\(\frac{x+110}{100}+\frac{x+8}{102}+\frac{x+6}{104}=\frac{x+4}{106}+\frac{x+2}{108}\)

\(\Leftrightarrow\frac{x+110}{100}+\left(\frac{x+8}{102}+1\right)+\left(\frac{x+6}{104}+1\right)=\left(\frac{x+4}{106}+1\right)+\left(\frac{x+2}{108}+1\right)\)

\(\Leftrightarrow\frac{x+110}{100}+\frac{x+110}{102}+\frac{x+110}{104}=\frac{x+110}{106}+\frac{x+110}{108}\)

\(\Leftrightarrow\frac{x+110}{100}+\frac{x+110}{102}+\frac{x+110}{104}-\frac{x+110}{106}-\frac{x+110}{108}=0\)

\(\Leftrightarrow\left(x+110\right)\left(\frac{1}{100}+\frac{1}{102}+\frac{1}{104}+\frac{1}{106}+\frac{1}{108}\right)=0\)

\(\Leftrightarrow x+110=0\) (vì \(\frac{1}{100}+\frac{1}{102}+\frac{1}{104}-\frac{1}{106}-\frac{1}{108}>0\))

\(\Leftrightarrow x=-110\)

Vậy \(x=-110\)

mn giải hộ mik ik

Ko hiểu đề bài!

đầu tiên ta tìm số hạng của dãy :

       ( 110 - 1 ) : 1 + 1 = 110 ( số )

tiếp ta tìm tổng :

       ( 110 + 1 ) x 110 : 2 = 6105 

 ĐS : 6105

=============> CHÚC HỌC GIỎI <==============

`4)x^2-5x+6`

`=x^2-2x-3x+6`

`=x(x-2)-3(x-2)=(x-2)(x-3)`

`5)x^2+7x+10`

`=x^2+5x+2x+10`

`=x(x+5)+2(x+5)=(x+5)(x+2)`

`6)x+7\sqrt{x}+10`    `ĐK: x >= 0`

`=(\sqrt{x})^2+5\sqrt{x}+2\sqrt{x}+10`

`=\sqrt{x}(\sqrt{x}+5)+2(\sqrt{x}+5)=(\sqrt{x}+5)(\sqrt{x}+2)`

`7)3x^4+7x^2+4`

`=3x^4+3x^2+4x^2+4`

`=3x^2(x^2+1)+4(x^2+1)=(x^2+1)(3x^2+4)`

`8)x^2-x-2`

`=x^2-2x+x-2`

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

`9)x^6-x^3-2`

`=x^6+x^3-2x^3-2`

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

`=(x^3+1)(x^3-2)`.

Phân tích đa thức thành nhân tử:

a, x = 27              

b, x = 324

a) \(P\left(-1\right)=\left(-1\right)^2+\left(-1\right)^4+...+\left(-1\right)^{106}\)

\(=1+1+...+1\)

=53

b) \(Q\left(-1\right)=\left(-1\right)+\left(-1\right)^3+\left(-1\right)^5+...+\left(-1\right)^{107}\)

\(=-1\cdot54=-54\)

Bài 1: 

a: \(x^3-6x^2+11x-6\)

\(=x^3-x^2-5x^2+5x+6x-6\)

\(=\left(x-1\right)\left(x^2-5x+6\right)\)

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

b: \(x^3-6x^2-9x+14\)

\(=x^3-7x^2+x^2-7x-2x+14\)

\(=\left(x-7\right)\left(x^2+x-2\right)\)

\(=\left(x-7\right)\left(x+2\right)\left(x-1\right)\)

c: \(x^3+6x^2+11x+6\)

\(=x^3+3x^2+3x^2+9x+2x+6\)

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

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

1.

\(8\times8=64\)

\(6\times4=24\)

\(4\times4=16\)

2. 

Ta có:

\(4\times6=24\)              \(7\times5=35\)

Vì \(24< 35\) nên \(4\times6< 7\times5\)

Vậy...

_____

Ta có:

\(5\times7=35\)              \(8\times9=72\)

Vì \(35< 72\) nên \(5\times7< 8\times9\)

Vậy....

_____

Ta có: 

\(3\times5=15\)              \(6\times9=54\)

Vì \(15< 54\) nên \(3\times5< 6\times9\)

Vậy...

\(#Wendy.Dang\)

8 x 8 =64

6 x 4 =24

4 x 4=16

Bài 2 : < , > , = 

4 x 6  <  7 x 5

5 x 7  < 8 x 9

3 x 5 <  6 x 9 

Em cảm ơn POP POP ạ