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\(a,\left(x+3\right)^2\)
\(b,\left(x+\frac{1}{2}\right)^2\)
\(c,\left(xy^2+1\right)^2\)
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) \(9x^2+6x+1=\left(3x+1\right)^2\)
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
1, \(x^2+2xy+y^2=\left(x+y\right)^2\)
2, \(4x^2+12x+9=\left(2x\right)^2+2\cdot3\cdot2x+3^2=\left(2x+3\right)^2\)
3, \(x^2+5x+\dfrac{25}{4}=x^2+2\cdot\dfrac{5}{2}\cdot x+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)
4, \(16x^2-8x+1=\left(4x\right)^2-2\cdot4x\cdot1+1^2=\left(4x-1\right)^2\)
5, \(x^2+x+\dfrac{1}{4}=x^2+2\cdot\dfrac{1}{2}\cdot x+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)
1: =(x+y)^2
2: =(2x+3)^2
3: =(x+5/2)^2
4: =(4x-1)^2
5: =(x+1/2)^2
6: =(x-3/2)^2
7: =(x+1)^3
8: =(1/2x+1)^2
9: =(3y-1/3)^3
10: =(2x+y)^3
a) \(x^2+6x+9=\left(x+3\right)^2\)
b) \(2xy^2+x^2y^4+1=x^2y^4+2xy^2+1=\left(xy^2+1\right)^2\)
c) \(x^2+x+\frac{1}{4}=x^2.2.\frac{1}{2}x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)
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
này mình có vài câu không làm được, xin lỗi bạn nha
\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)
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+3)^2
b,(x+1/2)^2