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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
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 ...
\(a,\left(3+xy^2\right)^2=9+6xy^2+x^2y^4\)
\(b,\left(10-2m^2n\right)^2=100-40m^2n+4m^4n^2\)
\(c,\left(a-b^2\right)\left(a+b^2\right)=a^2-b^4\)
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
\(\left(3+xy\right)^2=9+6xy+xy^2\)
\(\left(10-m^2n\right)^2=100-20m^2n^2+m^4n^2\)
\(\left(a-b^2\right)\left(a+b^2\right)=a^2-b^4\)
Bài1:
\(\left(3+xy^2\right)^2=81+6xy^2+x^2y^4\)
Các câu sau tương tự
Bài2:
\(a,\left(4x^2+4xy+y^2\right)\)
=\(\left(2x+y\right)^2\)
b)\(9m^2+n^2-6mn=\left(3m-n\right)^2\)
c)\(16a^2+25b^2+40ab=\left(4a+5b\right)^2\)
d)\(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
Bài3:
\(a,301^2=\left(300+1\right)^2=900+600+1=1501\)
b/\(499^2=\left(500-1\right)^2=2500-1000+1=1501\)
c/\(68.72=\left(70-2\right)\left(70+2\right)=70^2-2^2=4900-4=4896\)
Ta có:\(x+y=a\)
=>\(x^2+2xy+y^2=a^2\)
=>\(x^2+y^2=a^2-2xy=a^2-2b\left(đpcm\right)\)
Ta lại có:\(x^3+3x^2y+3xy^2+y^3=a^3\)
=>\(x^3+y^3+3xy\left(x+y\right)=a^3\)
=>\(x^3+y^3=a^3-3xy\left(x+y\right)=a^3-3ab\left(đpcm\right)\)
b)\(a+b+c=0\) =>\(a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3a^2c+6abc=0\) =>\(a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\) =>\(a^3+b^3+c^3+3\left(-a\right)\left(-b\right)\left(-c\right)=0\) =>\(a^3+b^3+c^3=3abc\left(đpcm\right)\)
\(a,4b^2c^2-\left(b^2+c^2-a^2\right)^2\\ =\left(2bc\right)^2-\left(b^2+c^2-a^2\right)^2\\ =\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\\ =-\left(\left(b-c\right)^2-a^2\right).\left(\left(b+c\right)^2-a^2\right)\\ =\left(-b+c+a\right)\left(b-c+a\right)\left(b+c-a\right)\left(b+c+a\right)\)
Các câu sau hầu như bn dùng HĐT số 2 nhóm vào
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 ?
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 ~
a) (3+xy2)2
=32+2.3.xy2+(xy2)2
=9+6xy2+x2y4
Vậy ...
b) (10−2m2n)2
=102−2.10.2m2n+(2m2n)2
=100−40m2n+4m4n2
Vậy ...
c) (a−b2)(a+b2)
=a2−(b2)2
=a2−b4
Vậy ...