\(\frac{2^{12}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}\)

\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)

\(=\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6+3^5\right)}\)

\(=\frac{3^5-3^4}{3^6+3^5}=\frac{3^4.\left(3-1\right)}{3^5\left(3+1\right)}\)

\(=\frac{3^4.2}{3^5.4}=\frac{3^4.2}{3^4.3.4}=\frac{2}{12}=\frac{1}{6}\)

P/s: Hoq chắc ạ (: Ms lp 6 lm đại

\(\frac{x}{2}=\frac{y}{3}\)

\(\Leftrightarrow\frac{x}{8}=\frac{y}{12}\)(1)

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

\(\Leftrightarrow\frac{y}{12}=\frac{z}{15}\)(2)

Từ (1) (2)

 \(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)

\(\Rightarrow\hept{\begin{cases}x=2.8\\y=2.12\\z=2.15\end{cases}\Rightarrow}\hept{\begin{cases}x=16\\y=24\\z=30\end{cases}}\)

1) Ta có: \(\frac{\left(5^4-5^3\right)^3}{125^4}\)

\(=\frac{\left[5^3\cdot\left(5-1\right)\right]^3}{5^{12}}\)

\(=\frac{5^9\cdot4^3}{5^{12}}\)

\(=\frac{4^3}{5^3}=\frac{64}{125}\)

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

\(=\frac{2^{12}\cdot3^{10}+3^{10}\cdot2^{12}\cdot5}{6^{12}-6^{11}}\)

\(=\frac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{6^{11}\cdot\left(6-1\right)}\)

\(=\frac{6^{11}\cdot2^2}{6^{11}\cdot5}=\frac{2^2}{5}=\frac{4}{5}\)

a) \(\frac{3}{7}x-\frac{1}{35}=\frac{3}{5}\)

\(\frac{3}{7}x=\frac{3}{5}+\frac{1}{35}\)

\(\frac{3}{7}x=\frac{22}{35}\)

\(x=\frac{49}{35}=1,4\)

b) \(1,5-x:\frac{1}{2}=\frac{1}{4}\)

\(x:\frac{1}{2}=1,5-\frac{1}{4}\)

\(x:\frac{1}{2}=\frac{5}{4}\)

\(x=\frac{5}{4}.\frac{1}{2}\)

\(x=\frac{5}{8}\)

Vậy ..

a.\(\frac{3}{7}.x-\frac{1}{35}=\frac{3}{5}\)

\(\frac{3}{7}.x=\frac{3}{5}+\frac{1}{35}\)

\(\frac{3}{7}.x=\frac{11}{35}\)

\(x=\frac{22}{35}+\frac{3}{7}\)

\(x=\frac{37}{35}\)

a. \(1,5-x:\frac{1}{2}=\frac{1}{4}\)

                \(x:\frac{1}{2}=1,5-\frac{1}{4}\)

               \(x:\frac{1}{2}=\frac{5}{4}\)

              \(x=\frac{5}{4}.\frac{1}{2}\)

               \(x=\frac{5}{8}\)

qwertyuiopasdfgggggghjkllzxcvbnmm,.//234567890-=`

\(a,\) \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\frac{2^{12}.3^{10}+\left(2.3\right)^9.2^3.3.5}{2^{12}.3^{12}-\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{\left(2^{12}.3^{10}\right)\left(1+5\right)}{\left(2^{11}.3^{11}\right)\left(2.3-1\right)}\)
\(=\frac{\left(2^{12}.3^{10}\right).6}{\left(2^{11}.3^{11}\right).5}\)
\(=\frac{2.6}{3.5}\)
\(=\frac{2.2}{5}\)
\(=\frac{4}{5}\)
\(b,\) \(\frac{2^{15}.9^4}{6^3.8^3}\)
\(=\frac{2^{15}.3^8}{2^3.3^3.2^9}\)
\(=\frac{2^{15}.3^8}{2^{12}.3^3}\)
\(=2^3.3^5\)
\(=8.243\)
\(=1944\)
Chúc bạn học tốt ^^


 

a) \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+6^9.120}{\left(2^3\right)^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}+6^9.120}{2^{12}.3^{12}-6^{11}}=\frac{6^{10}.4+6^{10}.20}{6^{12}-6^{11}}=\frac{6^{10}.\left(4+20\right)}{6^{11}.\left(6-1\right)}=\frac{6^{11}.4}{6^{11}.5}=\frac{4}{5}\)

b) \(\frac{2^{15}.9^4}{6^3.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^3.3^3.2^9}=\frac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5=1944\)

c) \(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(4.2\right)^{10}+4^{10}}{\left(2^3\right)^4+4^6.4^5}=\frac{4^{10}.2^{10}+4^{10}}{2^{12}+4^6.4^5}=\frac{4^{10}.\left(2^{10}+1\right)}{4^6+4^6.2^{10}}=\frac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(1+2^{10}\right)}=\frac{4^{10}}{4^6}=4^4=256\)