\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)

\(=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)

\(=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)

\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}\)

\(=\frac{2.6}{3.5}\)

\(=\frac{4}{5}\)

Ta có: \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)

\(=\frac{2^{12}\cdot3^{10}+2^3\cdot3\cdot5\cdot2^9\cdot3^9}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=\frac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}\)

\(=\frac{2\cdot6}{3\cdot5}=\frac{4}{5}\)

\(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-6^{11}}\)

                   \(=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{6^{12}-6^{11}}\)

                   \(=\frac{2^{12}3^{10}\left(1+5\right)}{6^{11}\left(6-1\right)}\)

                   \(=\frac{2^{10}\cdot3^{10}\cdot5\cdot2^2}{6^{10}\cdot6\cdot5}\)

                   \(=\frac{6^{10}\cdot20}{6^{10}\cdot30}\)

                   \(=\frac{2}{3}\)

\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\) (Sau đó phân tách ra)

=\(\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.\left(2^3.3.5\right)}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)

=\(\frac{2^{12}.3^{10}+2^9.3^9.2^33.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)

=\(\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\) (Gộp và giải như biểu thức thường)

=\(\frac{2^{12}.3^{10}.\left(1+1.5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}\)

=\(\frac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}\) (Rút gọn giữa tử và mẫu)

=\(\frac{2.1.6}{1.3.5}=\frac{2.1.2}{1.1.5}=\frac{4}{5}\)

\(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{-\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{-2^{12}.3^{12}-2^{11}.3^{11}}\)

\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{23}.3^{23}}=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{23}.3^{23}}=\frac{6}{2^{11}.3^{13}}=\frac{2.3}{2^{11}.3^{12}}=\frac{1}{2^{10}.3^{11}}=\frac{1}{6^{10}.3}\)

\(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3-1\right)}\)
\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.5}\)
\(=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}=\dfrac{2.6}{3.5}=\dfrac{4}{5}\)

Giải giúp em với mn

Tính giá trị của biểu thức :

\(B=\dfrac{4^6\cdot9^5+6^9\cdot120}{-8^4\cdot3^{12}+6^{11}}\\ =\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot3\cdot5\cdot2^3}{-2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\\ =\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{-2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\\ =\dfrac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\cdot\left(-2\cdot3+1\right)}\\ =\dfrac{2\cdot6}{3\cdot\left(-5\right)}\\ =\dfrac{12}{-15}\\ =\dfrac{-4}{5}\)

\(A=\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{8}{11}-\dfrac{5}{7}.\dfrac{2}{11}\)

\(A=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{8}{11}-\dfrac{2}{11}\right)\)

\(A=\dfrac{5}{7}.\dfrac{5+8-2}{11}\)

\(A=\dfrac{5}{7}.\dfrac{11}{11}\)

\(A=\dfrac{5}{7}.1=\dfrac{5}{7}\)

\(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{35}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{63}\)

\(B=\dfrac{95}{72}\)

\(C=\dfrac{4^6.9^5+6^9.120}{8^4-3^{12}-6^{11}}\)

\(C=\dfrac{\left(2^2\right)^3.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)

\(C=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)

\(C=\dfrac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.5}\)

\(C=\dfrac{2.6}{5.3}=\dfrac{12}{15}=\dfrac{4}{5}\)

\(=\dfrac{3}{4}-\dfrac{5}{6}\times\dfrac{7}{24}\times\dfrac{12}{7}=\dfrac{3}{4}-\dfrac{5}{12}=\dfrac{1}{3}\)

\(\dfrac{3}{4}-\dfrac{5}{6}\left(\dfrac{1}{6}+\dfrac{1}{8}\right):\dfrac{7}{12}\)

\(=\dfrac{3}{4}-\dfrac{5}{6}\cdot\dfrac{7}{24}\cdot\dfrac{12}{7}\)

\(=\dfrac{3}{4}-\dfrac{5}{12}\)

\(=\dfrac{4}{12}=\dfrac{1}{3}\)

\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)

\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)

a: \(A=\left(\frac{136}{15}-\frac{28}{5}+\frac{62}{10}\right)\cdot\frac{21}{24}\)

\(=\left(\frac{272}{30}-\frac{168}{30}+\frac{186}{30}\right)\cdot\frac78\)

\(=\frac{290}{30}\cdot\frac78=\frac{29}{3}\cdot\frac78=\frac{203}{24}\)

b: \(B=\frac56+6\frac56\left(11\frac{5}{20}-9\frac14\right):8\frac13\)

\(=\frac56+\frac{41}{6}\cdot\left(11+\frac14-9-\frac14\right):\frac{25}{3}=\frac56+\frac{41}{6}\cdot2\cdot\frac{3}{25}\)

\(=\frac56+\frac{41}{25}=\frac{125}{150}+\frac{246}{150}=\frac{371}{150}\)

c: \(C=1+3+6+\cdots+1225\)

\(=\frac12\left(2+6+12+\cdots+2450\right)=\frac12\cdot\left(1\cdot2+2\cdot3+\cdots+49\cdot50\right)\)

\(=\frac12\cdot\left\lbrack1\left(1+1\right)+2\left(2+1\right)+\cdots+49\left(49+1\right)\right\rbrack=\frac12\cdot\left\lbrack\left(1+2+\cdots+49\right)+\left(1^2+2^2+\cdots+49^2\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack49\cdot\frac{50}{2}+\frac{49\left(49+1\right)\left(2\cdot49+1\right)}{6}\right\rbrack\)

\(=\frac12\cdot\left\lbrack49\cdot25+49\cdot50\cdot\frac{99}{6}\right\rbrack=\frac12\cdot\left\lbrack49\cdot25+49\cdot25\cdot33\right\rbrack=\frac12\cdot49\cdot25\cdot\left(33+1\right)\)

\(=49\cdot25\cdot\frac{34}{2}=49\cdot25\cdot17=20825\)