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a) x2+6x+9=x2+2.x.3+32=(x+3)2
b) x2+x+\(\dfrac{1}{4}\)=x2+2.x.\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)=(x+\(\dfrac{1}{2}\))2
c) 2xy2+x2y4+1=(xy2)2+2.xy2+1=(xy2+1)2
a) \(x^2+6x+9=x^2+2.3x+3^2=\left(x+3\right)^2\)
b) \(x^2+x=\text{ }\left[x^2+2.\frac{1}{2}x+\left(\frac{1}{2}\right)^2\right]-\left(\frac{1}{2}\right)^2=\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\)
c) \(2xy^2+x^2y^4=\left[\left(xy^2\right)^2+2.xy^2+1^2\right]-1^2=\left(xy^2+1\right)^2-1^2\)
a) \(x^2+6x+9=\left(x+3\right)^2\)
b) \(x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)
c) \(2xy^2+x^2y^4+1=\left(xy^2+1\right)^2\)
a, \(\left(x+3\right)^2\)
b,\(\left(x+\frac{1}{2}\right)^2\)Mik giải thích tí nè, cái này =\(x^2+2.x.\frac{1}{2}+\frac{1}{4}\)=\(x^2+x+\frac{1}{4}\)
c,thì mik chịu.
Bài 8:
Ta có: \(A=-x^2+2x+4\)
\(=-\left(x^2-2x-4\right)\)
\(=-\left(x^2-2x+1-5\right)\)
\(=-\left(x-1\right)^2+5\le5\forall x\)
Dấu '=' xảy ra khi x=1
a) \(9x^2+6x+1=\left(3x+1\right)^2\)
b)\(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)
c)\(x^2y^4-2xy^2+1=\left(xy^2-1\right)^2\)
d) \(x^2+\frac{2}{3}x+\frac{1}{9}=\left(x+\frac{1}{3}\right)^2\)
a) 9x2 + 6x + 1 = ( 3x + 1 )2
b) x2 - x + 1/4 = ( x - 1/2)2
c) x2 . y4 - 2xy2 + 1 = ( xy2 - 1 ) 2
d) x2 + 2/3x + 1/9 = (x+1/3)2
a. (x + y)2 = x2 + 2xy + y2
b. (x - 2y)2 = x2 - 4xy - 4x2
c. (xy2 + 1)(xy2 - 1) = x2y4 - 1
d. (x + y)2(x - y)2 = (x2 + 2xy + y2)(x2 - 2xy + y2) = x4 - (2xy + y2)2 = x4 - (4x2y2 + y4) = x4 - 4x2y2 - y4
Chucs hocj toots
Câu 2:
a: \(x^2-4x+4=\left(x-2\right)^2\)
b: \(x^2+10x+25=\left(x+5\right)^2\)
d: \(9\left(x+1\right)^2-6\left(x+1\right)+1=\left(3x+2\right)^2\)
e: \(\left(x-2y\right)^2-8\left(x-2xy\right)+16x^2=\left(x-2y+4x\right)^2=\left(5x-2y\right)^2\)
\(2xy^2+x^2y^4+1\)
\(=\left(xy^2\right)^2+2.xy^2+1^2\)
\(=\left(xy^2+1\right)^2\)
a=(x+3)
b=(x+1/2)
c=(xy^2+1)
Good luck!
\(a,\left(x+3\right)^2\)
\(b,\left(x+\frac{1}{2}\right)^2\)
\(c,\left(xy^2+1\right)^2\)