Giải:

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

Ta có:

\(\left|x\right|+\left|x+1\right|+\left|x+2\right|\ge0\) (do mỗi số hạng đều \(\ge0\) )

\(\Rightarrow6x\ge0\) 

\(\Rightarrow x\ge0\) 

Vì \(x\ge0\) 

\(\Rightarrow x+x+1+x+2=6x\) 

\(\Rightarrow3x+3=6x\) 

\(\Rightarrow3x-6x=-3\) 

\(\Rightarrow-3x=-3\) 

\(\Rightarrow x=-3:-3\) 

\(\Rightarrow x=1\) 

Chúc bạn học tốt!

Bài 2:

a) \(2015.2015\) và \(2012.2018\)

Ta có:

\(2015.2015=2015^2\)

\(2012.2018\) 

\(=\left(2015-3\right).\left(2015+3\right)\) 

\(=2015^2-9\) 

Vì \(2015^2>2015^2-9\) nên \(2015.2015>2012.2018\) 

Giải:

a) Số vải xanh của cửa hàng là:

           \(120.\dfrac{2}{5}=48\left(m\right)\) 

Số vải đỏ của cửa hàng là:

           \(\left(120-48\right).\dfrac{1}{3}=24\left(m\right)\) 

Số vải trắng của cửa hàng là:

           \(120-\left(48+24\right)=48\left(m\right)\) 

b) Tỉ số % số mét vải trắng so vs số mét vải của cửa hàng là:

           \(\dfrac{48}{120}.100\%=40\%\) 

Giải:

\(A=\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\) 

\(A=\dfrac{1}{1.7}+\dfrac{1}{7.13}+\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+\dfrac{1}{31.37}\) 

\(A=\dfrac{1}{6}.\left(\dfrac{6}{1.7}+\dfrac{6}{7.13}+\dfrac{6}{13.19}+\dfrac{6}{19.25}+\dfrac{6}{25.31}+\dfrac{6}{31.37}\right)\) 

\(A=\dfrac{1}{6}.\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{37}\right)\) 

\(A=\dfrac{1}{6}.\left(1-\dfrac{1}{37}\right)\) 

\(A=\dfrac{1}{6}.\dfrac{36}{37}\) 

\(A=\dfrac{6}{37}\)

ừm nhưng rất tiếc mk ko có face :( 

Fill in each blank in the sentences with one hobby or one action verb from the box below

I always _____  to Ngoc's songs. I love the sweet melodies. At home I have to use my headphones because my parents don't like loud noise. ________is my favourite hobby.

Giải:

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\) 

Ta có:

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\) 

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\) 

\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\) 

\(...\)

\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{8.9}\) 

\(\Rightarrow B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\) 

\(\Rightarrow B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\) 

\(\Rightarrow B< 1-\dfrac{1}{9}< 1\) 

\(\Rightarrow B< 1\left(đpcm\right)\) 

Vậy \(B< 1\)

\(\dfrac{47}{53}.\left(\dfrac{17}{3}-\dfrac{53}{47}\right)+\dfrac{17}{3}.\left(\dfrac{6}{17}-\dfrac{47}{53}\right)\) 

\(=\dfrac{47}{53}.\dfrac{17}{3}-\dfrac{47}{53}.\dfrac{53}{47}+\dfrac{17}{3}.\dfrac{6}{17}-\dfrac{17}{3}.\dfrac{47}{53}\) 

\(=\left(\dfrac{47}{53}.\dfrac{17}{3}-\dfrac{17}{3}.\dfrac{47}{53}\right)+\left(\dfrac{-47}{53}.\dfrac{53}{47}+\dfrac{17}{3}.\dfrac{6}{17}\right)\) 

\(=\left(1-1\right)+\left(-1+2\right)\) 

\(=0+1\) 

\(=1\)