4517 nha tớ thử rồi

SỐ LƯỢNG MÁY TÍNH LÀ : 35 + 4000 + 5 +5 + 220 + 240 + 12 = 4517w nha bạn 

\(x^3\left(x^2-7\right)^2-36x\)

\(=x.\left[x^2.\left(x^2-7\right)^2-36\right]\)

\(=x.\left[\left(x^3-7x\right)^2-6^2\right]\)

\(=x.\left(x^3-7x-6\right).\left(x^3-7x+6\right)\)

\(=x.\left(x+1\right)\left(x^2-x-6\right).\left(x-1\right).\left(x^2+x-6\right)\)

\(=x.\left(x+1\right).\left(x+2\right).\left(x+3\right).\left(x-1\right).\left(x-2\right).\left(x-3\right)\)

Ta có : \(x^3\left(x^2-7\right)^2-36x\)

= \(x^3\left(x^4-14x^2+49\right)-36x\)

= \(x\left(x^6-14x^4+49x^2-36\right)\)

= \(x\left(x^2-1\right)\left(x^2-4\right)\left(x^2-9\right)\)---- chỗ này tắt 

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

Ta có : 1 < a + b

Mà 1 > 1/8

Nên a + b > 1/8 đúng mà  

Thiếu Đề!

1. 

      3x² + 2x - 1

=   3x² - x + 3x - 1

=   x(3x-1) + (3x-1)

=  (3x-1)(x+1)

2.

    x³ + 6x² + 11x + 6

= x³ + 3x² + 3x² + 9x + 2x + 6

= x²(x+3) + 3x(x+3) + 2(x+3)

= (x+3)(x² + 3x + 2)

= (x+3)(x² + 2x + x + 2)

= (x+3)[x(x+2) + (x+2)]

= (x+3)(x+2)(x+1)

3.

    x⁴ + 2x² - 3

= x⁴ + 2x² + 1 - 4

= (x² + 1)² - 2² 

= (x² +3)(x²-1)

= (x² + 3)(x-10(x+1)

4.

     ab + ac + b² + 2bc + c²

=   a(b+c) + (b+c)²

=   (b+c)(a+b+c)

5. 

      a³ - b³ + c³ + 3abc

=   (a+c)³ - 3ac(a+c) - b³ + 3abc

=   (a+c-b) [(a+c)² + b(a+c) + b²] -3ac (a+c-b)

=   (a+c-b) (a² + 2ac + c² + ab+ bc + b²) - 3ac(a+c-b)

=  (a+c-b) (a² + b² + c² + 2ac + ab + bc - 3ac)

=  (a+c-b) (a² + b² + c² - ac + ab + bc)

tại sao chỉ có lớp 8 thui 

:(

Câu 1:

\(3x\left(12x+4\right)+9x\left(4x+3\right)\)

\(\Leftrightarrow3x\left(12x+4\right)+3x\left[3.\left(4x+3\right)\right]\)

\(\Leftrightarrow3x\left(12x+4\right)+3x\left(12x+6\right)\)

\(\Leftrightarrow3x\left[12x+4+12x+6\right]\)

\(\Leftrightarrow3x.\left(24x+10\right)\)

\(\Leftrightarrow72x^2+30x\)

Câu 2:

\(x\left(5+2x\right)+2x^2\left(x-1\right)\)

\(\Leftrightarrow5x+2x^2+2x^3-2x^2\)

\(\Leftrightarrow2x^3+5x\)

Đề 1

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Input:

Inequality plot:

 

 

Alternate forms:

Expanded form:

Solution:

• Approximate form

Integer solution:

0 x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12

= ( 0 x 1 ) x ( 2 x 3 ) x ( 4 x 5 ) x ( 6 x 7 ) x ( 8 x 9 ) x ( 10 x 11 ) x 12

= 0 x 6 x 20 x 42 x 72 x 110 x 12

= 0 x 20 x 42 x 72 x 110 x 12

= 0 x 42 x 72 x 110 x 12

= 0 x 72 x 110 x 12

= 0 x 110 x 12

= 0 x 12

= 0

0 k nha