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MC
1
TT
3
9 tháng 7 2020
Bài làm:
Ta có: \(9\left(x-y\right)^2-4\left(x+y\right)^2=\left[3\left(x-y\right)\right]^2-\left[2\left(x+y\right)\right]^2\)
\(=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)=\left(x-5y\right)\left(5x-y\right)\)
Học tốt!!!!
9 tháng 7 2020
Ta có :
\(9\left(x-y\right)^2-4\left(x+y\right)^2=9x^2-18xy+9y^2-4x^2-8xy-4y^2\)
\(=5x^2-26xy+5y^2==\left(5x-y\right)\left(x-5y\right)\)
HL
1
KK
14 tháng 9 2021
x2(x - y) + 49(y - x)
= x2(x - y) - 49(x - y)
= (x2 - 49)(x - y)
= (x - 7)(x + 7)(x - y)
15 tháng 7 2019
1)\(36\left(x-y\right)^2-49\left(x+y\right)^2\)
\(=\left(6x-6y+7x+7y\right)\left(6x-6y-7x-7y\right)\)
\(=\left(13x+y\right)\left(-x-13y\right)\)
\(=-\left(13x+y\right)\left(x+13y\right)\)
2)\(16x^2-9\left(x+y\right)^2=\left(4x+3x+3y\right)\left(4x-3x-3y\right)=\left(7x+3y\right)\left(x-3y\right)\)
12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
CH
1
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 8 2021
\(=\left(3x-3y\right)^2-\left(2x+2y\right)^2=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\)
\(=\left(x-5y\right)\left(5x-y\right)\)
NH
3
TA
1 tháng 8 2021
`9(x-y)^2-4(x+y)^2`
`=[3(x-y)]^2-[2(x+y)]^2`
`=(3x-3y)^2-(2x+2y)^2`
`=(3x-3y+2x+2y)(3x-3y-2x-2y)`
`=(5x-y)(x-5y)`
1 tháng 8 2021
\(9\left(x-y\right)^2-4\left(x+y\right)^2\\ =\left[3\left(x-y\right)\right]^2-\left[2\left(x+y\right)\right]^2\\ =\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\)
19 tháng 9 2016
a)27x2.(y-1)9x3.(1-y)
=27x2.(y-1)+9x3.(y-1)
=9x2(y-1)[3+x]
b)8x3 + 1/27
=(2x)3 + (\(\frac{1}{3}\))3
= (2x+\(\frac{1}{3}\))(\(\left(2x\right)^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2\)
= (2x+\(\frac{1}{3}\))(\(\left(2x\right)^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2\)
c)49.(y-4)2-9.(y+2)2
=[7(y-4)]2-[3(y+2)]2
=[7(y-4)+3(y+2)][7(y-4)-3(y+2)]
=(7y-28+3y+6)(7y-28-3y-6)
=(10y-22)(4y-34)
19 tháng 9 2016
a) 27x2 ( y - 1) - 9x3 ( 1 - y)
=27x2 (y-1) + 9x3 ( y - 1 )
= (27x2 + 9x3) ( y -1 )
=9x2 ( x + 3) ( y - 1)
b)8x3+1/27
\(=\left(2x\right)^3+\left(\frac{1}{3}\right)^3\)
\(=\left(\frac{2x}{9}+\frac{1}{27}\right)\left(36x^2-6x+1\right)\)
c)49 ( y - 4 )2 - 9 ( y + 2)2
= [7(y - 4)]2 - [3(y + 2)]2
= (7y - 28 + 3y + 6)(7y - 28 - 3y - 6)
= (10y - 22)(4y - 34)
= 4(5y - 11)(2y - 34)
\(49\left(y-4\right)^2-9\left(y+2\right)^2\)
\(=7^2\left(y-4\right)^2-3^2\left(y+2\right)^2\)
\(=\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2\)
\(=\left[7\left(y-4\right)-3\left(y+2\right)\right]\left[7\left(y-4\right)+3\left(y+2\right)\right]\)
\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)\)
\(=\left(4y-34\right)\left(10y-22\right)\)
`49(y-4)²-9(y+2)²`
`= [7(y-4)]²-[3(y+2)]²`
`= [7(y-4)+3(y+2)]-[7(y-4)-3(y+2)]`
`= (7y-28+3y+6)(7y-28-3y-6)`
`= (10y-22)(4y-34)`