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a) \(\left(1,25\right)^3.8^3=\left(1,25.8\right)^3=1000\)
b) \(\left(\dfrac{-11}{9}\right)^4.\left(\dfrac{27}{22}\right)^4=\left(\dfrac{-11}{9}.\dfrac{27}{22}\right)^4=\left(\dfrac{-11.9.3}{9.2.\left(-11\right)}\right)^4\)
\(=\left(\dfrac{3}{2}\right)^4=\dfrac{81}{16}\)
c) \(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2=\left(\dfrac{13}{14}\right)^2=\dfrac{169}{196}\)
d) \(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{\left(5.20\right)^4}{\left(25.4\right)^5}=\dfrac{100^4}{100^5}=100^{-1}=0,01\)
a,
\(\dfrac{\left(3^3\right)^{15}.5^3.\left(2^3\right)^4}{\left(5^2\right)^2.\left(3^4\right)^{11}.2^{11}}=\dfrac{3^{45}.5^3.2^{12}}{5^4.3^{44}.2^{11}}=\dfrac{6}{5}\)
b, \(\left(-\dfrac{14}{25}\right)^2.\dfrac{125}{49}+\left(-3\dfrac{11}{36}\right).2\dfrac{2}{17}=\dfrac{4}{5}.\left(-7\right)=-\dfrac{28}{5}\)
c, \(\dfrac{1}{3}-2.1=-\dfrac{5}{3}\)
d) \(\left|x-1\right|+\left|x-5\right|+\left|2x+5\right|\)
\(=\left|1-x\right|+\left|5-x\right|+\left|2x+5\right|\)
\(\ge\left|1-x+5-x\right|+\left|2x+5\right|\)
\(\ge\left|6-2x+2x+5\right|=11\)
Dấu \(=\)khi \(\hept{\begin{cases}\left(1-x\right)\left(5-x\right)\ge0\\\left(6-2x\right)\left(2x+5\right)\ge0\end{cases}}\Leftrightarrow-\frac{5}{2}\le x\le1\).
e) \(\left|x+2\right|+\left|x-1\right|+\left|x-4\right|+\left|x+5\right|=12\)
\(\Leftrightarrow\left|x+2\right|+\left|1-x\right|+\left|4-x\right|+\left|x+5\right|=12\)
Có \(\left|x+2\right|+\left|1-x\right|+\left|4-x\right|+\left|x+5\right|\ge\left|x+2+1-x\right|+\left|4-x+x+5\right|=3+9=12\)
Dấu \(=\)khi \(\hept{\begin{cases}\left(x+2\right)\left(1-x\right)\ge0\\\left(4-x\right)\left(x+5\right)\ge0\end{cases}}\Leftrightarrow-2\le x\le1\).
f) \(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|3x-10\right|\)
\(\ge\left|x-1+x-2\right|+\left|3-x+3x-10\right|\)
\(=\left|2x-3\right|+\left|2x-7\right|\)
\(\ge\left|2x-3+7-2x\right|=4\)
Dấu \(=\)khi \(\hept{\begin{cases}\left(x-1\right)\left(x-2\right)\ge0\\\left(3-x\right)\left(3x-10\right)\ge0\\\left(2x-3\right)\left(7-2x\right)\ge0\end{cases}}\Leftrightarrow3\le x\le\frac{10}{3}\).
a) \(\frac{3^{17}.81^{11}}{27^{10}.9^{15}}=\frac{3^{17}.\left(3^4\right)^{11}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\frac{3^{17}.3^{44}}{3^{30}.3^{30}}=\frac{3^{61}}{3^{60}}=3\)
b) \(\frac{9^2.2^{11}}{16^2.6^3}=\frac{\left(3^2\right)^2.2^{11}}{\left(2^4\right)^2.2^3.3^3}=\frac{3^4.2^{11}}{2^8.2^3.3^3}=\frac{3^4.2^{11}}{2^{11}.3^3}=3\)
c) \(\frac{2^{10}.3^{31}+2^{40}.3^6}{2^{11}.3^{31}+2^{41}.3^6}=\frac{2^{10}.3^{31}+2^{40}.3^6}{2.\left(2^{10}.3^{31}+2^{40}.3^6\right)}=\frac{1}{2}\)
5) \(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}\)
=\(4+6-3+5\)
=\(12\)
2) \(\dfrac{11}{25}.\left(-24,8\right)-\dfrac{11}{25}.75,2\)
=\(\dfrac{11}{25}.\left(-24,8-75,2\right)\)
=\(\dfrac{11}{25}.\left(-100\right)\)
=\(-44\)
\(A=\dfrac{\left(-3\right)^{45}\cdot5^3\cdot2^{12}}{5^4\cdot3^{44}\cdot\left(-2\right)^{11}}=\dfrac{\left(-3\right)^{45}\cdot\left(-2\right)^{12}}{5\cdot\left(-3\right)^{44}\cdot\left(-2\right)^{11}}=\dfrac{\left(-3\right)\cdot\left(-2\right)}{5}=\dfrac{6}{5}\)
Thank u very much!!