bạn xem lại đề:

Có \(\frac{3}{2}<\frac{7}{3}\) nhưng \(\frac{3}{2}>\frac{3+7}{2+7}=\frac{10}{9}\)

X x X x X x X x X = X⁵ = 2¹⁰.10⁵ = 40⁵

=> X = 40

=> Y = 40 - 30 = 10

=> Z = 20 : 10 = 2

Ta có: A + B + C = 119 (1)

           B= 2+C

           A - C = C x 10

           A = 10  xC +C

           A = 11x C

Thay B= 2+C; A =11xC vào (1) ta được:

         11xC + 2+C +C = 119

               13 x C +2 =119

                  13 x C = 119 - 2

                   13 x C = 117

                          C  = 117 : 13

                          C  = 9

       =>           B = 2 +C = 9+2 = 11

       =>           A = 11 x 9

       =>           A = 99

A:B=C NEN A=BXC

A-C=BXC-C=C(B-1)=CX10

SUY RA B-1=10

B=10+1

B=11

C+2=B

C+2=11

C=11-2

C=9

A:B=C

A:11=9

A=9X11

A=99

VAY A=99,B=11,C=9

****

\(B=1-\frac{1}{1+2}-\frac{1}{1+2+3}-...-\frac{1}{1+2+3+...+2016}\)

\(=1-\left(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2016}\right)\)

\(=1-\left(\frac{1}{2.3:2}+\frac{1}{3.4:2}+...+\frac{1}{2016.2017:2}\right)\)

\(=1-\left(\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{2016.2017}\right)\)

\(=1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)\)

\(=1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)

\(=1-2\left(\frac{1}{2}-\frac{1}{2017}\right)=1-\left(1-\frac{2}{2017}\right)=1-1+\frac{2}{2017}=\frac{2}{2017}\)

\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\cdot...\cdot\left(\frac{1}{2015}-1\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2014}{2015}=\frac{1\cdot2\cdot3\cdot...\cdot2013\cdot2014}{2\cdot3\cdot...\cdot2014\cdot2015}=\frac{1}{2015}\)

\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2015}-1\right)=-\frac{1}{2}.\frac{-2}{3}...\frac{-2014}{2015}=\frac{1}{2015}\)

\(=3uv^2-\frac{1}{5}uv^2-367\frac{1}{4}uv^2-\frac{19}{5}uv^2+317\frac{1}{4}uv^2\)

\(=3uv^2-\frac{1}{5}uv^2-\frac{19}{5}uv^2=3uv^2-4uv^2=-uv^2\)

\(3uv^2-\left(\frac{1}{5}uv^2+367\frac{1}{4}uv^2\right)+\left(\frac{-19}{5}uv^2\right)+\left(367\frac{1}{4}uv^2\right)\)

=\(3uv^2-\frac{1}{5}uv^2-367\frac{1}{4}uv^2-\frac{19}{5}uv^2+317\frac{1}{4}uv^2\)

=\(3uv^2-\frac{1}{5}uv^2-\frac{19}{5}uv^2\)

=\(3uv^2-4uv^2\)

=\(-uv^2\)

A=( 1/2-2/2 )(1/3-3/3)....(1/2015-1)

A=(-1/2)(-2/3)....(-2014/2015)

A=\(\frac{\left(-1\right)\left(-2\right).....\left(-2014\right)}{2.3....2015}\)

A=\(\frac{\left(-1\right)\left(-1\right)...\left(-1\right)\left(2014số-1\right)}{1.1.....1.2015}\)

A=1/2015

A \(=\frac{1}{2016}\)

Mình chỉ biết A=5;B=2 còn N thì =25

Ta có:

AxBxAxNxBxN=A^2xB^2xN^2=(AxBxN)^2=10x25x50=12500

=> AxBxN=\(\sqrt{12500}=???????kochiadc\)