(x+1)+(x+2)+...+13+14=14
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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. \(\dfrac{3}{14}\)
\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+14\right)=14\)
\(x+x+1+x+2+x+3+...+x+14=14\)
\(15x+\left(1+2+3+...+14\right)=14\)
\(15x+\frac{\left(14+1\right).14}{2}=14\)
\(15x+105=14\)
\(15x=-91\)
\(x=-\frac{91}{15}=-6,06\)
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
a: \(=\dfrac{2}{5}+\dfrac{3}{5}\cdot\dfrac{10}{3}\cdot\dfrac{1}{2}=\dfrac{2}{5}+\dfrac{10}{5}\cdot\dfrac{1}{2}=\dfrac{2}{5}+1=\dfrac{7}{5}\)
\(x+\left(x+1\right)+\left(x+2\right)+....+13+14=14\Leftrightarrow x+\left(x+1\right)+....+13\Leftrightarrow x=-13\)
\(25+24+23+....+x+\left(x-1\right)+\left(x-2\right)=25\Leftrightarrow24+....+\left(x-2\right)=0\Leftrightarrow x-2=-24\)
\(\Leftrightarrow x=-22\)
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