Bài 2: 

x=13 nên x+1=14

\(f\left(x\right)=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+14\)

\(=x^{14}-x^{14}-x^{13}+x^{13}-...+x^3+x^2-x^2-x+14\)

=14-x=1

x=13 nên x+1=14

f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14

=x14−x14−x13+x13−...+x3+x2−x2−x+14=x14−x14−x13+x13−...+x3+x2−x2−x+14

=14-x=1

X+\(\dfrac{2}{7}\)=\(\dfrac{1}{2}\)

X     = \(\dfrac{1}{2}\)-\(\dfrac{2}{7}\)
X     = \(\dfrac{3}{14}\)
Chọn đáp án A. \(\dfrac{3}{14}\)

A

\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+14\right)=14\)

\(x+x+1+x+2+x+3+...+x+14=14\)

\(15x+\left(1+2+3+...+14\right)=14\)

\(15x+\frac{\left(14+1\right).14}{2}=14\)

\(15x+105=14\)

\(15x=-91\)

\(x=-\frac{91}{15}=-6,06\)

Ko chắc lắm

x+(x+1)+(x+2)+(x+3)+...+(x+13)+(x+14)=14

x+x+1+x+2+x+3+...+x+13+x+14=14

(x+x+x+x+...+x+x)+(1+2+3+...+14)=14

15x + 14 = 14

15x = 14 - 14

15x = 0

    x = 0:15

    x = 0

Vậy x=0

Hoàng hôn ( Cool Team )   bạn thông minh quá (1+2+3+4+...+14)=14

a: =>14x+2*(14*13/2)=14

=>14x+14*13=14

=>14x=-12*14

=>x=-12

b: =>3x+(0-1-2-3+1+2+...+25)=25

=>3x+4+...+25=25

=>3x+4+...+24=0

=>3x+(24-4+1)*28/2=0

=>3x+21*28/2=0

=>3x=-21*14

=>x=-7*14=-98

\(x+\left(x+1\right)+\left(x+2\right)+....+13+14=14\Leftrightarrow x+\left(x+1\right)+....+13\Leftrightarrow x=-13\)

\(25+24+23+....+x+\left(x-1\right)+\left(x-2\right)=25\Leftrightarrow24+....+\left(x-2\right)=0\Leftrightarrow x-2=-24\)

\(\Leftrightarrow x=-22\)

bạn làm rõ hơn đi

a: \(=\dfrac{2}{5}+\dfrac{3}{5}\cdot\dfrac{10}{3}\cdot\dfrac{1}{2}=\dfrac{2}{5}+\dfrac{10}{5}\cdot\dfrac{1}{2}=\dfrac{2}{5}+1=\dfrac{7}{5}\)

x+(x+1) +(x+2) + ... + (x+14 )=14

(x+x+x+...+x) + (1+2+3+4+ ...+14)=14

15x + [(14-1):1+1].(1+14) :2=14

15x + 105=14

15x=-91

suy ra x=-91/15

Bài 1:

ta có 13¹⁴<13¹⁵

13¹³<13¹⁴

=> dấu cân điền là"<"(13¹⁴ -13¹³ <13¹⁵-13¹⁴)

còn bài 2 thì sao bạn