1. (1 + 3 + 5 + ... + 2005) . ( 125125.127 - 127127.125)
Ta có 125125.127 - 127127.125
=125.1001.127 - 127.1001.125
=0
=> (1 + 3 + 5 + ... + 2005) . ( 125125.127 - 127127.125)
= (1 +3 + 5 +...+ 2005) . 0
=0
Bài 1:
\(\left(1+3+5+...+2005\right)\cdot\left(125125.127-127127.125\right).\)
\(=\left(1+3+5+...+2005\right)\left(125.1001.127-127127.125\right).\)
\(=\left(1+3+5+...+2005\right)\left(125.127127-127127.125\right).\)
\(=\left(1+3+5+...+2005\right).0=0.\)
Vậy.....
Bài 2:
\(\left(7.13+8.13\right):\left(\dfrac{29}{3}-x\right)=39.\)
\(\left[\left(7+8\right)13\right]:\left(\dfrac{29}{3}-x\right)=39.\)
\(\left[15.13\right]:\left(\dfrac{29}{3}-x\right)=39.\)
\(195:\left(\dfrac{29}{3}-x\right)=39.\)
\(\dfrac{29}{3}-x=195:39.\)
\(\dfrac{29}{3}-x=5.\)
\(\Rightarrow x=\dfrac{29}{3}-5.\)
\(\Rightarrow x=\dfrac{29}{3}-\dfrac{15}{3}=\dfrac{14}{3}.\)
Vậy.....
~ Hok tốt!!! ~