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13 tháng 8 2016

a) = (3x)^2 + 2.3x.5+ 5^2 = (3x+5)^2

b) = (2/3x)^2-(4y)^2=(2/3x-4y)(2/3x+4y)

c) = -(9x^4-12/5x^2y^2+4/25y^4) = -[(3x^2)^2 - 2.3x^2.2/5y^2 + (2/5y^2)^2]= -(3x^2-2/5y^2)^2

d) = (x-5)^2 -  4^2= (x-5+4)(x-5-4) = (x-1)(x-9)

e) = (2x)^3 + 3.(2x)^2.(5y) + 3.(2x).(5y)^2 + (5y)^3 = (2x+5y)^3

f) = (8x)^2 - (8a+b)^2 = (8x-8a-b)(8x+8a+b)

g) = (7x-4-2x-1)(7x-4+2x+1) = (5x-5)(9x-3) = 5(x-1).3(x-3)=15(x-1)(x-3)

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

# Chúc bạn học tốt #

13 tháng 8 2016

Cảm ơn bạn!

16 tháng 10 2016

hoa mắt chóng mặt

16 tháng 10 2016

Nhờ bạn làm cho mik ít câu cũng dc

16 tháng 10 2016

sao nhiều thế bạn

26 tháng 8 2017

quá nhiều

3 tháng 8 2023

a) 9x4+16y6-24x2y3

=(3x2)2-2.3x2.4y3+(4y3)2

=(3x2-4y3)2

b) 16x2-24xy+9y2

=(4x)2-2.4x.3y+(3y)2

=(4x-3y)2

c) 36x2-(3x-2)2

=(36x-3x+2)(36x+3x-2)

=(33x+2)(39x-2)

d) 27x3+54x2y+36xy2+8y3

=(3x)3+3.(3x)2.2y+3.3x.(2y)2+(2y)3

=(3x+2y)3

e) y9-9x2y6+27x4y3-27x6

=(y3)3-3.(y3)2.3x2+3.y3.(3x2)2-(3x2)3

=(y3-3x2)3

f) 64x3+1

= (4x)3+13

=(4x+1)[(4x)2-4x.1+12]

=(4x+1)(16x2-4x+1)

e) 27x6-8x3  *sửa đề*

=(3x2)3-(2x)3

=(3x2-2x)[(3x)2+3x2.2x+(2x)2]

=(3x2-2x)(9x2+6x3+4x2)

~~~

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 11 2023

\(a,3x-15xy=3x\left(1-5y\right)\\ ---\\ 8x^2+6x-4=2\left(4x^2+3x-2\right)\\ ---\\ 5x^2+25xy+10y^2=5\left(x^2+5xy+2y^2\right)\\ ---\\ 9x^2y^2+6x^2y-\dfrac{1}{2}xy^2=\dfrac{1}{2}xy\left(18xy+12x-y\right)\)

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

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

c: \(\dfrac{12}{5}x^2y^2-9x^4-\dfrac{4}{25}y^4\)

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

\(=-\left(3x^2-\dfrac{2}{5}y^2\right)^2\)

a: =(6x)^2-(3x-2)^2

=(6x-3x+2)(6x+3x-2)

=(9x-2)(3x+2)

d: \(=\left[\left(x+1\right)^2-\left(x-1\right)^2\right]\left[\left(x+1\right)^2+\left(x-1\right)^2\right]\)

\(=4x\cdot\left[x^2+2x+1+x^2-2x+1\right]\)

=8x(x^2+1)

e: =(4x)^2-2*4x*3y+(3y)^2

=(4x-3y)^2

f: \(=-\left(\dfrac{1}{4}x^4-2\cdot\dfrac{1}{2}x^2\cdot2y^3+4y^6\right)\)

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

g: =(4x)^3+1^3

=(4x+1)(16x^2-4x+1)

k: =x^3(27x^3-8)

=x^3(3x-2)(9x^2+6x+4)

l: =(x^3-y^3)(x^3+y^3)

=(x-y)(x+y)(x^2-xy+y^2)(x^2+xy+y^2)

17 tháng 12 2023

Bài 1

a) 5x²y - 20xy²

= 5xy(x - 4y)

b) 1 - 8x + 16x² - y²

= (1 - 8x + 16x²) - y²

= (1 - 4x)² - y²

= (1 - 4x - y)(1 - 4x + y)

c) 4x - 4 - x²

= -(x² - 4x + 4)

= -(x - 2)²

d) x³ - 2x² + x - xy²

= x(x² - 2x + 1 - y²)

= x[(x² - 2x+ 1) - y²]

= x[(x - 1)² - y²]

= x(x - 1 - y)(x - 1 + y)

= x(x - y - 1)(x + y - 1)

e) 27 - 3x²

= 3(9 - x²)

= 3(3 - x)(3 + x)

f) 2x² + 4x + 2 - 2y²

= 2(x² + 2x + 1 - y²)

= 2[(x² + 2x + 1) - y²]

= 2[(x + 1)² - y²]

= 2(x + 1 - y)(x + 1 + y)

= 2(x - y + 1)(x + y + 1)

17 tháng 12 2023

Bài 2:

a: \(x^2\left(x-2023\right)+x-2023=0\)

=>\(\left(x-2023\right)\left(x^2+1\right)=0\)

mà \(x^2+1>=1>0\forall x\)

nên x-2023=0

=>x=2023

b: 

ĐKXĐ: x<>0

\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)

=>\(-x\left(x-4\right)+2x^2-4x-9=0\)

=>\(-x^2+4x+2x^2-4x-9=0\)

=>\(x^2-9=0\)

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

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

c: \(x^2+2x-3x-6=0\)

=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)

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

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

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

d: 3x(x-10)-2x+20=0

=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)

=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)

=>\(\left(x-10\right)\left(3x-2\right)=0\)

=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)

Câu 1:

a: \(5x^2y-20xy^2\)

\(=5xy\cdot x-5xy\cdot4y\)

\(=5xy\left(x-4y\right)\)

b: \(1-8x+16x^2-y^2\)

\(=\left(16x^2-8x+1\right)-y^2\)

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

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

c: \(4x-4-x^2\)

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

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

d: \(x^3-2x^2+x-xy^2\)

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

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

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

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

e: \(27-3x^2\)

\(=3\left(9-x^2\right)\)

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

f: \(2x^2+4x+2-2y^2\)

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

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

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

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