\(\frac{a-b}{b}=\frac{3}{5}\). Tính \(\frac{a}{b}\)
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a, A = \(\frac{1}{2}.\frac{3}{4}.\frac{4}{5}...\frac{99}{100}\)
\(A=\frac{1}{2}.\left(\frac{3.4....99}{4.5...100}\right)\)
\(A=\frac{1}{2}.\left(\frac{3}{100}\right)\)\(\)\(A=\frac{3}{200}\)
\(B=\frac{2}{3}.\frac{4}{5}.\frac{5}{6}...\frac{100}{101}\)
\(B=\frac{2}{3}.\left(\frac{4.5...100}{5.6...101}\right)\)
\(B=\frac{2}{3}.\left(\frac{4}{101}\right)\)
\(B=\frac{8}{303}\)
\(A.B=\frac{8}{303}.\frac{3}{200}\)
\(A.B=\frac{1}{2525}\)
b, A = 1/2 x 3/100
B = 2/3 x 4/101
Ta có : 1 - 2/3 = 1/3; 1 - 1/2 = 1/2
MÀ 1/3 < 1/2 => 2/3 > 1/2 (1)
Ta có : 1 - 3/100 = 97/100
1 - 4/101 = 97/101
Mà 97/101 < 97/100 => 4/101 > 3/100 (2)
Từ (1) và (2) => B > A
a,
\(AB=\left[\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right]\cdot\left[\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\right]\)
\(AB=\frac{\left[1\cdot3\cdot5\cdot...\cdot99\right]\left[2\cdot4\cdot6\cdot...\cdot100\right]}{\left[2\cdot4\cdot6\cdot8\cdot...\cdot100\right]\left[3\cdot5\cdot7\cdot...\cdot101\right]}=\frac{1\cdot3\cdot5\cdot...\cdot99}{3\cdot5\cdot7\cdot...\cdot101}=\frac{1}{101}\)
b,
1/2 < 2/3
3/4 < 4/5
.............
99/100 < 100/101
=> \(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}< \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\Leftrightarrow A< B\)
a)
\(A=\left(\frac{19}{24}-\frac{7}{24}\right)-\left(\frac{1}{2}+\frac{1}{3}\right)\)
\(A=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}\)
\(A=\frac{1}{3}\)
\(B=\left(\frac{7}{12}-\frac{5}{12}\right)+\left(\frac{5}{6}+\frac{1}{4}-\frac{3}{7}\right)\)
\(B=\left(\frac{1}{6}+\frac{5}{6}\right)+\frac{1}{4}-\frac{3}{7}\)
\(B=\frac{5}{4}-\frac{3}{7}\)
\(B=\frac{23}{28}\)
b)
\(x=A-B\)
\(x=\frac{1}{3}-\frac{23}{28}\)
\(x=\frac{-41}{84}\)
1.
\(\frac{a^5}{b^3}+ab\ge2\sqrt{\frac{a^5}{b^3}.ab}=2.\frac{a^3}{b}\)
Tương tự và cộng lại:
\(\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\ge2\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)-\left(ab+bc+ca\right)\)(1)
Lại có: \(\frac{a^3}{b}+ab\ge2\sqrt{\frac{a^3}{b}.ab}=2a^2\)
\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)\ge2\left(ab+bc+ca\right)-\left(ab+bc+ca\right)\)
\(=ab+bc+ca\)
\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}-\left(ab+bc+ca\right)\ge0\)
Vậy từ (1) ta có đpcm.
2.
\(\frac{a^5}{bc}+abc\ge2\sqrt{\frac{a^5}{bc}.abc}=2a^3\)
Tương tự và cộng lại
\(A=\frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}\ge2\left(a^3+b^3+c^3\right)-3abc\ge a^3+b^3+c^3+3abc-3abc\)
\(\Rightarrow A\ge a^3+b^3+c^3=VP\)
a)
\(\begin{array}{l}\frac{4}{{ - 3}} + \frac{{ - 22}}{5} =\frac{-4}{{3}} + \frac{{ - 22}}{5}= \frac{{-4.5}}{{3.5}} + \frac{{ - 22.3}}{{5.3}}\\ = \frac{{-20}}{{15}} + \frac{{-66}}{{15}} = \frac{{ - 86}}{{ 15}}\end{array}\)
b)
\(\begin{array}{l}\frac{{ - 5}}{{ - 6}} + \frac{7}{{ - 8}} = \frac{{5}}{{6}} + \frac{-7}{{8}}= \frac{{5.4}}{{6.4}} + \frac{{-7.3}}{{8.3}}\\ = \frac{{20}}{{24}} + \frac{{-21}}{{24}} = \frac{-1}{{24}}\end{array}\).
\(A=-\frac{550}{9}\)\(\Rightarrow\)12.5% của A là \(\frac{-275}{36}\)
\(B=\frac{2}{5}\Rightarrow\)5% của B là \(\frac{1}{8}\)
\(\begin{array}{l}a)A = (2 - \frac{1}{2} - \frac{1}{8}):(1 - \frac{3}{2} - \frac{3}{4})\\ = (\frac{{16}}{8} - \frac{4}{8} - \frac{1}{8}):(\frac{4}{4} - \frac{6}{4} - \frac{3}{4})\\ = \frac{{11}}{8}:\frac{{ - 5}}{4}\\ = \frac{{11}}{8}.\frac{4}{{ - 5}}\\ = \frac{{ - 11}}{{10}}\\b)B = 5 - \frac{{1 + \frac{1}{3}}}{{1 - \frac{1}{3}}}\\ = 5 - \frac{{\frac{3}{3} + \frac{1}{3}}}{{\frac{3}{3} - \frac{1}{3}}}\\ = 5 - \frac{{\frac{4}{3}}}{{\frac{2}{3}}}\\ = 5 - \frac{4}{3}:\frac{2}{3}\\ = 5 - \frac{4}{3}.\frac{3}{2}\\ = 5 - 2\\ = 3\end{array}\)
Chú ý:
Khi thực hiện phép cộng hai phân số, nếu phân số thu được chưa tối giản thì ta rút gọn thành phân số tối giản.
\(\frac{a-b}{b}=\frac{3}{5}\Leftrightarrow\frac{a-b}{3}=\frac{b}{5}=\frac{a-b+b}{3+5}=\frac{a}{8}\left(\text{tính chất dãy tỉ số bằng nhau}\right)\Leftrightarrow\frac{b}{5}=\frac{a}{8}\Leftrightarrow\frac{a}{b}=\frac{8}{5}\)
Ta có: \(\frac{a-b}{b}=\frac{a}{b}-1=\frac{3}{5}\)
\(\Rightarrow\frac{a}{b}=\frac{3}{5}+1=\frac{8}{5}\)
Vậy \(\frac{a}{b}=\frac{8}{5}\)