\(M=\frac{1.3.5...2011.2013}{1008.1009.1010...2013.2014}\)

\(M=\frac{1.2.3.4.5.6...2011.2012.2013.2014}{\left(2.4.6...2014\right).1008.1009.1010....2013.2014}\)

\(M=\frac{1.2.3.4.5.6...2011.2012.2013.2014}{2^{1007}.\left(1.2.3...1007\right).1008.1009.1010...2013.2014}\)

\(M=\frac{1}{2^{1007}}\)

C = \(\frac{1.3.5.......2011.2013}{1008.1009......2013.2014}\)

C =  \(\frac{1.2.3.4.5......2013.2014}{\left(2.4.6....2012.2014\right).\left(1008.1009.....2014\right)}\)

C = \(\frac{1.2.3.......2013.2014}{2^{1007}.\left(1.2.3.4.5.....1007\right).1008.1009....2014}\)

C = \(\frac{1.2.3.4.5.......2014}{2^{1007}.1.2.3.4.5.......2014}\)

C = \(\frac{1}{2^{1007}}\)

c,

Tổng P có số số hạng là

   (113-2):3+1=38(số)

Có số cặp là

    38:2=19(cặp)

Ta có: P=2-5+8-11+14-17+...+110-113

= (2-5)+(8-11)+(14-17)+...+(110-113)

= (-3)+(-3)+(-3)+...+(-3)

= (-3).19

= -57

\(B=\frac{1.3.5....2011.2013}{1008.1009....2013.2014}\)

\(\Rightarrow B=\frac{1.3.5...1007.1009...2013}{1008.1009...2013.2014}\)

\(\Rightarrow B=\frac{1.3.5...1006}{1008.1010...2014}\)

=1/2^1007  nha

Khi x >0

A=\(\frac{|x-|x||}{x}=\frac{|x-x|}{x}=\frac{0}{x}=\)0

Khi x <0

A=\(\frac{|x-|x||}{x}=\frac{|x--x|}{x}=\frac{|2x|}{x}=\frac{-2x}{x}=-2\)

Vậy A\(\in\){-2;0}

\(A=\frac{\left|x-|x|\right|}{x}\)

Ta có : \(\left|x\right|\ge0\) và \(\left|x-|x|\right|\ge0\)

=> |x-|x||=|x-x|=0

<=> \(\frac{0}{x}=0\)

Vậy A=0

Bạn bắt tờ cô sai rồi nhé Bạn phá dấu trị tuyệt đối theo kiểu vô lí

\(A=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)

\(A=\frac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}\)

\(A=\frac{11.3^{29}-3^{30}}{4.3^{28}}\)

\(A=\frac{3^{29}.\left(11-3\right)}{4.3^{28}}\)

\(A=\frac{3^{29}.8}{4.3^{28}}=3.2=6\)