1+1=3 :)))

Tách 100 thành 100 số 1

Ta có: TS=\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=100-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{100}=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)

=\(0+\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}=\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}\)=MS

=> Phân số trên=1

haha!dungs rois!

trả lời: \(\frac{1}{100}\) nha

😁 😁 😁

\(F=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)

\(F=\left(\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\right)\)

\(F=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)-2.\left(\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\right)\)

\(F=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{50}}\right)\)

\(F=\frac{1}{2^{51}}+\frac{1}{2^{52}}+...+\frac{1}{2^{100}}\)

\(E=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(2E=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(2E-E=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(E=1-\frac{1}{2^{100}}\)

a)S=2+2²+2³+...+2¹⁰⁰

2S=2(2+2²+2³+...+2¹⁰⁰)

2S=2²+2³+...+2¹⁰¹

2S-S=(2²+2³+...+2¹⁰¹)-(2+2²+2³+...+2¹⁰⁰)

S=2¹⁰¹-2

b)\(P=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)

\(3P=3\left(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{100}}\right)\)

\(3P=1+\frac{1}{3}+...+\frac{1}{3^{99}}\)

\(3P-P=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)\)

\(2P=1-\frac{1}{3^{100}}\)

\(P=\left(1-\frac{1}{3^{100}}\right):2\)

ngài Kiệt ღ ๖ۣۜLý๖ۣۜ   đúng là không ái sánh bằng sự gian xảo này