Question

[Nguyễn Mỹ Chi]

\(\dfrac{x-2}{5}=\dfrac{x}{3}\)

\(\dfrac{x+23}{x+40}=\dfrac{3}{4}\)

A=\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.........+\dfrac{1}{2^{2017}}\)

B=1+2+2²+2³+.............+2²⁰¹⁷

Answer 1

[Nguyễn Thanh Hằng]

a, \(\dfrac{x-2}{5}=\dfrac{x}{3}\)

\(\Leftrightarrow3\left(x-2\right)=5x\)

\(\Leftrightarrow3x-6=5x\)

\(\Leftrightarrow5x-3x=6\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

b, \(\dfrac{x+23}{x+40}=\dfrac{3}{4}\)

\(\Leftrightarrow4\left(x+23\right)=3\left(x+40\right)\)

\(\Leftrightarrow4x+92=2x+80\)

\(\Leftrightarrow4x-2x=80-92\)

\(\Leftrightarrow2x=-12\)

\(\Leftrightarrow x=-6\)

c, \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...........+\dfrac{1}{2^{2017}}\)

\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...........+\dfrac{1}{2^{2016}}\)

\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2016}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2017}}\right)\)

\(\Leftrightarrow A=1-\dfrac{1}{2^{2017}}\)

d, \(B=1+2+2^2+........+2^{2017}\)

\(\Leftrightarrow2B=2+2^2+2^3+......+2^{2018}\)

\(\Leftrightarrow2B-B=\left(2+2^2+.....+2^{2018}\right)-\left(1+2+....+2^{2017}\right)\)

\(\Leftrightarrow B=2^{2018}-1\)

Answer 2

[Dật Hàn Bạch]

\(\dfrac{x-2}{5}=\dfrac{x}{3}=>3\left(x-2\right)=5x\)

\(< =>3x-6=5x=>x=-3\)

\(\dfrac{x+23}{x+40}=\dfrac{3}{4}=>4\left(x+23\right)=3\left(x+40\right)\)

\(4x+92=3x+120=>x=28\)