\(\dfrac{5,313.10^{-23}}{1,66.10^{-24}}\approx32\left(S\right)\) Lưu huỳnh

\(2016^{2017}=\left(2016^{2,017}\right)^{1000}\)

\(2017^{2016}=\left(2017^{2,016}\right)^{1000}\)

Ta có \(2016^{2,017}-2017^{2,016}=30469,58759=>2016^{2,017}>2017^{2,016}\)

=> \(2016^{2017}>2017^{2016}\)

a,\(\sqrt{4\left(a-5\right)^2}=\sqrt{4}.\sqrt{\left(a-5\right)^2}=2.\left|a-5\right|=2\left(a-5\right)\left(a\ge5\right)\)

b,\(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3=-1}\)

c,Mạn phép sửa đề ,nếu ko thì kết quả ko đẹp

\(\sqrt{8+2\sqrt{15}}-\sqrt{5}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{5}=\sqrt{5}+\sqrt{3}-\sqrt{5}=\sqrt{3}\)

d,\(\sqrt{\left(3-2\sqrt{3}\right)^2}-\sqrt{\left(3+2\sqrt{3}\right)^2}=2\sqrt{3}-3-3-2\sqrt{3}=-6\)

e,\(\sqrt{24\left(b-3\right)}^2=\sqrt{24^2}.\sqrt{\left(b-3\right)^2}=24.\left(3-b\right)\left(b< 3\right)\)

\(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}=\dfrac{\left(1-a\sqrt{a}\right)\left(1+\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}=\dfrac{1+\sqrt{a}-a\sqrt{a}-a^2}{1-a}=\dfrac{\left(1-a\right)\left(\sqrt{a}+a+1\right)}{1-a}\)

=> \(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}=a+2\sqrt{a}+1=\left(\sqrt{a}+1\right)^2\)

Tương tự \(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}=\left(\sqrt{a}-1\right)^2\)

biểu thức trong dấu ngoặc vuông = \(\left[\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)\right]^2=\left(a-1\right)^2\)

\(E=\dfrac{1-a^2}{\left(a-1\right)^2}\)