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2) (a-1)2+(b-2)2+(2c-1)2=0
do (a-1)2, (b-2)2 và (2c-1)2 lớn hơn hoặc bằng 0 nên để thỏa mãn biểu thức trên thì (a-1)2, (b-2)2 và (2c-1)2 đồng thời bằng 0
suy ra a=1, b=2, c=1/2
a. x2 + 6x + 9 = (x + 3)2
b. 25 + 10x + x2 = (5 + x)2
c. x2 + 8x + 16 = (x + 4)2
d. x2 + 14x + 49 = (x + 7)2
e. 4x2 + 12x + 9 = (2x + 3)2
f. 9x2 + 12x + 4 = (3x + 2)2
h. 16x2 + 8 + 1 = (4x + 1)2
i. 4x2 + 12xy + 9y2 = (2x + 3y)2
k. 25x2 + 20xy + 4y2 = (5x + 2y)2
a) \(=\left(x+3\right)^2\)
b) \(=\left(x+5\right)^2\)
c) \(=\left(x+4\right)^2\)
d) \(=\left(x+7\right)^2\)
e) \(=\left(2x+3\right)^2\)
f) \(=\left(3x+2\right)^2\)
h) \(=\left(4x+1\right)^2\)
i) \(=\left(2x+3y\right)^2\)
k) \(=\left(5x+2y\right)^2\)
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) \(25x^2-20xy+4y^2\)
\(=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\)
\(=\left(5x-2y\right)^2\)
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) Ta có: \(\left(4x^2-12x+9\right)-1\)
\(=\left(2x-3\right)^2-1^2\)
\(=\left(2x-3-1\right)\left(2x-3+1\right)\)
\(=\left(2x-4\right)\left(2x-2\right)\)
\(=4\left(x-2\right)\left(x-1\right)\)
b) Ta có: \(\left(\frac{x^2}{4}+2xy+4y^2\right)-25\)
\(=\left[\left(\frac{x}{2}\right)^2+2\cdot\frac{x}{2}\cdot2y+\left(2y\right)^2\right]-5^2\)
\(=\left(\frac{x}{2}+2y\right)^2-5^2\)
\(=\left(\frac{x}{2}+2y-5\right)\left(\frac{x}{2}+2y+5\right)\)
c) Ta có: \(1+12x+35x^2\)
\(=35x^2+12x+1\)
\(=35x^2+5x+7x+1\)
\(=5x\left(7x+1\right)+\left(7x+1\right)\)
\(=\left(7x+1\right)\left(5x+1\right)\)
d) Ta có: \(9x^2-24xy+15y^2\)
\(=9x^2-9xy-15xy+15y^2\)
\(=9x\left(x-y\right)-15y\left(x-y\right)\)
\(=\left(x-y\right)\left(9x-15y\right)\)
\(=3\left(x-y\right)\left(3x-5y\right)\)
e) Ta có: \(25x^2-20xy+3y^2\)
\(=25x^2-15xy-5xy+3y^2\)
\(=5x\left(5x-3y\right)-y\left(5x-3y\right)\)
\(=\left(5x-3y\right)\left(5x-y\right)\)
f) Ta có: \(24x^4-10x^2y+y^2\)
\(=24x^4-4x^2y-6x^2y+y^2\)
\(=4x^2\left(6x^2-y\right)-y\left(6x^2-y\right)\)
\(=\left(6x^2-y\right)\left(4x^2-y\right)\)
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,9a2-30a+25
=(3a)2-30a+52
=(3a-5)2
b,1+4x+4x2
=4x2+4x+1
=(2x)2+4x+12
=(2x+1)2
c,a2+16+8a
=a2+8a+16
=a2+8a+42
=(a+2)2
d,25x2+4y2-20xy
=25x2-20xy+4y2
=(5x)2-20xy+(2y)2
=(5x-2y)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. $x^2-x+\dfrac14$
$=x^2-2\cdot x\cdot \dfrac12+\Bigg(\dfrac12\Bigg)^2$
$=\Bigg(x-\dfrac12\Bigg)^2$
