Rút gọn (toán lớp 9)
a,√405+3√27/3√3+√45
b,√6-2√5/√5-1
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a. \(\dfrac{\sqrt{2}.\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7}.\left(\sqrt{3}+\sqrt{5}\right)}=\dfrac{\sqrt{2}}{\sqrt{7}}=\sqrt{\dfrac{2}{7}}\)
d. \(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{5-2\sqrt{5}+1}}{\sqrt{5}-1}=\dfrac{\left(\sqrt{5}-1\right)^2}{\sqrt{5}-1}=\sqrt{5}-1\)
d: \(D=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\)
\(=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{3\left(x+y\right)}{2}\)
\(=\dfrac{3}{x-y}\)
a: \(=3\sqrt{3}-2\sqrt{3}+4\sqrt{3}-5\sqrt{3}=2\sqrt{3}\)
\(\frac{1}{4}+\frac{3}{5}=\frac{17}{20}\)
\(\frac{5}{2}+\frac{7}{9}=\frac{59}{18}\)
\(\frac{3}{2}+\frac{2}{3}=\frac{13}{6}\)
\(\frac{4}{5}+\frac{3}{2}=\frac{23}{10}\)
\(\frac{4}{5}+\frac{3}{15}=\frac{4}{5}+\frac{1}{5}=1\)
\(\frac{2}{3}+\frac{32}{24}=\frac{2}{3}+\frac{4}{3}=2\)
\(\frac{5}{6}+\frac{15}{18}=\frac{5}{6}+\frac{5}{6}=\frac{10}{6}=\frac{5}{3}\)
\(\frac{8}{15}+\frac{2}{3}=\frac{8}{15}+\frac{10}{15}=\frac{18}{15}=\frac{6}{5}\)
\(\frac{3}{7}+\frac{4}{8}=\frac{24}{56}+\frac{28}{56}=\frac{52}{56}=\frac{13}{14}\)
_HT_