a) Nêu phân số đảo ngược của mỗi phân số sau: \(\dfrac{3}{4},\dfrac{10}{3},\dfrac{2}{5},\dfrac{1}{4}.\)
b) Tính:
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a) Các phân số đảo ngược là:
\(\dfrac{5}{8}\rightarrow\dfrac{8}{5};\dfrac{3}{4}\rightarrow\dfrac{4}{3};\dfrac{1}{2}\rightarrow\dfrac{2}{1}=2\)
b) \(\dfrac{3}{7}:\dfrac{5}{8}=\dfrac{3}{7}\times\dfrac{8}{5}=\dfrac{24}{35}\)
\(\dfrac{8}{7}:\dfrac{3}{4}=\dfrac{8}{7}\times\dfrac{4}{3}=\dfrac{32}{21}\)
\(\dfrac{1}{3}:\dfrac{1}{2}=\dfrac{1}{3}\times2=\dfrac{2\times1}{3}=\dfrac{2}{3}\)
a) \(\dfrac{20}{12}\) b) \(\dfrac{15}{15}\) c) \(\dfrac{24}{42}\)
d) \(\dfrac{26}{39}\) e) \(\dfrac{0}{8}\) g) \(\dfrac{15}{3}\)
a) \(\dfrac{15}{3}\)
b)\(\dfrac{15}{15}\)
c) \(\dfrac{24}{42}\)
d) \(\dfrac{26}{39}\)
e)\(\dfrac{0}{8}\)
f) \(\dfrac{15}{15}\)
a) \(A=\left\{x\in N|x=3k+1;0\le k\le3;k\in z\right\}\)
b) \(B=\left\{x\in Q^+|x=\dfrac{k}{k^2-1};2\le k\le6;k\in N\right\}\)
a) Ta có \(A=\dfrac{n-5}{n-3}=\dfrac{n-3-2}{n-3}=1-\dfrac{2}{n-3}\). Để \(A\inℤ\) thì \(\dfrac{2}{n-3}\inℤ\) hay \(n-3\) là ước của 2. Suy ra \(n-3\in\left\{\pm1;\pm2\right\}\).
Nếu \(n-3=1\Rightarrow n=4\); \(n-3=-1\Rightarrow n=2\); \(n-3=2\Rightarrow n=5\); \(n-3=-2\Rightarrow n=1\). Vậy để \(A\inℤ\) thì \(n\in\left\{1;2;4;5\right\}\)
\(A=\dfrac{n+4}{n+1}\) làm tương tự.
b) Dễ thấy các số ở mẫu có thể viết dưới dạng:
\(10=1+2+3+4=\dfrac{4\left(4+1\right)}{2}=\dfrac{4.5}{2}\)
\(15=1+2+3+4+5=\dfrac{5\left(5+1\right)}{2}=\dfrac{5.6}{2}\)
\(21=1+2+...+6=\dfrac{6\left(6+1\right)}{2}=\dfrac{6.7}{2}\)
...
\(120=1+2+...+15=\dfrac{15\left(15+1\right)}{2}=\dfrac{15.16}{2}\)
Do đó \(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\)
\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(A=2\left(\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{16-15}{15.16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(A=\dfrac{3}{8}\)
a) \(\dfrac{2}{5}\times?=\dfrac{3}{10}\)
\(?=\dfrac{3}{10}:\dfrac{2}{5}=\dfrac{3}{4}\)
b) \(\dfrac{1}{8}:?=\dfrac{1}{5}\)
\(?=\dfrac{1}{8}:\dfrac{1}{5}=\dfrac{5}{8}\)
a: Phân số cần tìm là: \(\dfrac{3}{10}:\dfrac{2}{5}=\dfrac{3}{10}\cdot\dfrac{5}{2}=\dfrac{15}{20}=\dfrac{3}{4}\)
b: Phân số cần tìm là \(\dfrac{1}{8}:\dfrac{1}{5}=\dfrac{5}{8}\)
Tham khảo:
Như vậy,\(\dfrac{2}{5} = \dfrac{4}{{10}}\)
\(\dfrac{1}{3} = \dfrac{3}{9}\)
Bài 1:
a)
\(\dfrac{1}{2}=\dfrac{1\times6}{2\times6}=\dfrac{6}{12}\)
\(\dfrac{2}{3}=\dfrac{2\times4}{3\times4}=\dfrac{8}{12}\)
\(\dfrac{3}{4}=\dfrac{3\times3}{4\times3}=\dfrac{9}{12}\)
b)
\(\dfrac{1}{3}=\dfrac{1\times15}{3\times15}=\dfrac{15}{45}\)
\(\dfrac{2}{15}=\dfrac{2\times3}{15\times3}=\dfrac{6}{45}\)
\(\dfrac{4}{45}\) (giữ nguyên)
c)
\(\dfrac{1}{8}=\dfrac{1\times3}{8\times3}=\dfrac{3}{24}\)
\(\dfrac{2}{3}=\dfrac{2\times8}{3\times8}=\dfrac{16}{24}\)
\(\dfrac{5}{2}=\dfrac{5\times12}{2\times12}=\dfrac{60}{24}\)
d)
\(\dfrac{2}{7}=\dfrac{2\times4}{7\times4}=\dfrac{8}{28}\)
\(\dfrac{9}{4}=\dfrac{9\times7}{4\times7}=\dfrac{63}{28}\)
\(\dfrac{5}{28}\) (giữ nguyên)
Bài 2:
a)
\(4=\dfrac{4}{1}=\dfrac{4\times12}{1\times12}=\dfrac{48}{12}\)
\(\dfrac{9}{4}=\dfrac{9\times3}{4\times3}=\dfrac{27}{12}\)
b)
\(\dfrac{5}{8}=\dfrac{5\times30}{8\times30}=\dfrac{150}{240}\)
\(\dfrac{25}{30}=\dfrac{5}{6}=\dfrac{5\times40}{6\times40}=\dfrac{200}{240}\)
\(2=\dfrac{2}{1}=\dfrac{2\times240}{1\times240}=\dfrac{480}{240}\).
a) Phân số đảo ngược của các phân số $\frac{3}{4};\frac{{10}}{3};\frac{2}{5};\frac{1}{4}$lần lượt là $\frac{4}{3};\frac{3}{{10}};\frac{5}{2};\frac{4}{1}$
b) $\frac{7}{4}:\frac{3}{4} = \frac{7}{4} \times \frac{4}{3} = \frac{{28}}{{12}} = \frac{7}{3}$
$\frac{{10}}{9}:\frac{{10}}{3} = \frac{{10}}{9} \times \frac{3}{{10}} = \frac{{30}}{{90}} = \frac{3}{9}$
$\frac{6}{{35}}:\frac{2}{5} = \frac{6}{{35}} \times \frac{5}{2} = \frac{{30}}{{70}} = \frac{3}{7}$
$\frac{1}{{20}}:\frac{1}{4} = \frac{1}{{20}} \times \frac{4}{1} = \frac{4}{{20}} = \frac{1}{5}$