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TH
2
6 tháng 9 2017
a) ( 3+ xy2 )2
= 32 + 2.3.xy2 + ( xy2 )2
= 9 + 6xy2 + x2y4
b) ( 10 - 2m2n2 )
= 2 ( 5 - m2n2 )
c) ( a - b2 )( a + b2 )
= a2 + ab2 - ab2 - b4
= a2 - b4
28 tháng 6 2018
a) (3+\(xy^2\))\(^2\)= \(3^2\)+2*3*\(xy^2\)+\(\left(xy^2\right)\)\(^2\)
=9+6\(xy^2\)+\(x^2\)\(y^4\)
b) (a+\(b^2\))(a-\(b^2\))
=a(a-\(b^2\))+\(b^2\)(a-\(b^2\))
=a*a+a*-\(b^2\)+\(b^2\)*a+\(b^2\)*-\(b^2\)
=\(a^2\)-a\(b^2\)+\(b^2\)a-\(b^4\)
=\(a^2\)+(-a\(b^2\)+a\(b^2\))-\(b^4\)
=\(a^2\)-\(b^4\)
c)(2y-1)\(^2\)=(2y)\(^2\)-2*2y*1+1\(^{^2}\)
=4y\(^2\)-4y+1
d)(10-m\(^2\))\(^2\)=10\(^2\)-2*10*m\(^2\)+(m\(^2\))\(^2\)
=100-20m\(^2\)+m\(^4\)
câu e bạn tự làm nha tương tự như câu b
chúc bạn học tốt
NM
2
HH
10 tháng 6 2018
Giải:
a) \(\left(3+xy^2\right)^2\)
\(=3^2+2.3.xy^2+\left(xy^2\right)^2\)
\(=9+6xy^2+x^2y^4\)
Vậy ...
b) \(\left(10-2m^2n\right)^2\)
\(=10^2-2.10.2m^2n+\left(2m^2n\right)^2\)
\(=100-40m^2n+4m^4n^2\)
Vậy ...
c) \(\left(a-b^2\right)\left(a+b^2\right)\)
\(=a^2-\left(b^2\right)^2\)
\(=a^2-b^4\)
Vậy ...
QN
1
DN
25 tháng 8 2018
\(a,\left(3+xy^2\right)^2=9+6xy^2+x^2y^4\)
\(b,\left(10-2m^2n\right)^2=100-40m^2n+4m^4n^2\)
\(c,\left(a-b^2\right)\left(a+b^2\right)=a^2-b^4\)
GP
1
2 tháng 9 2017
\(\left(3+xy\right)^2=9+6xy+xy^2\)
\(\left(10-m^2n\right)^2=100-20m^2n^2+m^4n^2\)
\(\left(a-b^2\right)\left(a+b^2\right)=a^2-b^4\)
8 tháng 10 2017
\(a,4b^2c^2-\left(b^2+c^2-a^2\right)^2\\ =\left(2bc\right)^2-\left(b^2+c^2-a^2\right)^2\\ =\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\\ =-\left(\left(b-c\right)^2-a^2\right).\left(\left(b+c\right)^2-a^2\right)\\ =\left(-b+c+a\right)\left(b-c+a\right)\left(b+c-a\right)\left(b+c+a\right)\)
Các câu sau hầu như bn dùng HĐT số 2 nhóm vào
8 tháng 10 2017
Cacs câu hầu như đều là dùng hằng đẳng thưc shieeuj hai bình phương ,mk lm mẫu ba câu đầu nha bn,nếu mà các câu sau ko lm đc ,thì bn bảo mk nha ?
