\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{a+b+c}=1\)

\(\Rightarrow\frac{a}{b}=1;\frac{b}{c}=1;\frac{c}{a}=1\)

\(a=b=c\)

\(\Rightarrow a+a+a=2007\Leftrightarrow3a=2007\Rightarrow a=669\)

Vậy \(a=b=c=669\)

We have \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=\frac{2007}{2007}=1\)

\(\Rightarrow\frac{a}{b}=1\Rightarrow a=b\)

\(\frac{b}{c}=1\Rightarrow b=c\)

\(\frac{c}{a}=1\Rightarrow c=a\)

Then a = b = c ; 2007 : 3 = ?????????

Wrong topic

Ta có : \(\left|2x+3\right|+\left|2x-5\right|\ge\left|2x+3+2x-5\right|=\left|4x-2\right|\)

\(\Rightarrow\left|4x-2\right|\le8\)

\(\Leftrightarrow-8\le4x-2\le8\)

\(\Leftrightarrow-6\le4x\le10\)

\(\Rightarrow-1\le x\le2\)

\(\Rightarrow x=-1;0;1;2\)

=—11/4xy^2+2x^2y+6/5xy

A=\(\left(3xy^2-2xy^2-4xy^2\right)+\left(2x^2y+\frac{1}{4}x^2y\right)+\left(xy+\frac{1}{5}xy\right)\)

A=\(-3xy^2+\frac{9}{4}x^2y+\frac{6}{5}xy\)

dài thế ai mà làm được

a) =1+1+1+....+1     (50 số bạn nhé)

    =1*50=50

b)TH x=-1: => a-b+c

   TH x=1   => a+b+c

Nếu \(x=\frac{1}{2}\Rightarrow S=-2\)

       \(x=\frac{-1}{2}\Rightarrow S=\frac{-14}{5}\)

Nhớ k đúng nhé !

Chúc các bạn học tốt !!!!

Vì | x| = 1/2 suy ra x = 1/2 hoặc x=- 1/2

* Tại x = 1/2:Thay x = 1/2 vào S ta có: S=\(\frac{5\cdot\left(\frac{1}{2}\right)^2-7\cdot\frac{1}{2}+1}{3\cdot\left(\frac{1}{2}\right)-1}=\frac{5\cdot\frac{1}{4}-\frac{7}{2}+1}{\frac{3}{2}-1}=-\frac{5}{4}\cdot2=-\frac{5}{2}\)

* Tại  x= - 1/2: Thay x= - 1/2 vào S ta có: S=\(\frac{5\cdot\left(-\frac{1}{2}\right)^2-7\left(-\frac{1}{2}\right)+1}{3\cdot\left(-\frac{1}{2}\right)-1}=\frac{5\left(\frac{1}{4}\right)+\frac{7}{2}+1}{-\frac{3}{2}-1}=\frac{23}{4}\cdot\left(-\frac{2}{5}\right)=-\frac{23}{10}\)

Ta thấy 1 +3=4 chia hết cho 4 ; 3+3^2 =36 chia hết cho 4; ...đến 3^99 +3^100 chia hết cho 4

suy ra 1+3+3^2+...+3^100 chia hết cho 4(điều phải chứng minh)

k mình nhé

\(1+3+3^2+................+3^{100}\)

\(=4+3^2+3^3+.......+3^{100}\)

\(=4+3^2\left(1+3\right)+.......+3^{99}\left(1+3\right)\)

\(=4+3^2\cdot4+.......+3^{99}\cdot4\)

\(=4\left(1+3^2+.....+3^{99}\right)⋮4\)

\(\Rightarrow1+3+3^2+3^3+......+3^{100}⋮4\)