Rút gọn:
\(\dfrac{8}{9}\)
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a) \(\dfrac{2}{4}\times\dfrac{9}{5}=\dfrac{1}{2}\times\dfrac{9}{5}=\dfrac{1\times9}{2\times5}=\dfrac{9}{10}\)
b) \(\dfrac{13}{8}\times\dfrac{5}{15}=\dfrac{13}{8}\times\dfrac{1}{3}=\dfrac{13}{24}\)
c) \(\dfrac{3}{9}\times\dfrac{6}{12}=\dfrac{1}{3}\times\dfrac{1}{2}=\dfrac{1}{3\times2}=\dfrac{1}{6}\)
a: =18/20=9/10
b: \(=\dfrac{13\cdot5}{8\cdot15}=\dfrac{13}{8}\cdot\dfrac{1}{3}=\dfrac{13}{24}\)
c: \(=\dfrac{1}{3}\cdot\dfrac{1}{2}=\dfrac{1}{6}\)
\(B=\dfrac{x^2+39}{\left(x-3\right)\left(x+3\right)}+\dfrac{8}{x+3}-\dfrac{1}{x-3}\)
\(=\dfrac{x^2+39+8x-24-x-3}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+7x+12}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x+4}{x-3}\)
\(B=\dfrac{x^2+39}{x^2-9}+\dfrac{8}{x+3}+\dfrac{1}{3-x}\) (ĐK: \(x\ne3,x\ne-3\))
\(B=\dfrac{x^2+39}{\left(x+3\right)\left(x-3\right)}+\dfrac{8\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{1}{x-3}\)
\(B=\dfrac{x^2+39+8x-24}{\left(x+3\right)\left(x-3\right)}-\dfrac{1}{x-3}\)
\(B=\dfrac{x^2+8x+15}{\left(x+3\right)\left(x-3\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-3\right)}\)
\(B=\dfrac{x^2+8x+15-x-3}{\left(x+3\right)\left(x-3\right)}\)
\(B=\dfrac{x^2+7x+12}{\left(x+3\right)\left(x-3\right)}\)
\(B=\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x-3\right)\left(x+3\right)}\)
\(B=\dfrac{x+4}{x-3}\)
a) \(\dfrac{2}{5}-\dfrac{3}{15}\)
\(=\dfrac{2}{5}-\dfrac{3:3}{15:3}\)
\(=\dfrac{2}{5}-\dfrac{1}{5}\)
\(=\dfrac{1}{5}\)
b) \(\dfrac{9}{27}-\dfrac{2}{9}\)
\(=\dfrac{9:3}{27:3}-\dfrac{2}{9}\)
\(=\dfrac{3}{9}-\dfrac{2}{9}\)
\(=\dfrac{1}{9}\)
c) \(\dfrac{18}{24}-\dfrac{4}{8}\)
\(=\dfrac{18:6}{24:6}-\dfrac{4:2}{8:2}\)
\(=\dfrac{3}{4}-\dfrac{2}{4}\)
\(=\dfrac{1}{4}\)
d) \(\dfrac{6}{16}-\dfrac{10}{64}\)
\(=\dfrac{6\times2}{16\times2}-\dfrac{10:2}{64:2}\)
\(=\dfrac{12}{32}-\dfrac{5}{32}\)
\(=\dfrac{7}{32}\)
a: \(\dfrac{1}{8}+\dfrac{5}{8}=\dfrac{1+5}{8}=\dfrac{6}{8}=\dfrac{3}{4}\)
b: \(\dfrac{1}{15}+\dfrac{4}{15}=\dfrac{1+4}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
c: \(\dfrac{5}{9}+\dfrac{7}{9}=\dfrac{5+7}{9}=\dfrac{12}{9}=\dfrac{4}{3}\)
d: \(\dfrac{23}{100}+\dfrac{27}{100}=\dfrac{23+27}{100}=\dfrac{50}{100}=\dfrac{1}{2}\)
a) 18+58=1+58=68=3418+58=1+58=68=34
b) 115+415=1+415=515=13115+415=1+415=515=13
c) 59+79=5+79=129=4359+79=5+79=129=43
d) 23100+27100=23+27100=50100=12
a) \(\dfrac{25}{60}+6\times\dfrac{1}{8}=\dfrac{25}{60}+\dfrac{3}{4}=\dfrac{25}{69}+\dfrac{45}{60}=\dfrac{25+45}{60}=\dfrac{70}{60}=\dfrac{7}{6}\)
b) \(\dfrac{15}{9}\times\left(6:\dfrac{30}{9}\right)=\dfrac{15}{9}\times\left(6\times\dfrac{30}{9}\right)=\dfrac{15}{9}\times20=\dfrac{100}{3}\)
ĐK: \(x\ne\pm3\)
Khi đó:
\(C=\dfrac{2\left(x-3\right)}{x^2-9}+\dfrac{1\left(x+3\right)}{x^2-9}-\dfrac{8}{x^2-9}\\ =\dfrac{2x-6}{x^2-9}+\dfrac{x+3}{x^2-9}-\dfrac{8}{x^2-9}\\ =\dfrac{2x-6+x+3-8}{x^2-9}\\ =\dfrac{3x-11}{x^2-9}\)
Thế x = 4 vào C được:
\(C=\dfrac{3.4-11}{4^2-9}=\dfrac{12-11}{16-9}=\dfrac{1}{7}\)
\(\frac{10}{5}=\frac{10:5}{5:5}=\frac{2}{1}\)
\(\frac{12}{9}=\frac{12:3}{9:3}=\frac{4}{3}\)
\(\frac{8}{4}=\frac{8:4}{4:4}=\frac{2}{1}\)
\(\frac{5}{6}\)giữ nguyên
a: \(\dfrac{15}{8}-\dfrac{13}{8}=\dfrac{15-13}{8}=\dfrac{2}{8}=\dfrac{1}{4}\)
b: \(\dfrac{7}{15}-\dfrac{2}{15}=\dfrac{7-2}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
c: \(\dfrac{11}{12}-\dfrac{2}{12}=\dfrac{11-2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
d: \(\dfrac{19}{7}-\dfrac{5}{7}=\dfrac{19-5}{7}=\dfrac{14}{7}=2\)
a: \(\dfrac{3}{9}\times\dfrac{5}{4}=\dfrac{5}{4}\times\dfrac{1}{3}=\dfrac{5\times1}{4\times3}=\dfrac{5}{12}\)
b: \(\dfrac{10}{15}\times\dfrac{3}{5}=\dfrac{3}{5}\times\dfrac{2}{3}=\dfrac{3\times2}{5\times3}=\dfrac{2}{5}\)
c: \(\dfrac{5}{8}\times\dfrac{4}{12}=\dfrac{5}{8}\times\dfrac{1}{3}=\dfrac{5}{3\times8}=\dfrac{5}{24}\)
d: \(\dfrac{9}{27}\times\dfrac{3}{21}=\dfrac{1}{7}\times\dfrac{1}{3}=\dfrac{1\times1}{7\times3}=\dfrac{1}{21}\)
\(\dfrac{8}{9}\) là phân số tối giản
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