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a)\(x^2+2x+1=x^2+2x1+1^2=\left(x+1\right)^2\)
b)\(9x^2+y^2+6xy=3^2x^2+y^2+2.3x.y=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)
c)\(25a^2+4b^2-20ab=5^2a^2+2^2b^2-2.5a.2b=\left(5a\right)^2-2.5a.2b+\left(2b\right)^2=\left(5a-2b\right)^2\)
d)\(x^2-x+\frac{1}{4}=x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=\left(x-\frac{1}{2}\right)^2\)
a) \(x^2+2x+1=\left(x+1\right)^2\)
b) \(9x^2+y^2+6xy=\left(3x+y\right)^2\)
c) \(25a^2+4b^2-20ab=\left(5a-2b\right)^2\)
Câu d thì biểu thức là \(\frac{x^2-1}{2x+\frac{1}{10}}\) hay là \(\frac{x^2-1}{\frac{2x+1}{10}}\) z bạn???
Bài 1:
a) \(x^2\left(5x^3-x-6\right)\)
\(=x^2.5x^3-x^2.x-x^2.6\)
\(=5x^5-x^3-6x^2\)
b) \(\left(x^2-2xy+y^2\right)\left(x-y\right)\)
\(=x^2\left(x-y\right)-2xy\left(x-y\right)+y^2\left(x-y\right)\)
\(=x^3-x^2y-2x^2y+2xy^2+y^2x-y^3\)
\(=x^3-3x^2y+3xy^2-y^3\)
Bài 2:
a) \(y^2+2y+1\)
\(=\left(y+1\right)^2\)
b) \(9x^2+y^2-6xy\)
\(=\left(3x\right)^2-2.3x.y+y^2\)
\(=\left(3x-y\right)^2\)
c) \(25a^2+4b^2+20ab\)
\(=\left(5a\right)^2+2.5a.2b+\left(2b\right)^2\)
\(=\left(5a+2b\right)^2\)
d) \(x^2-x+\dfrac{1}{4}\)
\(=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\)
\(=\left(x-\dfrac{1}{2}\right)^2\)
d) x^2 - x + 1/4
= x^2 - 2.x + 1/4 + (1/2)^2
= ( x - 1/2)^2
a) x2 + 2x + 1 = x2 + 2.x.1+ 12 = ( x + 1)2
b) 9x2 + y2 + 6xy = (3x)2 + 2.3.x.y + y2 = (3x + y)2
c) 25a2 + 4b2 – 20ab = (5a)2 – 2.5.a.2b. + (2b)2 = (5a – 2b)2
Hoặc 25a2 + 4b2 – 20ab = (2b)2 – 2.2b.5a. + (5a)2 = (2b – 5a)2
d) x2 – x + \(\dfrac{1}{4}\) = x2 – 2.x. \(\dfrac{1}{2}\) + ( \(\dfrac{1}{2}\))22 = ( x - \(\dfrac{1}{2}\) )2
Hoặc x2 – x + \(\dfrac{1}{4}\) = \(\dfrac{1}{4}\) - x + x2 = (\(\dfrac{1}{2}\))2 – 2. \(\dfrac{1}{2}\).x + x2 = (\(\dfrac{1}{2}\) - x)2
a) x2 + 2x + 1 = x2+ 2 . x . 1 + 12
= (x + 1)2
b) 9x2 + y2+ 6xy = (3x)2 + 2 . 3 . x . y + y2 = (3x + y)2
c) 25a2 + 4b2– 20ab = (5a)2 – 2 . 5a . 2b + (2b)2 = (5a – 2b)2
Hoặc 25a2 + 4b2 – 20ab = (2b)2 – 2 . 2b . 5a + (5a)2 = (2b – 5a)2
d) x2 – x + 1414 = x2 – 2 . x . 1212 + (12)2(12)2= (x−12)2(x−12)2
Hoặc x2 – x + 1414 = 1414 - x + x2 = (12)2(12)2 - 2 . 1212 . x + x2 = (12−x)2
a/ (x+1)2
b/ (3x+y)2
c/ (5a-2b)2
d/ câu d hình như sai đề nhé
a) ; b) ; c) Bạn Nguyễn Võ Nhiệt My làm rồi
d) \(x^2-x+\frac{1}{4}\)
\(=x^2-2.\frac{1}{2}x+\left(\frac{1}{2}\right)^2\)
\(=\left(x-\frac{1}{2}\right)^2\)
Đap số \(\left(x-\frac{1}{2}\right)^2\)
It's khai triển :)
a) \(\left(5x-x^2\right)\left(5x+x^2\right)=25x^2-x^4\)
b) \(\left(2x-y\right)\left(4x^2+2xy+y^2\right)=8x^3-y^3\)
c) \(\left(x+3\right)\left(x^2-3x+9\right)=x^3-27\)
d) \(-x^3+3x^2-3x+1=\left(1-x\right)^3\)
e) \(x^2-2x+9=\left(x-1\right)^2+8??\) ko ra gì cả-.-
g) \(\left(x+1\right)\left(x-1\right)=x^2-1\)
h) \(\left(x-2y\right)\left(x+2y\right)=x^2-4y^2\)
i) \(25a^2+4b^2-20ab=\left(5a-2b\right)^2\)
\(a,y^2+2y+1=y^2+2.y.1+1^2=\left(y+1\right)^2\)
\(b,9x^2+y^2-6xy=\left(3x\right)^2-2\cdot3x\cdot y+y^2=\left(3x-y\right)^2\)
\(c,25a^2+4b^2+20ab=\left(5a\right)^2+2\cdot5a\cdot2b+\left(2b\right)^2=\left(5a+2b\right)^2\)
\(d,x^2-x+\dfrac{1}{4}=x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2\)
`y^2 + 2y + 1 = y^2 + 2y .1 + 1^2 + (y+1)^2`
`9x^2 + y^2 - 6xy = (3x)^2 - 2.3x.y + y^2 = (3x + y)^2`
`25a^2 + 4b^2 + 20ab = (5a)^2 + 2.5a .2b + (2b)^2 = (5a + 2b)^2`
`x^2 - x + 1/4 = x^2 - 2x .1/2 +(1/2)^2 = (x-1/2)^2`