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\(\frac{x+2014}{2}+\frac{2x+4028}{7}=\frac{x+2014}{5}+\frac{x+2014}{6}\)

<=> \(\frac{x+2014}{2}+\frac{2\left(x+2014\right)}{7}=\frac{x+2014}{5}+\frac{x+2014}{6}\)

<=> \(\frac{x+2014}{2}+\frac{x+2014}{\frac{7}{2}}=\frac{x+2014}{5}+\frac{x+2014}{6}\)

<=> \(\frac{x+2014}{2}+\frac{x+2014}{\frac{7}{2}}-\frac{x+2014}{5}-\frac{x+2014}{6}=0\)

<=> \(\left(x+2014\right)\left(\frac{1}{2}+\frac{1}{\frac{7}{2}}-\frac{1}{5}-\frac{1}{6}\right)=0\)

Vì \(\frac{1}{2}+\frac{1}{\frac{7}{2}}-\frac{1}{5}-\frac{1}{6}\ne0\)

=> x + 2014 = 0 <=> x = -2014

Bài làm :

\(\frac{x+2014}{2}+\frac{2x+4028}{7}=\frac{x+2014}{5}+\frac{x+2014}{6}\)

\(\Rightarrow\frac{x+2014}{2}+\frac{2x+4028}{7}-\frac{x+2014}{5}-\frac{x+2014}{6}=0\)

\(\Rightarrow\frac{x+2014}{2}+\frac{2.\left(x+2014\right)}{7}-\frac{x+2014}{5}-\frac{x+2014}{6}=0\)

\(\Rightarrow\left(x+2014\right).\left(\frac{1}{2}+\frac{2}{7}-\frac{1}{5}-\frac{1}{6}\right)=0\)

\(\Rightarrow x+2014=0:\left(\frac{1}{2}+\frac{2}{7}-\frac{1}{5}-\frac{1}{6}\right)\)

\(\Rightarrow x+2014=0\)

\(\Rightarrow x=-2014\)

Vậy x = - 2014 .

Học tốt nhé

\(\dfrac{x+2014}{2}+\dfrac{2\left(x+2014\right)}{7}=\dfrac{x+2014}{5}+\dfrac{x+2014}{6}\)

\(\left(x+2014\right)\left(\dfrac{1}{2}+\dfrac{2}{7}\right)=\left(x+2014\right)\left(\dfrac{1}{5}+\dfrac{1}{6}\right)\)

\(\left(x+2014\right)\dfrac{11}{14}=\left(x+2014\right)\dfrac{11}{30}\)

Dấu ''=''↔x=-2014

bạn có thể ghi rõ câu hỏi k

Đặt \(g\left(x\right)=2014x\).

Ta có \(f\left(1\right)-g\left(1\right)=0;f\left(2\right)-g\left(2\right)=0;f\left(3\right)-g\left(3\right)=0\).

Do đó \(f\left(x\right)-g\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)Q\left(x\right)\).

\(f\left(x\right)=2014x+\left(x-1\right)\left(x-2\right)\left(x-3\right)Q\left(x\right)\).

Do f(x) có bậc 4, hệ số cao nhất là 1 nên Q(x) là đa thức có dạng x + m.

Từ đó \(f\left(x\right)=2014x+\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+m\right)\)

\(\Rightarrow f\left(-1\right)+f\left(5\right)=2014.\left(-1\right)+\left(-2\right).\left(-3\right).\left(-4\right)\left(m-1\right)+2014.5+4.3.2\left(m+5\right)=12228\).

2014 x 3 + 4028 x 2 + 2014 x 5 - 2014 x 2

= 2014 x 3 + 2014 x 2 x 2 + 2014 x 5 - 2014 x 2

= 2014 x 3 + 2014 x 4 + 2014 x 5 - 2014 x 2

= 2014 x ( 3 + 4 + 5 - 2 )

= 2014 x 10 = 20140

\(2014\text{×}3+4028\text{×}2+2014\text{×}5-2014\text{×}2\\ =2014\text{×}3+2014\text{×}2\text{×}2+2014\text{×}5-2014\text{×}2\\ =2014\text{×}3+2014\text{×}4+2014\text{×}5-2014\text{×}2\\ =2014\text{×}\left(3+4+5-2\right)=2014\text{×}10=20140\)

hack à

Có : 

A = (2014/1-x + 2014/1+x) + 4028/1+x^2 + 8056/1+x^4 + 16112/1+x^8 + 2,1314

   = 4028/1-x^2 + 4028/1+x^2 + 8056/1+x^4 + 16112/1+x^8 + 2,1314

   = 8056/1-x^4 + 8056/1+x^4 + 16112/1+x^8 + 2,1314

   = 16112/1-x^8 + 16112/1+x^8 + 2,1314

   = 32224/1-x^16 + 2,1314

Tk mk nha

Đề bài là gì vậy bạn

Sửa lại đề đi rùi báo cho mk để mk làm cho

Nhớ đó nha 

Aaaaaaaa.......Sorry mình thiếu đề bài phần thiếu là Rút gọn rrooif tính giá trị của A khi x=1,1.Mh không cố í