\(A=\frac{2^{30}.5^7+2^{13}.5^{27}}{2^{27}.5^7+2^{10}.5^{27}}\)

\(=\frac{2^3\left(2^{27}.5^7+2^{10}.5^{27}\right)}{2^{27}.5^7+2^{10}.5^{27}}\)

\(=2^3=8\)

\(\frac{2^{30}.5^7+2^{13}.5^{27}}{2^{27}.5^7+2^{10}.5^{27}}+\frac{2^{13}.5^7\left(2^{17}+5^{20}\right)}{2^{10}.5^7\left(2^{17}+5^{20}\right)}=2^3=8\)

\(a)\frac{2}{{15}} + \left( {\frac{{ - 5}}{{24}}} \right) = \frac{{16}}{{120}} + \left( {\frac{{ - 25}}{{120}}} \right) = \frac{{ - 9}}{{120}} = \frac{{ - 3}}{{40}}\)          

 b) \(\left( {\frac{{ - 5}}{9}} \right) - \left( { - \frac{7}{{27}}} \right) = \left( {\frac{{ - 15}}{{27}}} \right) + \frac{7}{{27}} = \frac{{ - 8}}{{27}}\)          

 c)\(\left( { - \frac{7}{{12}}} \right) + 0,75 = \left( { - \frac{7}{{12}}} \right) + \frac{75}{100} \\= \left( { - \frac{7}{{12}}} \right) + \frac{3}{4} \\= \left( { - \frac{7}{{12}}} \right) + \frac{9}{{12}} = \frac{2}{{12}} = \frac{1}{6}\)

d)\(\left( {\frac{{ - 5}}{9}} \right) - 1,25 =\left( {\frac{{ - 5}}{9}} \right) - \frac{125}{100} = \left( {\frac{{ - 5}}{9}} \right) - \frac{5}{4}\\ = \left( {\frac{{ - 20}}{{36}}} \right) - \frac{{45}}{{36}} = \frac{{ - 65}}{{36}}\)       

 e)\(0,34.\frac{{ - 5}}{{17}} =\frac{{34}}{{100}}.\frac{{ - 5}}{{17}} = \frac{{17}}{{50}}.\frac{{ - 5}}{{17}} = \frac{{ - 1}}{{10}}\)                       

g) \(\frac{4}{9}:\left( { - \frac{8}{{15}}} \right) = \frac{4}{9}.\left( { - \frac{{15}}{8}} \right) = \frac{{ - 5}}{6}\)

h)\(\left( {1\frac{2}{3}} \right):\left( {2\frac{1}{2}} \right) = \frac{5}{3}:\frac{5}{2} = \frac{5}{3}.\frac{2}{5} = \frac{2}{3}\)       

i) \(\frac{2}{5}.\left( { - 1,25} \right) = \frac{2}{5}.\frac{{ - 125}}{100} = \frac{2}{5}.\frac{{ - 5}}{4} = \frac{{ - 1}}{2}\)                     

k) \(\left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).3\frac{1}{9} = \left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).\frac{{28}}{9}\\ = \frac{{ - 3.3.5.7.4}}{{5.\left( { - 7} \right).3.3}} = 4\)

a: \(A=3^{\dfrac{2}{5}}\cdot3^{\dfrac{1}{5}}\cdot3^{\dfrac{1}{5}}=3^{\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{1}{5}}=3^{\dfrac{4}{5}}\)

b: \(B=\left(-27\right)^{\dfrac{1}{3}}=\left[\left(-3\right)^3\right]^{\dfrac{1}{3}}=\left(-3\right)^{\dfrac{1}{3}\cdot3}=\left(-3\right)^1=-3\)

c: \(C=\sqrt[3]{-64}\cdot\left(\dfrac{1}{2}\right)^3\)

\(=\sqrt[3]{\left(-4\right)^3}\cdot\dfrac{1}{2^3}=-4\cdot\dfrac{1}{8}=-\dfrac{4}{8}=-\dfrac{1}{2}\)

d: \(D=\left(-27\right)^{\dfrac{1}{3}}\cdot\left(\dfrac{1}{3}\right)^4\)

\(=\left[\left(-3\right)^3\right]^{\dfrac{1}{3}}\cdot\dfrac{1}{3^4}\)

