CÓ VẺ LÀ HƠI KHÓ HỈ CẬU Ạ

\(2\times\frac{2}{4}\div\frac{4}{5}\)

\(\Rightarrow\frac{2}{1}\times\frac{2}{4}\div\frac{4}{5}\)

\(\Rightarrow\frac{4}{4}\div\frac{4}{5}\)

\(\Rightarrow\frac{4}{4}\times\frac{5}{4}\)

\(\Rightarrow\frac{5}{4}\)

=3,4  ✖   0,1

=0,34

\(\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)

\(=\sqrt{7-2\sqrt{21}+3}+\sqrt{7+2\sqrt{21}+3}\)

\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}.\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{7}\right)^2+2.\sqrt{7}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

\(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\)

\(=\left|\sqrt{7}-\sqrt{3}\right|+\left|\sqrt{7}+\sqrt{3}\right|\)

\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)

\(=\sqrt{7}+\sqrt{7}=2\sqrt{7}\)

Ta có: \(\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)

\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)

\(=2\sqrt{7}\)

đặt \(A=\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)

\(=>A^2=10-2\sqrt{21}+10+2\sqrt{21}+2\sqrt{\left(10-2\sqrt{21}\right)\left(10+2\sqrt{21}\right)}\)

\(=>A^2=20+2\sqrt{10^2-\left(2\sqrt{21}\right)^2}=20+2\sqrt{16}=20+2.4=28\)

\(=>A=\sqrt{28}=2\sqrt{7}\)

4/7 *1/3+5/7:2/3

\(\frac{4}{21}\)+ \(\frac{15}{14}\)= \(\frac{53}{42}\)

=4/21+15/14

=53/42

cho hỏi 4/21,15/14 đâu ra vậy,hai bạn làm ko đúng,sao có người làm bằng 1 vậy

=(198 : 2) x 444 x 2 x 132) : 25) / 33) x 8888) : 5) x 66) : 125) x 5 =66009449.5949

66009449.5949 

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