Tìm x biết
a) x +(x + 1) + (x + 2) + .... + (x +30) = 620
b) 2+4+6+8+....+2x = 210
c) (x-1) +(x-2)+(x-3)+...+(x-20) = -610
d) (x-3)(2x-7) = 0
e) |3-x| = x-5
g) |5x-2| \(^{_{ }\le}\)13
h) x+|2-x|=6
t) \(3^x+3^{x+1}+3^{x+2}=1053\)
i) | x+1| +|x-2| = 3
Nhanh lên! Mình cần gấp. Nguyễn Huy Tú
a) \(x +(x + 1) + (x + 2) + ... + (x +30) = 620\)
\(=\left(x+x+...+x+x\right)+\left(1+2+...+30\right)\)
\(=31x+465=620\)
\(=31x=620-465\)
\(=31x=155\)
\(=x=155\div31\)
\(x=5\)
b) \(2+4+6+8+....+2x = 210\)
\(\Rightarrow2.1+2.2+2.3+2.4+...+2.x\)
\(\Rightarrow2.\left(2+4+6+8+...+x\right)=210\)
\(\Rightarrow2+4+6+8+x=210\div2\)
\(\Rightarrow2+4+6+8+...+x=105\)
\(\Rightarrow x=14\)
c) \(\left(x-1\right)+\left(x-2\right)+...+\left(x-20\right)=-610\)
\(\Rightarrow\left(x+x+...+x\right)-\left(1+2+...+20\right)=-610\) ( 20 số x )
\(\Rightarrow20x-210=-610\)
\(\Rightarrow20x=-400\)
\(\Rightarrow x=-20\)
Vậy x = -20
d) \(\left(x-3\right)\left(2x-7\right)=0\)
\(\Rightarrow\left[\begin{matrix}x-3=0\\2x-7=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=3\\x=3,5\end{matrix}\right.\)
Vậy \(x\in\left\{3;3,5\right\}\)
t) \(3^x+3^{x+1}+3^{x+2}=1053\)
\(\Rightarrow3^x+3^x.3+3^x.3^2=1053\)
\(\Rightarrow3^x\left(1+3+3^2\right)=1053\)
\(\Rightarrow3^x.13=1053\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
Vậy x = 4
