(x-y-3\right)\left(x+y+3\right)\)

5: \(\left(3x-y\right)^2-4y^2\)

\(=\left(3x-y\right)^2-\left(2y\right)^2\)

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

\(=\left(3x-3y\right)\left(3x+y\right)\)

\(=3\left(x-y\right)\left(3x+y\right)\)

6: \(4x^2-9y^2-4x+1\)

\(=\left(4x^2-4x+1\right)-9y^2\)

\(=\left(2x-1\right)^2-\left(3y\right)^2\)

\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)

8: \(x^2y-xy^2-2x+2y\)

\(=xy\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(xy-2\right)\)

9: \(x^2-y^2-2x+2y\)

\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)

\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-2\right)\)

HQ
Hà Quang Minh
Giáo viên
12 tháng 1

a)      \({x^3} + 512 = \left( {x + 8} \right)\left( {{x^2} - 8x + 64)} \right)\)

b)      \(27{x^3} - 8{y^3} = \left( {3x - 2y} \right)\left( {9{x^2} + 6xy + 4{y^2}} \right)\)

15 tháng 7 2021

a. \(9x^2+30x+25=\left(3x+5\right)^2\)

b. \(\dfrac{4}{9}x^4-16x^2=\left(\dfrac{2}{3}x^2-4x\right)\left(\dfrac{2}{3}x^2+4x\right)=x^2\left(\dfrac{2}{3}x-4\right)\left(\dfrac{2}{3}x+4\right)\)

c. \(a^2y^2+b^2x^2-2axby=\left(ay-bx\right)^2\)

d. \(100-\left(3x-y\right)^2=\left(10-3x+y\right)\left(10+3x-y\right)\)

e. \(\dfrac{12}{5}x^2y^2-9x^4-\dfrac{4}{25}y^4=-\left(9x^4-\dfrac{12}{5}x^2y^2+\dfrac{4}{25}y^4\right)=-\left(3x^2-\dfrac{2}{5}y^2\right)^2\)

f. \(64x^2-\left(8a+b\right)^2=\left(8x-8a-b\right)\left(8x+8a+b\right)\)

g. \(27x^3-a^3b^3=\left(3x-ab\right)\left(9x^2+3xab+a^2b^2\right)\)

28 tháng 6 2016

1)  2xy2+x2y4+1=(xy2)2+2xy2.1+12=(xy2+1)2

2)

a)2(x-y)(x+y)+(x+y)2+(x-y)2=(x+y+x-y)2=(2x)2=4x2

b)(x-y+z)2+(z-y)2+2(x-y+z)(y-z)

=(x-y+z)2+(y-z)2+2(x-y+z)(y-z)

=(x-y+z+y-z)2

=x2

3 tháng 12 2016

chịch chịch chịch

22 tháng 7 2023

a) \(\left(3x-5\right)\left(3x+5\right)\)

\(=\left(3x\right)^2-5^2\)

\(=9x^2-25\)

b) \(\left(x-2y\right)\left(x+2y\right)\)

\(=x^2-\left(2y\right)^2\)

\(=x^2-4y^2\)

c) \(\left(-x-\dfrac{1}{2}y\right)\left(-x+\dfrac{1}{2}y\right)\)

\(=\left(-x\right)^2-\left(\dfrac{1}{2}y\right)^2\)

\(=x^2-\dfrac{1}{4}y^2\)

`a, (3x-5)(3x+5) = 9x^2 - 25`

`b, (x-2y)(x+2y) = x^2 -4y^2`

`c, (-x-1/2y)(-x+1/2y) = x^2 - 1/4y^2`

f: \(x^2y^2+2xy+1=\left(xy+1\right)^2\)

g: \(\left(3x-2y\right)^2+2\left(3x-2y\right)+1=\left(3x-2y+1\right)^2\)

h: \(\left(x-3y\right)^2-8\left(x-3y\right)+16=\left(x-3y-4\right)^2\)

i: \(\left(x+y\right)^2+2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)

\(=\left(x+y+x-y\right)^2=4x^2\)

12 tháng 7 2017

a) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24\)

\(=24-11x\)

b) \(\left(4x^2-3y\right)\cdot2y-\left(3x^2-4y\right)\cdot3y\)

\(=8x^2y-6y^2-9x^2y+12y^2\)

\(=6y^2-x^2y\)

c) \(3y^2\left[\left(2x-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)

\(=3y^2\cdot\left(2x-1+y+1\right)-y\cdot\left(1-y-y^2\right)+y\)

\(=6xy^2-3y^2+3y^3+3y^2-y+y^2+y^3+y\)

\(=4y^3+y^2+6xy^2\)