\(\left(x\times1\right)+\left(x\times2\right)+...+\left(x\times9\right)=405\)
\(\Rightarrow x\left(1+2+3+...+9\right)=405\)
\(\Rightarrow x\times45=405\)
\(\Rightarrow x=405:45\)
\(\Rightarrow x=9\)
Xyz sao mình tính được là 405:50 là có dư
( X x 1 ) + ( X x 2 ) + ( X x 3 ) +.....+ ( X x 9 ) = 405
X x ( 1 + 2 +3 + .....+ 9 ) = 405
X x 45 = 405
X = 405: 45
X = 9
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\(\left(x\times1\right)+\left(x\times2\right)+...+\left(x\times9\right)=405\)
=> \(x\left(1+2+3+...+9\right)=405\)
=>\(x\times45=405\)
=> \(x=9\)
Hok tot
(x . 1) + (x . 2) + (x . 3) + ... + (x . 9) = 405
<=> x . 1 + x .2 + x . 3 + ... + x . 9 = 405
<=> x(1 + 2 + 3 + ... + 9) = 405
<=> x . 45 = 405
<=> x = 405/45 = 9