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\(\sqrt{\frac{9}{25}}+\sqrt{\frac{1}{6}}-\sqrt{\frac{4}{49}}\)\(=\sqrt{\left(\frac{3}{5}\right)^2}+\sqrt{\frac{1}{6}}-\sqrt{\left(\frac{2}{7}\right)^2}\)\(=\frac{3}{5}+\sqrt{\frac{1}{6}}-\frac{2}{7}\)\(=\left(\frac{3}{5}-\frac{2}{7}\right)+\sqrt{\frac{1}{6}}\)\(=\left(\frac{21}{35}-\frac{10}{35}\right)+\sqrt{\frac{1}{6}}\)\(=\frac{11}{35}+\sqrt{\frac{1}{6}}\)\(=\sqrt{\frac{1.6}{6.6}}+\frac{11}{35}\)\(=\frac{\sqrt{6}}{6}+\frac{11}{35}\)\(=\frac{35\sqrt{6}}{210}+\frac{66}{210}\)\(=\frac{35\sqrt{6}+66}{210}\)
\(a,\sqrt{25}-\sqrt{16}+\sqrt{1}=\sqrt{5^2}-\sqrt{4^2}+\sqrt{1^2}=5-4+1=2\)
\(b,\sqrt{\frac{4}{9}}+\sqrt{\frac{25}{4}}+\sqrt{\left(-3\right)^4}=\sqrt{\left(\frac{2}{3}\right)^2}+\sqrt{\left(\frac{5}{2}\right)^2}+\sqrt{\left[\left(-3\right)^2\right]^2}\)
\(=\frac{2}{3}+\frac{5}{2}+\left(-3\right)^2=\frac{2}{3}+\frac{5}{2}+9=\frac{4}{6}+\frac{15}{6}+\frac{54}{6}=\frac{73}{6}\)
\(c,\frac{7}{5}+\sqrt{49}+\sqrt{\left(-3\right)^2}=\frac{7}{5}+\sqrt{7^2}+\sqrt{3^2}=\frac{7}{5}+7+3\)
\(=\frac{7}{5}+\frac{35}{5}+\frac{15}{5}=\frac{57}{5}\)
con lạy cha nào làm được hết bài này và giải trình tự ra
\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)
\(B=\frac{1}{4}\)
\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{\dfrac{8}{2}-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1}{4}\)