a/ \(=\left(9x^2+30x+25\right)+\left(x^2+10x+25\right)=\)

\(=\left(3x+5\right)^2+\left(x+5\right)^2\)

b/ \(=\left(16x^2+8x+1\right)+\left(y^2-4y+4\right)=\left(4x+1\right)^2+\left(y-2\right)^2\)

c/ 

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) \(=\left(5x+2y\right)^2\)

\(1,\)\(4x^2-4x+y^2+2y+2\)

\(=4x^2+4x+1+y^2+2y+1\)

\(=\left[\left(2x\right)^2-2.2x+1\right]+\left(y^2+2.y.1+1^2\right)\)

\(=\left(2x-1\right)^2+\left(y+1\right)^2\)

\(2,\)\(a^2-4ab+5b^2-4bc+4c^2\)

\(=a^2-4ab+4b^2+b^2-4bc+4c^2\)

\(=\left[a^2-2.a.2b+\left(2b\right)^2\right]+\left[b^2-2.b.2c+\left(2c\right)^2\right]\)

\(=\left(a-2b\right)^2+\left(b-2c\right)^2\)

\(3,\)\(16x^2+5+8x-4y+y^2\)

\(=16x^2+8x+1+y^2-4y+4\)

\(=\left[\left(4x\right)^2+2.4x.1+1^2\right]+\left[y^2-2.y.2+2^2\right]\)

\(=\left(4x+1\right)^2+\left(y-2\right)^2\)

1, (\(x\) + 3y)²

= \(x^2\) + 2.3\(xy\) + (3y)²

= \(x^2\) + 6\(xy\) + 9y²

2, (4a + b)²

= (4a)² + 2.4.a.b + b²

= 16a² + 8ab + b²

3, (6a + 7b)²

= (6a)² + 2.6.7ab + (7b)²

= 36a² + 84ab + 49b²

\(a,=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

\(b,=9x^2+2.3x.\frac{1}{2}+\frac{1}{4}+\frac{19}{4}=\left(3x+\frac{1}{2}\right)^2+\frac{19}{4}\)

\(c,=16x^2-2.4x+1+6=\left(4x-1\right)^2+6\)

Mình không ghi lại đề:

\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}=\frac{1}{5}\)

\(\frac{2}{\left(x+1\right)\left(x+3\right)}+...+\frac{2}{\left(x+7\right)\left(x+9\right)}=\frac{2}{5}\)

\(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+...+\frac{1}{x+7}-\frac{1}{x+9}=\frac{2}{5}\)

\(\frac{1}{x+1}-\frac{1}{x+9}=\frac{2}{5}\)

\(\frac{8}{\left(x+1\right)\left(x+9\right)}=\frac{2}{5}\)

<=>40=2(x+1)(x+9)

<=>\(x^2+10x-11=0\)

<=>\(\left(x-1\right)\left(x+11\right)=0\)

<=>x=1 hoặc x=-11

Ta có:

\(1^2+\left(-11\right)^2=122\)

Ai thấy mình làm đúng thì tích nha.Ai tích mình mình tích lại

mk nghĩ là = 122 đó bn

\(P=\left(6x^2+16x+3m\right):6\\ =6x^2:6+16x:6+3m:6\\ =x^2+\dfrac{8}{3}x+\dfrac{1}{2}m\\ =\left(x^2+2.x.\dfrac{4}{3}+\dfrac{16}{9}\right)=\left(x+\dfrac{4}{3}\right)^2\\ Vậy:\dfrac{1}{2}m=\dfrac{16}{9}\\ Vậy:m=\dfrac{16}{9}:\dfrac{1}{2}=\dfrac{16}{9}.2=\dfrac{32}{9}\)

x^2 + 4y^2 -6x + 4y +10

=x²-6x+9+4y²+4y+1

=(x-3)²+(2y+1)²

x^2-4y^2-6x+4y+10

= x^2- 2.x.3 + 9 + 4y^2-2.2y+1

= ( x - 3)^2+ (2y-1) ^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

6, \(x^2-3x+\dfrac{9}{4}=x^2-2\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2=\left(x-\dfrac{3}{2}\right)^2\)

7, \(x^3+3x^2+3x+1=x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=\left(x+1\right)^3\)

8, \(\dfrac{x^2}{4}+x+1=\left(\dfrac{x}{2}\right)^2+2\cdot\dfrac{x}{2}\cdot1+1^2=\left(\dfrac{x}{2}+1\right)^2\)

9, \(27y^3-9y^2+y-\dfrac{1}{27}=\left(3y\right)^3-3\cdot\left(3y\right)^2\cdot\dfrac{1}{3}+3\cdot3y\cdot\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^3=\left(3y-\dfrac{1}{3}\right)^3\)

10, \(8x^3+12x^2y+6xy^2+y^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3=\left(2x+y\right)^3\)

`x^2-2x+4y^2+4y+2`

`=x^2-2x+1+4y^2+4y+1`

`=(x-1)^2+(2y+1)^2`

Ta có: \(x^2-2x+2+4y^2+4y\)

\(=x^2-2x+1+4y^2+4y+1\)

\(=\left(x-1\right)^2+\left(2y+1\right)^2\)