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) x²-4x+4=\(\left(X-2\right)^2\)

b)9x²-12x+4=\(\left(3x-2\right)^2\)

c)x²-6xy+9y^{2=\(\left(x-3y\right)^2\)}

d)x²/4+x+1=\(\left(\dfrac{x}{2}+1\right)^2\)

e)-8x+16+x² =\(X^2-8x+16=\left(x-4\right)^2\)

f)xy²+1/4x²y⁴+1 => sai đề

g)x²-1=\(\left(X-1\right)\left(x+1\right)\)

h)25x²-16=\(\left(5x-4\right)\left(5x+4\right)\)

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) \(-6xy+x^2+9y^2\)

\(=x^2-6xy+9y^2\)

\(=x^2-2.x.3y+\left(3y\right)^2\)

\(=\left(x-3y\right)^2\)

Nguyễn Bích Quỳnh: bạn chúc như một vị thần

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. $x^2-x+\dfrac14$

$=x^2-2\cdot x\cdot \dfrac12+\Bigg(\dfrac12\Bigg)^2$

$=\Bigg(x-\dfrac12\Bigg)^2$

`x^2 +4x+4`

`=x^2+2*x*2+2^2`

`=(x+2)^2`

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`x^2-6xy+9y^2`

`=x^2 - 2*x*3y+(3y)^2`

`=(x-3y)^2`

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`4x^2 +12x+9`

`=(2x)^2 +2*2x*3+3^2`

`=(2x+3)^2`

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`x^2-x+1/4`

`=x^2 - 2*x*1/2 +(1/2)^2`

`=(x+1/2)^2`