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a) \(x^2-6x+9=x^2-2.3.x+3^2=\left(x-3\right)^2\)
b)\(x^2+4x+4=x^2+2.2.x+2^2=\left(x+2\right)^2\)
c)\(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2=\left(2x+1\right)^2\)
d)\(4x^2+12xy+9y^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2=\left(2x+3y\right)^2\)
e)\(x^2-8x+16=x^2-2.4.x+4^2=\left(x-4\right)^2\)
a) x2 -6x +9 = (x-3)2
b) x2+4x +4= (x+2)2
c) 4x2+4x+1= (2x+1)2
d) 4x2+12xy+9y2 = (2x+3y)2
e) x2-8x+16 = (x-4)2
Đây chính là hằng đẳng thức nhé bn....
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/ 9x2-12xy+4y2 = (3x - 2y)2
b/ 25x2-10x+1 = (5x - 1)2
c/ 9x2-12x+4 = (3x - 2)2
d/ 4x2+20x+25 = (2x + 5)2
e/ x4-4x2+4 = (x2 - 2)2
a. $x^2+4x+4$
$=x^2+2\cdot x\cdot2+2^2$
$=(x+2)^2$
b. $x^2-6xy+9y^2$
$=x^2-2\cdot x\cdot3y+(3y)^2$
$=(x-3y)^2$
c. $4x^2+12x+9$
$=(2x)^2+2\cdot2x\cdot3+3^2$
$=(2x+3)^2$
d. $x^2-x+\dfrac14$
$=x^2-2\cdot x\cdot \dfrac12+\Bigg(\dfrac12\Bigg)^2$
$=\Bigg(x-\dfrac12\Bigg)^2$
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
Bài 3: Viết các biểu thức dưới dạng bình phương một tổng hoặc hiệu
a. 4x2+4x+1
=(2x)2+2.2x.1+12
=(2x+1)2
b. x2+16- 8x
=x2-2.4x+42
=(x-4)2
c. x2-x+\(\frac{1}{4}\)
=x2 - 2x\(\frac{1}{4}\) + (\(\frac{1}{2}\))2
=(x-\(\frac{1}{2}\))2
a) \(25x^2-20xy+4y^2\)
\(=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\)
\(=\left(5x-2y\right)^2\)
b) \(4x^2+9+12x\)
\(=\left(2x\right)^2+3^2+2.2x.3\)
\(=\left(2x+3\right)^2\)
c) \(\frac{1}{4}+3x+9x^2\)
\(=\left(\frac{1}{2}\right)^2+2.\frac{1}{2}.3x+\left(3x\right)^2\)
\(=\left(\frac{1}{2}+3x\right)^2\)
d) \(-6xy+x^2+9y^2\)
\(=x^2-6xy+9y^2\)
\(=x^2-2.x.3y+\left(3y\right)^2\)
\(=\left(x-3y\right)^2\)
a) \({x^2} + 4x + 4 = {x^2} + 2.x.2 + {2^2} = {\left( {x + 2} \right)^2}\)
b) \(16{a^2} - 16ab + 4{b^2} = {\left( {4a} \right)^2} - 2.4a.2b + {\left( {2b} \right)^2} = {\left( {4a - 2b} \right)^2}\)
a. \(x^2-x+\frac{1}{4}=x^2-2x\frac{1}{2}+\left(\frac{1}{2}\right)^2=\left(x-\frac{1}{2}\right)^2\)
b. \(4x^2-4x+1=\left(2x\right)^2-2.2x.1+1^2=\left(2x-1\right)^2\)
c. \(x^2-3x+\frac{9}{4}=x^2-2x\frac{3}{2}+\left(\frac{3}{2}\right)^2=\left(x-\frac{3}{2}\right)^2\)
TK MIK NHA~