alo 

b: \(1024:2^5+140:\left(38+2^5\right)+2\cdot3^2-7^{23}:7^{21}\)

\(=1024:32+140:\left(38+32\right)+2\cdot9-7^2\)

=32+140:70+18-49

=50-49+2

=1+2

=3

c: \(100:\left\lbrace300:\left\lbrack450-\left(4\cdot5^3-2^3\cdot25\right)\right\rbrack\right\rbrace\)

\(=100:\left\lbrace300:\left\lbrack450-\left(4\cdot125-8\cdot25\right)\right\rbrack\right\rbrace\)

\(=100:\left\lbrace300:\left\lbrack450-\left(500-200\right)\right\rbrack\right\rbrace\)

=100:{300:[450-300]}

=100:{300:150}

=100:2

=50

siuu

a: \(2^3-5^3:5^2+12\cdot2^2\)

\(=8-5+48\)

\(=51\)

b: \(5\cdot\left[\left(85-35:7\right):8+90\right]-5\)

\(=5\cdot\left[10+90\right]-5\)

=495

a: 23−53:52+12⋅2223−53:52+12⋅22

=8−5+48=8−5+48

=51=51

b: 5⋅[(85−35:7):8+90]−55⋅[(85−35:7):8+90]−5

=5⋅[10+90]−5=5⋅[10+90]−5

=495

300

465 

-3

=(51-52)+(53-54)+(55-56)

=(-1)+(-1)+(-1)

=-3

50) \(\sqrt{98-16\sqrt{3}}=4\sqrt{6}-\sqrt{2}\)

51) \(\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{\sqrt{6}-\sqrt{2}}{2}\)

52) \(\sqrt{4+\sqrt{15}}=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)

53) \(\sqrt{5-\sqrt{21}}=\dfrac{\sqrt{10-2\sqrt{21}}}{\sqrt{2}}=\dfrac{\sqrt{14}-\sqrt{6}}{2}\)

54) \(\sqrt{6-\sqrt{35}}=\dfrac{\sqrt{12-2\sqrt{35}}}{\sqrt{2}}=\dfrac{\sqrt{14}-\sqrt{10}}{2}\)

55) \(\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{6}+\sqrt{2}}{2}\)

56) \(\sqrt{4-\sqrt{15}}=\dfrac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)

Can bac 8