mk đang cần rất gấp mọi người  giải giúp mk nha

A=[(-1).(-1)³.(-1)⁵. ..... .(-1)²⁰¹⁹  ].[(-1)².(-1)⁴. ... .(-1)²⁰¹⁸  ]

  =          -1     .    1

  =    -1

B=1152-(374+1152)+(-65+374)

  = 1152- 374- 1152- 65+ 374

  = 1152- 1152 -374+ 374 - 65

   =0-65

   =-65

\(=>M=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{2020}{2019}\)

\(=>M=\frac{2020}{2}=1010\)

\(M=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{2018}\right)\left(1+\frac{1}{2019}\right)\)

    \(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{2019}{2018}.\frac{2020}{2019}\)

    \(=\frac{2020}{2}\)

    \(=1010\)

Study well ! >_<

\(A=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{1}{2018}\)

\(A=1+\left(1+\frac{2017}{2}\right)+\left(1+\frac{2016}{3}\right)+...+\left(1+\frac{1}{2018}\right)\)

\(A=\frac{2019}{2019}+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2018}\)

\(A=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)\)

Ta có: \(\frac{A}{B}=\frac{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}=2019\)

Theo bài ra, ta có: \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2017.2018.2019}\)

\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2017.2018.2019}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2018.2019}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2018.2019}\right)\)

Giải thích:

\(\frac{2}{1.2.3}=\frac{3}{1.2.3}-\frac{1}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\)

\(\frac{2}{2.3.4}=\frac{4}{2.3.4}-\frac{2}{2.3.4}=\frac{1}{1.2}-\frac{1}{3.4}\)

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

\(\frac{2}{2017.2018.2019}=\frac{2019}{2017.2018.2019}-\frac{2017}{2017.2018.2019}=\frac{1}{2017.2018}-\frac{1}{2018.2019}\)

2020/2019 x 2019/2018 x 2018/2017 x....................3/2
= 2020/2
= 1010 

\(A=1\dfrac{1}{2019}\times1\dfrac{1}{2018}\times1\dfrac{1}{2017}\times...\times1\dfrac{1}{2}\)

\(=\dfrac{2020}{2019}\times\dfrac{2019}{2018}\times\dfrac{2018}{2017}\times...\times\dfrac{3}{2}\)

\(=\dfrac{2020}{2}\)

\(=1010\)

a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)

=1-2/4=1/2

b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)

c: x-y=0 nên x=y

\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)

=2019

hình như cái này đâu phải toán lớp 5 đâu bạn

nhầm toán lớp 6

12+13×14

\(x=\frac{1}{2}\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=\frac{1}{2}\sqrt{\frac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}}=\frac{\sqrt{2}-1}{2}\)

\(\Leftrightarrow2x+1=\sqrt{2}\)

\(\Leftrightarrow4x^2+4x-1=0\)
Giờ thế vô A đi