19 tháng 9 2018
Đăng từng bài thui bn êi ~.~
\(h)\)\(\left(xy+1\right)^2-\left(x+y\right)^2\)
\(=\)\(\left(xy-x-y+1\right)\left(xy+x+y+1\right)\)
\(=\)\(\left[x\left(y-1\right)-\left(y-1\right)\right].\left[x\left(y+1\right)+\left(y+1\right)\right]\)
\(=\)\(\left(x-1\right)\left(y-1\right)\left(x+1\right)\left(y+1\right)\)
\(i)\)\(16b^2c^2-4\left(b^2+c^2-a^2\right)^2\)
\(=\)\(\left(4bc\right)^2-\left(2b^2+2c^2-2a^2\right)^2\)
\(=\)\(\left(4bc-2b^2-2c^2+2a^2\right)\left(4bc+2b^2+2c^2-2a^2\right)\)
\(=\)\(2\left[a^2-\left(b^2-2bc+c^2\right)\right].2\left[\left(b^2+2bc+c^2\right)-a^2\right]\)
\(=\)\(-4\left[a^2-\left(b-c\right)^2\right].\left[a^2-\left(b+c\right)^2\right]\)
\(=\)\(-4\left(a-b+c\right)\left(a+b-c\right)\left(a-b-c\right)\left(a+b+c\right)\)
Chúc bạn học tốt ~
1 tháng 8 2017
\(1.\)
\(x^3z+x^2yz-x^2z^2-xyz^2\)
\(=x^3z-x^2z^2+x^2yz-xyz^2\)
\(=x^2z\left(x-z\right)-xyz\left(x-z\right)\)
\(=\left(x^2z-xyz\right)\left(x-z\right)\)
\(=xz\left(x-y\right)\left(x-z\right)\)
\(2.\)
\(x^2-\left(a+b\right)xy+aby^2\)
\(=x^2-axy-bxy+aby^2\)
\(=x^2-bxy-axy+aby^2\)
\(=x\left(x-by\right)-ay\left(x-by\right)\)
\(=\left(x-ay\right)\left(x-by\right)\)
\(3.\)
\(ab\left(x^2+y^2\right)+xy\left(x^2+y^2\right)\)
\(=abx^2+aby^2+a^2xy+b^2xy\)
\(=abx^2+b^2xy+a^2xy+aby^2\)
\(=bx\left(ax+by\right)+ay\left(ax+by\right)\)
\(=\left(ax+by\right)\left(bx+ay\right)\)
\(4.\)
\(\left(xy+ab\right)^2+\left(ay-bx\right)^2\)
\(=x^2y^2+2abxy+a^2b^2+a^2y^2-2aybx+b^2x^2\)
\(=x^2y^2+a^2b^2+a^2y^2+b^2x^2\)
\(=x^2y^2+b^2x^2+a^2b^2+a^2y^2\)
\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)
\(=\left(a^2+x^2\right)\left(b^2+y^2\right)\)
\(5.\)
\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2b-a^2c+b^2c-ab^2+ac^2-bc^2\)
\(=a^2b-ab^2-a^2c-b^2c+ac^2-bc^2\)
\(=ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)\)
\(=ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)
\(=\left(a-b\right)\left(ab-bc-ac+c^2\right)\)
\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)
\(=\left(a-c\right)\left(b-c\right)\left(a-c\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
\(6.\)
\(16x^2-40xy+2y^2\)
\(=\left(4x\right)^2-2\cdot4\cdot5xy+\left(5y\right)^2\)
\(=\left(4x-5y\right)^2\)
\(7.\)
\(25x^4-10x^2y+y^2\)
\(=\left(5x^2\right)^2-2\cdot5x^2y+y^2\)
\(=\left(5x^2+y\right)^2\)
\(8.\)
\(-16x^4y^6-24x^5y^5-9x^6y^4\)
\(=-\left(4^2x^4y^6+2\cdot4\cdot3x^5y^5+3^2x^6y^4\right)\)
\(=-\left[\left(4x^2y^3\right)^2+2\left(4x^2y^3\right)\left(3x^3y^2\right)+\left(3x^3y^2\right)^2\right]\)
\(=\left(4x^2y^3+3x^3y^2\right)^2\)
\(9.\)
\(16x^2-4y^2-8x+1\)
\(=\left(4x\right)^2-\left(2y\right)^2-8x+1\)
\(=\left(4x\right)^2-8x+1-\left(2y\right)^2\)
\(=\left(4x+1\right)^2-\left(2y\right)^2\)
\(=\left(4x-2y+1\right)\left(4x+2y+1\right)\)
\(10.\)
\(49x^2-25+42xy+9y^2\)
\(=\left(7x\right)^2-5^2+2\cdot7\cdot3xy+\left(3y\right)^2\)
\(=\left(7x\right)^2+2\cdot7\cdot3xy+\left(3y\right)^2-5^2\)
\(=\left(7x+3y\right)^2-5^2\)
\(=\left(7x+5y+5\right)\left(7x+3y-5\right)\)
4 tháng 9 2017
\(1.a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
=\(a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)
=\(a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)
=\(a^4-b^4\)=\(\left(a^2-b^2\right)\left(a^2+b^2\right)\)
a)9+6xy^2+x^2y^4
b)100-400m^2n+160m^4n^2
c)a^2-b^4