a: Ta có: \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)

\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)

\(=\left(x^2+9x\right)^2+38\left(x^2+9x\right)+360+1\)

\(=\left(x^2+9x\right)^2+2\cdot\left(x^2+9x\right)\cdot19+19^2\)

\(=\left(x^2+9x+19\right)^2\)

b. \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

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

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

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

c. \(x^2-2x\left(y+2\right)+y^2+4y+4\)

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

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

d. \(x^2+2x\left(y+1\right)+y^2+2y+1\)

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

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

a, \(25+10x+x^2=5^2+2.5x+x^2=\left(5+x\right)^2\)
b, \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
c, \(x^2-10x+25=x^2-2.5x+5^2=\left(x-5\right)^2\)

1. \(25+10x+x^2\\ \Leftrightarrow5^2+2\cdot5\cdot x+x^2\\ \Leftrightarrow\left(5+x\right)^2\\ \Leftrightarrow\left(5+x\right)\left(5+x\right)\)

2. \(8x^3-\dfrac{1}{8}\\ \Leftrightarrow\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[4x^2+x+\dfrac{1}{4}\right]\)

3. \(x^2-10x+25\\ \Leftrightarrow x^2-2\cdot5\cdot x+5^2\\ \Leftrightarrow\left(x-5\right)^2\\ \Leftrightarrow\left(x-5\right)\left(x-5\right)\)

Đọc giúp em số này ạ 1000000000000

1000=10³

1000000=10⁶

1 tỉ =10⁹

1000000000000=10¹²

Có bao nhiêu chữ số 0 tận cùng thì lũy thừa có số mũ bấy nhiêu

\(a,=\left(x+\dfrac{5}{2}\right)^2\\ b,=\left(2x+3y\right)^2\\ c,=a^2+b^2+c^2+2ab-2bc-2ac\\ d,=\left(4x-1\right)^2\\ e,=a^2+b^2+c^2+2ab+2bc+2ac\\ f,=a^2+b^2+c^2-2ab+2bc-2ac\)

`a, 4x^2 - 25y^2 = (2x-5y)(2x+5y)`.

`b, 8x^3 +27 = (2x+3)(4x^2 - 6x + 9)`.

`c, 125x^3 - 64y^3 = (5x)^3 - (4y)^3 = (5x-4y)(25x^2 + 20xy + 16y^2)`.

\(a,\\ 4x^2-25y^2=\left(2x\right)^2-\left(5y\right)^2=\left(2x-5y\right)\left(2x+5y\right)\\ b,\\ 8x^3+27=\left(2x\right)^3+3^3=\left(2x+3\right)\left(4x^2+6x+9\right)\\ c,\\ 125x^3-64y^3=\left(5x\right)^3-\left(4y\right)^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)

a) (1/4)³ x (1/8)²

= [(1/2)²]³ x [(1/2)³]²

= (1/2)⁶ x (1/2)⁶

= (1/2)¹²

b) 4² x 32: 2³

= (2²)² x 2⁵: 2³

= 2⁴ x 2⁵: 2³

= 2⁴ x 2²

= 2⁶

c) 25 x 5³ x 1/625 x 5³

= 5²x 5³ x (1/5)⁴ x 5³ 

= (1/5)⁴ x 5⁸

= 1/5⁴ x 5⁸ (giải thích nếu ko hiểu: (1/5)⁴= 1⁴/5⁴= 1/5⁴)

= 5⁸/5⁴

= 5⁴ 

d) 5⁶ x 1/20 x 2² x 3² : 125

= 5⁶/20 x (2x3)² : 5³

= 5⁶/ (5x4) x 6²: 5³

= 5⁵/4 x 6²/5³ (6²/5³ là dạng phân số, bản chất vẫn là lấy 6² chia 5³)

= 5⁵ x 6²/ 4x 5³ (nhân phân số: tử nhân tử, mẫu nhân mẫu)
= 5²x 6²/ 2² (chia 5⁵ cho 5³ ra 5²)

= 30²/ 2²

= 15²

*Kiến thức áp dụng:
a^mx aⁿ = a^m+n

a^m: aⁿ= a^m-n

(a^m)ⁿ = a^{m x n}

a^m x b^m = (a x b)^m

 x - 4 + x - 4 + x - 4 + x - 4 =  4  . ( x - 4 ) =  4  . ( -5 - 4 ) =  -36 

A, \(\frac{9}{4}x^2+3x+4\)

= \(\left(\frac{3}{2}x^2\right)+2\cdot\frac{3}{2}x\cdot2+2^2\)

    = \(\left(\frac{3}{2}x+2\right)^2\)

B. ( 9\(x^2\)+ 12x+ 4) + 6(3x+2)+9

=[ (3x)\(^2\)+2.3x.2+\(2^2\) ) + 2.(3x+2).3 +3\(^2\)

= ( 3x+2)\(^2\)+2.(3x+2).3+3\(^2\)

= (3x+2+3)\(^2\)

Chúc bạn học tốt <3