dễ thấy vế trái luôn>0 nên 6x>0=> x>0

x>0, bỏ dấu trị tuyệt đối ra ta đc 4x+10=6x

x=5

chúc bạn học giỏi, ăn Tết đc ngon, hehe -_-

HYC-30/1/2022

Answer:

\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=6x\)

Có \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|\ge0\)

\(\Rightarrow6x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+1+x+2+x+3+x+4=6x\)

\(\Rightarrow4x+10=6x\)

\(\Rightarrow2x=10\)

\(\Rightarrow x=5\)

=6/11 nha

HT

chúc bn năm mới vui vẻ

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=0.54545454545 nha 

= \(\frac{11}{18}\)nha bạn

=11/8

chắc chắn

\(1-\dfrac{5}{14}-\dfrac{5}{84}-\dfrac{5}{204}-\dfrac{5}{374}\) =\(1-\left(\dfrac{1}{1.14}+\dfrac{1}{14.6}+\dfrac{1}{6.34}+\dfrac{1}{34.11}\right)\)

= \(1-\left(\dfrac{1}{1}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{11}\right)\)

= 1 - \(\left(1-\dfrac{1}{11}\right)\) = \(1-\dfrac{10}{11}=\dfrac{1}{11}\)

\(=1-\dfrac{5}{2\cdot7}-\dfrac{5}{7\cdot12}-\dfrac{5}{12\cdot17}-\dfrac{5}{17\cdot22}\)

\(=1-\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}\right)\)

\(=5-\dfrac{11-1}{22}=5-\dfrac{10}{22}=5-\dfrac{5}{11}=\dfrac{50}{11}\)

Chữ con trai gì mà đẹp thế