Question

[trần văn trung]

|x+1|+|x+2|+|x+3|+...+|x+100|=605x

Answer 1

[Tsukasa Hiiragi]

Ta có \(|x+1|\ge0\)

         \(|x+2|\ge0\)

        \(|x+3|\ge0\)

         \(.......\)

        \(|x+100|\ge0\)

\(\Rightarrow|x+1|+|x+2|+|x+3|+...+|x+100|=605x\)

\(\Rightarrow x+1+x+2+x+3+...+x+100=605x\)

\(\Rightarrow\left(x+x+x+x+...+x\right)+\left(1+2+3+...+100\right)=605x\)

\(\Rightarrow100x+\left(100+1\right)\cdot100:2=605x\)

\(\Rightarrow100x+5050=605x\)

\(\Rightarrow605x-100x=5050\)

\(\Rightarrow x\left(605-100\right)=5050\)

\(\Rightarrow x\cdot505=5050\)

\(\Rightarrow x=5050:505\)

\(\Rightarrow x=10\)

Vậy x=10

Answer 2

[Oh Nova]

Ta có:

|x+1|>=0 với mọi x

|x+2|>=0 với mọi x

.........................

|x+100|>=0 với mọi x

=> |x+1| + |x+2| + |x+3| + ......+ |x+100| >=0 với mọi x

Mà |x+1| + |x+2| + |x+3| + ......+ |x+100| = 605x

605x >=0 với mọi x

=> x>=0

Vậy |x+1|= x+1

 |x+2|= x+2

.............

 |x+100|= x+100

Vậy |x+1| + |x+2| + |x+3| + ......+ |x+100| = x+1+x+2 x+3 + ....+ x+100=605x

=100x+ 5050=605x

605x-100x=5050

505x=5050

=> x=10

Vậy X = 10