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\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)
a) \(x^2+2x+1=x^2+2\cdot x\cdot1+1^2=\left(x+1\right)^2\)
b) \(x^2-4x+4=x^2-2\cdot x\cdot2+2^2=\left(x-2\right)^2\)
c) \(x^2+6xy+9y^2=x^2+2\cdot x\cdot3y+\left(3y\right)^2=\left(x+3y\right)^2\)
d) \(z^2-z+\dfrac{1}{4}=z^2-2\cdot z\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(z-\dfrac{1}{2}\right)^2\)
e) \(25x^2-10x+1=\left(5x\right)^2-2\cdot5x\cdot1+1^2=\left(5x-1\right)^2\)
a)x2-6x+9
=x2-2.x.3+32
=(x-3)2
b)4x2+4x+1
=(2x)2+2.2x.1+12
=(2x+1)2
c)4x2+12xy+9y2
=(2x)2+2.2x.3y+(3y)2
=(2x+3y)2
d)4x4-4x2+4
=(2x2)2-2.2x2.2+22
=(2x2-2)2
a) \(=\left(x-2\right)^2\)
b) \(=\left(3x-2\right)^2\)
c) \(=\left(x-3y\right)^2\)
d) \(=\left(\dfrac{x}{2}+1\right)^2\)
e) \(=\left(x-4\right)^2\)
f) \(=\left(\dfrac{1}{2}xy^2+1\right)^2\)
g) \(=\left(x-1\right)\left(x+1\right)\)
h) \(=\left(5x-4\right)\left(5x+4\right)\)
`a)x^2-2x+2+4y^2+4y`
`=x^2-2x+1+4y^2+4y+1`
`=(x-1)^2+(2y+1)^2`
`b)4x^2+y^2+12x+4y+13`
`=4x^2+12x+9+y^2+4y+4`
`=(2x+3)^2+(y+2)^2`
`c)x^2+17+4y^2+8x+4y`
`=x^2+8x+16+4y^2+4y+1`
`=(x+4)^2+(2y+1)^2`
`d)4x^2-12xy+y^2-4y+13`
`=4x^2-12x+9+y^2-4y+4`
`=(2x-3)^2+(y-2)^2`
a) \(x^2-2x+2+4y^2+4y=\left(x-1\right)^2+\left(2y+1\right)^2\)
b) \(4x^2+y^2+12x+4y+13=\left(2x+3\right)^2+\left(y+2\right)^2\)
c) \(x^2+17+4y^2+8x+4y=\left(x+4\right)^2+\left(2y+1\right)^2\)
d) \(4x^2-12x+y^2-4y+13=\left(2x-3\right)^2+\left(y-2\right)^2\)
a) \(x^2+4x+4\)
\(=x^2+2\cdot2\cdot x+2^2\)
\(=\left(x+2\right)^2\)
b) \(4x^2-4x+1\)
\(=\left(2x\right)^2-2\cdot2x\cdot1+1^2\)
\(=\left(2x-1\right)^2\)
c) \(x^2-x+\dfrac{1}{4}\)
\(=x^2-2\cdot\dfrac{1}{2}\cdot x+\left(\dfrac{1}{2}\right)^2\)
\(=\left(x-\dfrac{1}{2}\right)^2\)
d) \(4\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=\left[2\left(x+y\right)\right]^2-2\cdot2\left(x+y\right)\cdot1+1^2\)
\(=\left[2\left(x+y\right)-1\right]^2\)
\(=\left(2x+2y-1\right)^2\)
a) \(x^4+4x^2+4=\left(x^2+2\right)^2\)
b) \(\left(2y-x\right)^2+2\left(2y-x\right)+1=\left(2y-x+1\right)^2\)
c) \(\left(2a-4b\right)^2+4a-8b+1=\left(2a-4b\right)^2+2\cdot\left(2a-4b\right)\cdot1+1^2=\left(2a-4b+1\right)^2\)
a) ( x + 1 ) 2 . b) ( x – 4 ) 2 .
c) x 2 4 + x + 1 ; d) ( 2 x – 2 y ) 2 .
a) Sửa đề: \(x^2+3x+1\rightarrow x^2+2x+1\)
\(x^2+2x+1=\left(x+1\right)^2\)
b) \(x^2+y^2+2xy=\left(x+y\right)^2\)
c) \(9x^2+12x+4=\left(3x+2\right)^2\)
d) \(-4x^2-9-12x=-\left(4x^2+12x+9\right)=-\left(2x+3\right)^2\)
\(a.x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(=\left(x+\frac{1}{4}\right)^2\)
b) \(x^2+12xy+36xy^2=x^2+2.x.y+y^2\)
\(a.x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(=\left(x+\frac{1}{4}\right)^2\)
b) \(x^2+12xy+36xy^2=x^2+2.x.y+y^2\)
\(a.x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(=\left(x+\frac{1}{4}\right)^2\)
b) \(x^2+12xy+36xy^2=x^2+2.x.y+y^2\)
\(a.x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(=\left(x+\frac{1}{4}\right)^2\)
b) \(x^2+12xy+36xy^2=x^2+2.x.y+y^2\)
\(=\left(x+y\right)^2\) c) \(4x^2-12xy+9y^2=\left(2x\right)^2-2.2x.3y+\left(3y\right)^2\) \(=\left(2x-3y\right)^2\)d) \(x^2-2x+4=x^2-2.x.4+4^2\)
\(=\left(x-4\right)^2\)
e) \(25x^2+4y^2-20xy=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\)
\(=\left(5x-2y\right)^2\)
^...^ ^_^ Bài làm có gì ko hiểu bạn cứ hỏi nhé ^_^
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