\(7^{x+1}=49^5\)

\(7^{x+1}=\left(7^2\right)^5\)

\(7^{x+1}=7^{10}\)

`=>x+1=10`

`x=10-1`

`x=9`

a) \(\left(2x+1\right)\left(2x-3\right)=0\)

\(\left[{}\begin{matrix}2x+1=0\\2x-3=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}2x=-1\\2x=3\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{1}{2};\dfrac{3}{2}\right\}\)

b) \(x^2-9=0\)

\(x^2=0+9\)

`x^2=9`

`x^2=3^2`

`=>`\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

a) \(\sqrt{9x+9}-\sqrt{25x+25}+\sqrt{36x+36}=9\)

\(3\sqrt{x+1}-5\sqrt{x+1}+6\sqrt{x+1}=8\)

\(4\sqrt{x+1}=8\)

\(\sqrt{x+1}=\dfrac{8}{4}\)

\(\sqrt{x+1}=2\)

\(x+1=2^2\)

\(x+1=4\)

`=>x=3`

\(\dfrac{72^3\cdot54^2}{108^4}\)

\(=\dfrac{36^3\cdot2^3\cdot54^2}{54^4\cdot2^4}\)

\(=\dfrac{36^3\cdot54^2\cdot1}{54^4\cdot2}\)

\(=\dfrac{36^3}{2^3\cdot3^6}\)

\(=\dfrac{18^3}{3^6}=\dfrac{18^3}{\left(3^2\right)^3}=\dfrac{18^3}{9^3}=2^3=8\)

\(\dfrac{2^{10}\cdot15+2^{10}\cdot65}{2^8\cdot104}\)

\(=\dfrac{2^{10}\cdot\left(15+65\right)}{2^8\cdot108}\)

\(=\dfrac{2^{10}\cdot80}{2^8\cdot104}\)

\(=\dfrac{2^2\cdot2^4\cdot5}{2^3\cdot13}\)

\(=\dfrac{2^6\cdot5}{2^3\cdot13}=\dfrac{2^3\cdot5}{13}=\dfrac{8\cdot5}{13}=\dfrac{40}{13}\)

Nhìn đc mỗi 2 câu này :(

\(\dfrac{1}{4}+\left(-x\right)=0,75\)

\(\dfrac{1}{4}-x=\dfrac{3}{4}\)

\(x=\dfrac{1}{4}-\dfrac{3}{4}\)

`=>x=-2/4=-1/2`

a) \(\dfrac{3}{5}+\dfrac{5}{7}=\dfrac{21}{35}+\dfrac{25}{35}=\dfrac{46}{35}\)

\(\dfrac{3}{4}+\dfrac{5}{6}=\dfrac{9}{12}+\dfrac{10}{12}=\dfrac{19}{12}\)

\(\dfrac{3}{4}+2\dfrac{5}{7}=\dfrac{3}{4}+\dfrac{19}{7}=\dfrac{21}{28}+\dfrac{76}{28}=\dfrac{97}{28}\)

\(1\dfrac{3}{4}+7\dfrac{5}{6}=\dfrac{7}{4}+\dfrac{47}{6}=\dfrac{21}{12}+\dfrac{94}{12}=\dfrac{115}{12}\)

b) \(\dfrac{19}{12}-\dfrac{3}{4}=\dfrac{19}{12}-\dfrac{9}{12}=\dfrac{10}{12}=\dfrac{5}{6}\)

\(\dfrac{19}{12}-\dfrac{5}{6}=\dfrac{19}{12}-\dfrac{10}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

\(1\dfrac{11}{35}-\dfrac{5}{7}=\dfrac{35}{35}+\dfrac{11}{35}-\dfrac{25}{35}=\dfrac{21}{35}=\dfrac{3}{5}\)

\(7\dfrac{19}{12}-4\dfrac{5}{6}=\dfrac{103}{12}-\dfrac{29}{6}=\dfrac{103}{12}-\dfrac{58}{12}=\dfrac{45}{12}=\dfrac{15}{4}\)

\(\dfrac{2}{3}-\dfrac{5}{3}x=\dfrac{7}{10}x+\dfrac{5}{6}\)

\(\dfrac{5}{3}x+\dfrac{7}{10}x=\dfrac{2}{3}-\dfrac{5}{6}\)

\(\dfrac{50+21}{30}x=\dfrac{4-5}{6}\)

\(\dfrac{71}{30}x=-\dfrac{1}{6}\)

\(x=\left(-\dfrac{1}{6}\right)\cdot\dfrac{30}{71}\)

`=>x=-5/71`