1. a) (x-2)² =1

=> x - 2 = \(\pm\sqrt{1}\)

=> x - 2 = 1 hoặc -1

=> x = 3 hoặc 1

b) 2x - 1= -8

=> 2x = -7

=>x = \(\dfrac{-7}{2}\)

c)thiếu đề

d) (x-1)^x+2 = (x-1)^x+4

(x-1)^x+2 = (x-1)^x+2+2

(x-1)^x+2 = (x-1)^x+2. (x-1)²

(x-1)^x+2 - (x-1)^x+2. (x-1)² = 0

=> (x-1)^x+2. [1 - (x-1)²] = 0

\(\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

2a) \(\dfrac{45^{10}.5^{10}}{75^{10}}\) = \(\dfrac{\left(3.3.5\right)^{10}.5^{10}}{\left(5.5.3\right)^{10}}\) = \(\dfrac{3^{10}.3^{10}.5^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}\) = \(3^{10}\)

b) \(\dfrac{2^{15}.9^4}{6^6.8^3}\)=\(\dfrac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}\)=\(\dfrac{2^{15}.3^8}{2^6.3^6.2^9}\)=\(3^2\)

c)\(\left(x-\dfrac{2}{9}^3\right)=\left(\dfrac{2}{3}\right)^6\)thank nhé

a, \(4^3.5^3=\left(4.5\right)^3=20^3=8000\)

b, \(6^3.5^3=\left(6.5\right)^3=30^3=27000\)

c, \(8^2.5^2=\left(8.5\right)^2=40^2=1600\)

d, \(125^3.8^3=\left(125.8\right)^3=1000^3\)

e, \(5^2.6^2.3^2=\left(5.6.3\right)^2=90^2\)

b) \(\dfrac{7}{15}-\dfrac{9}{19}\)\(-\dfrac{-8}{15}-\dfrac{10}{19}\)
=\(\left(\dfrac{7}{15}-\dfrac{8}{15}\right)\) \(-\left(\dfrac{9}{19}-\dfrac{10}{19}\right)\)
= \(-\dfrac{1}{15}\) - \(\left(-\dfrac{1}{19}\right)\)
\(=-\dfrac{1}{15}\) + \(\dfrac{1}{19}\)

= \(-\dfrac{4}{285}\)

c) \(1\dfrac{1}{3}\) \(\div\) \(\dfrac{4}{5}\) + 2\(\dfrac{2}{3}\) \(\div\)\(\dfrac{4}{5}\)
= \(\left(1\dfrac{1}{3}+2\dfrac{2}{3}\right)\) \(\div\dfrac{4}{5}\)
= \(\left[\left(1+2\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\right]\) \(\div\dfrac{4}{5}\)
= ( 3 + 1 ) \(\div\dfrac{4}{5}\)
= 4 \(\div\dfrac{4}{5}\)
= \(\dfrac{4.5}{4}\)
= 5

Câu 1 :

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

\(\text{b) }C=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\\ C=\dfrac{5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9}{5\cdot2^9\cdot\left(2\cdot3\right)^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{2^{29}\cdot3^{18}\left(10-9\right)}{2^{28}\cdot3^{18}\left(15-14\right)}\\ C=\dfrac{2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}}\\ C=2\\ \)

\(\text{c) }D=\dfrac{49^{24}\cdot125^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot4^5}{5^{29}\cdot16^2\cdot7^{48}}\\ D=\dfrac{\left(7^2\right)^{24}\cdot\left(5^3\right)^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot\left(2^2\right)^5}{5^{29}\cdot\left(2^4\right)^2\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{5^{29}\cdot2^8\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1-28\right)}{5^{29}\cdot2^8\cdot7^{48}}\\ D=5\cdot\left(-27\right)\\ D=-135\)

Câu 2 :

\(\text{a) }9^{x+1}-5\cdot3^{2x}=324\\ \Leftrightarrow9^x\cdot9-5\cdot9^x=81\cdot4\\ \Leftrightarrow9^x\left(9-5\right)=9^2\cdot4\\ \Leftrightarrow9^x\cdot4=9^2\cdot4\\ \Leftrightarrow9^x=9^2\\ \Leftrightarrow x=2\\ \text{Vậy }x=2\\ \)

Sorry . Mình chỉ biết đến đây thôi

a) \(\left(0,25\right)^3\cdot32=0,015625\cdot32=0,5\)

b) \(\left(-0,125\right)^3\cdot80^4=\dfrac{-1}{512}\cdot40960000=80000\)

c) \(\dfrac{8^2\cdot4^5}{2^{20}}=\dfrac{2^{3^2}\cdot2^{2^5}}{2^{20}}=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{2^{16}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\)

d) \(\dfrac{81^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\dfrac{3^{4^{11}}\cdot3^{17}}{3^{3^{10}}\cdot3^{2^{15}}}=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)

a: \(\Leftrightarrow4^x\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)=4^8\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)\)

=>4^x=4^8

=>x=8

b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)

=>2^x=2^11

=>x=11

c: =>1/6*6^x+6^x*36=6^15(1+6^3)

=>6^x=6*6^15

=>x=16

d: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)

=>x=9

bn ơi con b) có vấn đề