12 x 2 =
13 x 2 =
14 x 2 =
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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
Bạn xem lại đề nha \(\dfrac{x+2}{15}\)
\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
=> \(\left(x+2\right)\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)
=> x=-2 do \(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\)\(\ne0\)
\(\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)
\(\Rightarrow\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}-\frac{x+2}{14}-\frac{x+2}{15}=0\)
\(\Rightarrow\left(x+2\right)\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
\(\Rightarrow x+2=0\left(\text{Vì }\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\ne0\right)\)
=>x=-2
Vậy x=-2
\(\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)
\(\Leftrightarrow\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}-\frac{x+2}{14}-\frac{x+2}{15}=0\)
\(\Leftrightarrow\left(x+2\right)\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
\(\Leftrightarrow x+2=0\). Do \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\ne0\)
\(\Leftrightarrow x=-2\)
x+2/11+x+2/12+x+2/13=x+2/14+x+2/15
(x+x+x)+(2/11+2/12+2/13)= (x+x)+(2/14+2/15)
x.3+431/858 = x.2 + 29/105
x.3-x.2=-431/858 +29/105
x.1 =-6791/30030
x= -9791/30030:1
x= -9791/30030
vậy x= -9791/ 30030
= 24
= 26
=28
HT
12 x 2 = 24
13 x 2 = 26
14 x 2 = 28