Đặt \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}=k\)

\(\Rightarrow a=2003k;b=2004k;c=2005k\)

\(\Rightarrow4\left(a-b\right)\left(b-c\right)=4\left(2003k-2004k\right).\left(2004k-2005k\right)=4.\left(-k\right).\left(-k\right)=4k^2\)(1)

     \(\left(c-a\right)^2=\left(2006k-2004k\right)^2=\left(2k\right)^2=4k^2\)(2)

Từ (1) và (2) 

\(\Rightarrow4.\left(a-b\right).\left(b-c\right)=\left(c-a\right)^2\)

                                      đpcm

Tham khảo nhé~  

\(\left(5-x\right).\left(3x-\frac{1}{4}\right)>0\)

\(\Leftrightarrow\hept{\begin{cases}5-x>0\\3x-\frac{1}{4}>0\end{cases}}\) hoặc \(\hept{\begin{cases}5-x< 0\\3x-\frac{1}{4}< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x< 5\\x>\frac{1}{12}\end{cases}}\)  hoặc  \(\hept{\begin{cases}x>5\\x< \frac{1}{12}\end{cases}}\) (vô lí)

Vậy \(\frac{1}{12}< x< 5\)

a) \(\frac{x}{6}=\frac{y}{-7};\frac{x}{3}=\frac{z}{-8}\Rightarrow\frac{y}{-21}=\frac{x}{18}=\frac{z}{-48}\)

Áp dụng tính chất dãy tỉ số bằng nhau

Ta có: \(\frac{2x}{36}=\frac{2y}{-42}=\frac{3z}{-144}=\frac{2x-2y+3z}{36-\left(-42\right)+\left(-144\right)}=\frac{56}{-66}=\frac{-28}{33}\)

\(\Rightarrow2x=\frac{28}{33}.36=\frac{-336}{11}\Rightarrow x=\frac{-168}{11}\)

    \(2y=\frac{-28}{33}.\left(-42\right)=\frac{392}{11}\Rightarrow y=\frac{196}{11}\)

    \(3z=\frac{-28}{33}.\left(-144\right)=\frac{1344}{11}\Rightarrow z=\frac{448}{11}\)

b) \(3x=-4y=2z\Rightarrow\frac{x}{-4}=\frac{y}{3};\frac{y}{2}=\frac{z}{-4}\Rightarrow\frac{x}{-8}=\frac{y}{6}=\frac{z}{-12}\)

\(\Rightarrow\frac{2x}{-16}=\frac{2y}{12}=\frac{3z}{-36}=\frac{2x-2y+3z}{-16-12+\left(-36\right)}=\frac{56}{-64}=\frac{-7}{8}\)

\(\Rightarrow2x=\frac{-7}{8}.\left(-16\right)=14\Rightarrow x=7\)

     \(2y=\frac{-7}{8}.12=\frac{-21}{2}\Rightarrow y=\frac{-21}{4}\)

     \(3z=\frac{-7}{8}.\left(-36\right)=\frac{63}{2}\Rightarrow z=\frac{21}{2}\)

c) Tương tự

làm giúp mình với

 =>  \(3\frac{1}{3}:2,4=x\)

=> x = \(\frac{25}{18}\)

\(x^3< x^2\)
\(\Leftrightarrow x< 0\)

x là số nguyên âm

Để a+x < a

=> x < 0

\(a+x< a\)
\(\Leftrightarrow x< 0\)

Để x > 2x 

=> x <  0

x > 2x

x - 2x > 2x -2x

-x > 0

(-x).(-1) < 0.(-1)

=> x < 0

Thay x vào N ta có : N = 25.(-1/5)^2 +10.-1/5 +1

= 25.1/25 + (-10/5) +1

= 1 + (-2) +1

=0

\(N=25x^2+10x+1=\left(5x+1\right)^2\)

Tại   \(x=-\frac{1}{5}\)thì:   \(N=\left(5.\frac{-1}{5}+1\right)^2=0\)

p/s: chúc bạn học tốt

\(\frac{1}{7}\times\frac{1}{49}\times49^2=\frac{1}{7}\times\frac{49^2}{49}=\frac{1}{7}\times49=\frac{49}{7}=\frac{7\times7}{7}=7\)

\(\frac{1}{7}\cdot\frac{1}{49}\cdot49^2=\frac{1}{7}\cdot\left(\frac{1}{49}\cdot49^2\right)=\frac{1}{7}\cdot\frac{49^2}{49}=\frac{1}{7}\cdot49=\frac{49}{7}=7\)