Bài 1:

a) Ta có: \(\left(0.125\right)\cdot\left(-3\cdot7\right)\cdot\left(-2\right)^3\)

\(=\frac{1}{8}\cdot\left(-21\right)\cdot\left(-8\right)\)

\(=\frac{1}{8}\cdot168\)

\(=21\)

b) Ta có: \(\sqrt{36}\cdot\sqrt{\frac{25}{16}}+\frac{1}{4}\)

\(=\sqrt{36\cdot\frac{25}{16}}+\frac{1}{4}\)

\(=\sqrt{\frac{225}{4}}+\frac{1}{4}\)

\(=\frac{15}{2}+\frac{1}{4}\)

\(=\frac{31}{4}\)

c) Ta có: \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac{2}{5}\)

\(=\frac{2}{9}:\frac{5}{9}-\frac{7}{5}\)

\(=\frac{2}{5}-\frac{7}{5}=-1\)

d) Ta có: \(0,1\cdot\sqrt{225}\cdot\sqrt{\frac{1}{4}}\)

\(=0,1\cdot15\cdot\frac{1}{2}=\frac{3}{4}\)

hay quá  cảm ơn bạn nhé 

a) \(\sqrt{0,09}-\sqrt{0,64}=\frac{-1}{2}=-0,5\)

b) \(0,1\cdot\sqrt{225}-\sqrt{\frac{1}{4}}=0,1\cdot15-\frac{1}{2}=1\)

c) \(\sqrt{0,36}\cdot\sqrt{\frac{25}{16}+\frac{1}{4}}=\frac{3\sqrt{29}}{20}\)

d) đề baì có sai ko ban?

C = \(\frac{2}{3}\sqrt{144}-\left(-\frac{3}{4}\right)\div\sqrt{\frac{225}{144}}\)

C = \(\frac{2}{3}.12+\frac{3}{4}\div\frac{5}{4}\)

C = \(8+\frac{3}{5}\)

C = \(8\frac{3}{5}\)

D = \(\frac{4^6.25^5-2^{12}.25^4}{2^{12}.5^8-10^8.64}\)

D = \(\frac{\left(2^2\right)^6.\left(5^2\right)^5-2^{12}.\left(5^2\right)^4}{2^{12}.5^8-\left(2.5\right)^8.2^6}\)

D = \(\frac{2^{12}.5^{10}-2^{12}.5^8}{2^{12}.5^8-2^8.5^8.2^6}\)

D = \(\frac{2^{12}.5^8.\left(25-1\right)}{2^{12}.5^8.\left(1-2^2\right)}\)

D = \(\frac{24}{-3}\)

D = \(-8\)

\(C=\frac{2}{3}\sqrt{144}-\left(\frac{-3}{4}\right):\sqrt{\frac{225}{144}}\)

\(=\frac{2}{3}\cdot12+\frac{3}{4}:\frac{5}{4}\)

\(=8+\frac{3}{4}\cdot\frac{4}{5}\)

\(=8+\frac{3}{5}\)

\(=\frac{40}{5}+\frac{3}{4}=\frac{43}{5}\)

\(D=\frac{4^6\cdot25^5-2^{12}\cdot25^4}{2^{12}\cdot5^8-10^8\cdot64}=\frac{\left(2^2\right)^6\cdot\left(5^2\right)^5-2^{12}\cdot\left(5^2\right)^4}{2^{12}\cdot5^8-\left(2\cdot5\right)^8\cdot2^6}\)

\(=\frac{2^{12}\cdot5^{10}-2^{12}\cdot5^8}{2^{12}\cdot5^8-2^{14}\cdot5^8}=\frac{5^8\left(2^{12}\cdot5^2-2^{12}\right)}{5^8\left(2^{12}-2^{14}\right)}\)

\(=\frac{2^{12}\cdot5^2-2^{12}}{2^{12}-2^{14}}=\frac{2^{12}\left(5^2-1\right)}{2^{12}\left(1-2^2\right)}=\frac{24}{-3}=-8\)

Sorry mới lớp 6 chưa học

thông cảm 

no chửi 

Ta có:

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}\)

\(=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Thế vào bài toán ta được

\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{225\sqrt{224}+224\sqrt{225}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{224}}-\frac{1}{\sqrt{225}}\)

\(=1-\frac{1}{\sqrt{225}}=1-\frac{1}{15}=\frac{14}{15}\)

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