Question

Câu 1: Phân số \(\dfrac{a}{b}\) sau khi rút gọn được phân số \(\dfrac{-8}{11}\) . Biết b-a = 190, tìm phân số \(\dfrac{a}{b}\) .

Câu 2: Tính giá trị biểu thức:

P = \(\dfrac{2.3.4-2.3.4.9+2.3.4.11-2.3.4.13}{5.6.7-5.6.7.9+5.6.7.11-5.6.7.13}\)

Câu 3: Chứng minh rằng S =\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< 1\)

Answer 1

1,

Vì \(\dfrac{a}{b}=\dfrac{-8}{11}\)

=>a.11=b.(-8) (1)

Mà b-a=190

=>b=a+190 (2)

Từ (1), (2)

=>a.11=(a+190).(-8)

=>11a=(-8).a+190.(-8)

=>11a=-8.a+(-1520)

=>11a+8a=-1520

=>a.(11+8)=-1520

=>a.19=-1520

=>a=(-1520):19

=>a=-80

=>b=-80+190=110

Vậy a=-80;b=110

Answer 2

Câu 1:

Ta có:

\(\dfrac{a}{b}=\dfrac{-8}{11}\Leftrightarrow1-\dfrac{a}{b}=1-\dfrac{-8}{11}\)

Hay \(\dfrac{b-a}{b}=\dfrac{11+8}{11}\left(1\right)\)

Thay \(b-a=190\) vào \(\left(1\right)\) ta được:

\(\dfrac{190}{b}=\dfrac{19}{11}\Leftrightarrow b=110\Leftrightarrow a=-80\)

Vậy phân số \(\dfrac{a}{b}=\dfrac{-80}{110}\)

Câu 2: Ta có:

\(P=\dfrac{2.3.4-2.3.4.9+2.3.4.11-2.3.4.13}{5.6.7-5.6.7.9+5.6.7.11-5.6.7.13}\)

\(=\dfrac{2.3.4.\left(1-9+11-13\right)}{5.6.7.\left(1-9+11-13\right)}\)

\(=\dfrac{2.3.4}{5.6.7}=\dfrac{4}{35}\)

Vậy \(P=\dfrac{4}{35}\)

Câu 3:

\(S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}\)

\(\Rightarrow2S=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}\right)\)

\(=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{19}}\)

Do đó \(2S-S=1-\dfrac{1}{2^{20}}\) Hay:

\(S=1-\dfrac{1}{2^{20}}< 1\) (Đpcm)

Answer 3

\(P=\dfrac{2.3.4-2.3.4.9+2.3.4.11-2.3.4.13}{5.6.7-5.6.7.9+5.6.7.11-5.6.7.13}\)

=>\(P=\dfrac{2.3.4.\left(1-9+11-13\right)}{5.6.7.\left(1-9+11-13\right)}\)

=>\(P=\dfrac{2.3.4}{5.6.7}\)

=>\(P=\dfrac{2.3.4}{2.3.5.7}\)

=>\(P=\dfrac{4}{5.7}\)

=>\(P=\dfrac{4}{35}\)

Answer 4

Ta có \(S=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{20}}\)

=>\(2S=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{20}}\right)\)

=>\(2S=1+\dfrac{1}{2}+...+\dfrac{1}{2^{19}}\)

=>\(2S-S=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^{19}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{20}}\right)\)

=>S=1\(-\dfrac{1}{2^{20}}\)

Mà \(1-\dfrac{1}{2^{20}}\)<1

=>S<1

Vậy S<1