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a: \(x^2+6xy+9y^2=\left(x+3y\right)^2\)
b: \(4a^4-4a^2b^2+b^4=\left(2a^2-b^2\right)^2\)
\(x^6-2x^3y+y^2=\left(x^3-y\right)^2\)
b: \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left(x+y-x+y\right)\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\left(3x^2+y^2\right)\)
\(25x^4-10x^2y^2+y^4=\left(5x^2-y^2\right)^2\)
\(-a^2-2a-1=-\left(a+1\right)^2\)
mk ghi đáp án, ko phân tích đc thì IB mk
a) \(x^2+6xy+9y^2=\left(x+3y\right)^2\)
b) \(4a^4-4a^2b^2+b^4=\left(2a^2-b^2\right)^2\)
c) \(x^6+y^2-2x^3y=\left(x^3-y\right)^2\)
d) \(\left(x+y\right)^3-\left(x-y\right)^3=2y\left(3x^2+y^2\right)\)
e) \(25x^4-10x^2y^2+y^4=\left(5x^2-y^2\right)^2\)
f) \(-a^2-2a-1=-\left(a+1\right)^2\)
g) \(27b^3-8a^3=\left(3b-2a\right)\left(9b^2+6ab+4a^2\right)\)
h) \(x^3+9x^2y+27xy^2+27y^3=\left(x+3y\right)^3\)
i) \(16x^2-9\left(x+y\right)^2=\left(x-3y\right)\left(7x+3y\right)\)
a) ta có : \(x^2+6xy+9y^2=x^2+2.x.3y+\left(3y\right)^2=\left(x+3y\right)^2\)
b) ta có : \(4a^4-4a^2b^2+b^4=\left(2a^2\right)^2-2.2a^2.b^2+\left(b^2\right)^2=\left(2a^2-b^2\right)^2\)
c) ta có : \(x^6+y^2-2x^3y=\left(x^3\right)^2-2.x^3.y+y^2=\left(x^3-y^2\right)^2\)
d) ta có : \(\left(x+y\right)^3-\left(x-y\right)^3=\left(x+y-x+y\right)\left(\left(x+y\right)^2+2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right)\)
\(=2y\left(x^2+y^2+2xy+2x^2-2y^2+x^2+y^2-2xy\right)\)
\(=2y\left(4x^2\right)=8x^2y\)
e) ta có : \(25x^4-10x^2y^2+y^4=\left(5x^2\right)^2-2.5x^2.y^2+\left(y^2\right)^2=\left(5x^2-y^2\right)^2\)
f) ta có : \(-a^2-2a-1=-\left(a^2+2a+1\right)=-\left(a+1\right)^2\)
g) ta có : \(27b^3-8a^3=\left(3b\right)^3-\left(2a\right)^3=\left(3b-2a\right)\left(9b^2+6ab+4a^2\right)\)
i) ta có : \(16x^2-9\left(x+y\right)^2=\left(4x\right)^2-\left(3\left(x+y\right)\right)^2\)
\(=\left(4x-3x-3y\right)\left(4x+3x+3y\right)=\left(x-3y\right)\left(7x+3y\right)\)
Bài 1:
\(\frac{15ab+5b^2}{9a^2-b^2}=\frac{5b\left(3a+b\right)}{\left(3a\right)^2-b^2}=\frac{5b\left(3a+b\right)}{\left(3a-b\right)\left(3a+b\right)}=\frac{5b}{3a-b}\)
\(\frac{3x^2-3y^2}{9x+9y}=\frac{3\left(x^2-y^2\right)}{9\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)}{3\left(x+y\right)}=\frac{x-y}{3}\)
\(\frac{m^2-4m+4}{2x-4}=\frac{\left(x-2\right)^2}{2\left(x-2\right)}=\frac{x-2}{2}\)