\(=\left(-3\right)^{3\cdot\dfrac{1}{3}}\cdot\dfrac{1}{81}=\dfrac{-3}{81}=\dfrac{-1}{27}\)

e: \(E=\left(\sqrt{3}+1\right)^{106}\cdot\left(\sqrt{3}-1\right)^{106}\)

\(=\left[\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\right]^{106}\)

\(=\left(3-1\right)^{106}=2^{106}\)

f: \(F=360^{\sqrt{5}+1}\cdot20^{3-\sqrt{5}}\cdot18^{3-\sqrt{5}}\)

\(=360^{\sqrt{5}+1}\cdot\left(20\cdot18\right)^{3-\sqrt{5}}\)

\(=360^{\sqrt{5}+1}\cdot360^{3-\sqrt{5}}=360^{\sqrt{5}+1+3-\sqrt{5}}=360^4\)

g: \(G=2023^{3+2\sqrt{2}}\cdot2023^{2\sqrt{2}-3}\)

\(=2023^{3+2\sqrt{2}+2\sqrt{2}-3}\)

\(=2023^{4\sqrt{2}}\)

\(\left|x+\frac{1}{3}\right|+\frac{4}{5}=\left|-3,2+\frac{2}{5}\right|+\left(27-\frac{3}{5}\right)\left(27-\frac{3^2}{6}\right)...\left(27-\frac{3^5}{9}\right)...\left(27-\frac{3^{2010}}{2014}\right)\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}+\left(27-\frac{3^2}{6}\right)\left(27-\frac{3^3}{7}\right)...\left(27-27\right)...\left(27-\frac{3^{2010}}{2014}\right)\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|=2\)

\(\Rightarrow\hept{\begin{cases}x+\frac{1}{3}=2\\x+\frac{1}{3}=-2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=-\frac{7}{3}\end{cases}}}\)

bạn ơi, có một chỗ chưa chuẩn .bạn kiểm tra lại giú mình. chỗ vế trái bạn thiếu \(\left(27-\frac{3}{5}\right)\). bạn bổ sung vào cho đúng nhé. dù sao vẫn cảm ơn bạn.

a)\(\sqrt{400.0,81}=\sqrt{4.81}=\sqrt{2^2.9^2}=2.9=18\)

b)\(\sqrt{\dfrac{5}{27}.\dfrac{3}{20}}=\sqrt{\dfrac{5}{3^3}.\dfrac{3}{2^2.5}}=\sqrt{\dfrac{1}{3^2.2^2}}=\dfrac{1}{3.2}=\dfrac{1}{6}\)

c)\(\sqrt{\left(-5\right)^2.3^2}=\sqrt{5^2.3^2}=5.3=15\)

d)\(\sqrt{\left(2-\sqrt{5}\right)^2\left(2+\sqrt{5}\right)^2}=\sqrt{\left[2^2-\left(\sqrt{5}\right)^2\right]^2}=\sqrt{\left(-1\right)^2}=1\)

a) \(\left(x-10\right)^2-x\left(x+8\right)=-12x+100=-11,76+100=88,24\)

b) \(x^3-9x^2+27x-27=\left(x-3\right)^3=\left(5-3\right)^3=8\)

c) \(6x\left(2x-7\right)-\left(3x-5\right)\left(4x+7\right)=-43x+35=121\)

\(a)\) \(\left(x-10\right)^{^2}-x.\left(x+8\right)\) \(với\) \(x=0,98\)

\(=-12x+100\)

\(=-11,76+100\)

\(=88,24\)

\(b)\) \(x^3-9x^2+27.x-27\) \(với\) \(x=5\)

\(=\left(x-3\right)^3\)

\(=\left(5-3\right)^3\)

\(=8\)

\(c)\)\(6x.\left(2x-7\right)-\left(3x-5\right).\left(4x+7\right)\) \(tại\) \(x=-2\)

\(=-43+35\)

\(=121\)

Chúc bạn hôc tốt nha ❤

a: \(=\dfrac{2^{13}\cdot5^7\left(2^{17}+5^{20}\right)}{2^{10}\cdot5^7\left(2^{17}+5^{20}\right)}=2^3\)

b: \(M=\left(7-4\right)^{\left(7-5\right)^{\left(7-6\right)^{\left(7+6\right)^{\left(7+5\right)}}}}\)

\(=3^{2\cdot1\cdot13\cdot12}=3^{312}\)