and

7)Ta có:\(A=\dfrac{2021\times2021\times20202020-2020\times2020\times20212021}{2020\times20192019}\)

                \(=\dfrac{2021\times2021\times2020\times10001-2020\times2020\times2021\times10001}{2020\times2019\times10001}\)

                \(=\dfrac{2021\times2021\times2020-2020\times2020\times2021}{2020\times2019}\)

                \(=\dfrac{2021\times2020\times2020+2021\times2020-2020\times2020\times2021}{2020\times2019}\)

                \(=\dfrac{2021\times2020}{2020\times2019}=\dfrac{2021}{2019}\)

6)Ta có:\(\dfrac{18\times123+9\times4567\times2+3\times5310\times6}{\left(2+4+6+8+...+20+22\right)+48}\)

         \(=\dfrac{18\times\left(123+4567+5310\right)}{\left(2+22\right)+\left(4+20\right)+..+\left(10+14\right)+12+48}\)

         \(=\dfrac{18\times10000}{5\times24+60}=\dfrac{18\times10000}{5\times\left(24+12\right)}=\dfrac{18\times10000}{5\times36}=1000\) 

Bài 5.

Ta có:\(\dfrac{2022\times2021-1}{2021+2022\times2020}=\dfrac{2022\times2020+2022-1}{2021+2022\times2020}\)   

      \(=\dfrac{2022\times2020+2021}{2021+2022\times2020}=1\)

d,Ta có:\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)

        \(=3\sqrt{75\sqrt{2}}+5\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)

        \(=15\sqrt{3\sqrt{2}}+20\sqrt{3\sqrt{2}}-16\sqrt{3\sqrt{2}}\)

        \(=19\sqrt{3\sqrt{2}}\)

a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)

   \(=28.3+9.3-4.3=99\)

b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)

  \(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)

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