\(3^x=3^{10}\\ =>x=10\)

đk a > 0

\(=\dfrac{4+4\sqrt{a}+a-a+6\sqrt{a}-9}{\sqrt{a}\left(2\sqrt{a}-1\right)}\\ =\dfrac{10\sqrt{a}-5}{\sqrt{a}\left(2\sqrt{a}-1\right)}\\ =\dfrac{5\left(2\sqrt{a}-1\right)}{\sqrt{a}\left(2\sqrt{a}-1\right)}\\ =\dfrac{5}{\sqrt{a}}\)

\(=\dfrac{2}{5}\left(\dfrac{7}{4}-\dfrac{3}{7}\right)\\ =\dfrac{2}{5}\left(\dfrac{49-12}{28}\right)\\ =\dfrac{2}{5}.\dfrac{37}{28}\\ =\dfrac{37}{70}\)

\(a,x-35=120\\ x=120+35\\ x=155\\ b,118-x=310-217\\ 118-x=93\\ x=25\\ c,x+61=156-82\\ x+61=74\\ x=74-61\\ x=13\\ d,x-305=814-712\\ x-305=102\\ x=407\\ e,2x-138=72\\ 2x=72+138\\ 2x=210\\ x=105\\ g,x+35=515:5\\ x+35=103\\ x=68\)

\(\dfrac{1}{5}\) thế kỉ = \(20\) năm

\(\dfrac{1}{8}\times100=12,5\left(năm\right)\\ \dfrac{1}{3}\times100=33,3\left(năm\right)\\ \dfrac{1}{6}.60=10\left(giây\right)\)

\(a,2x+2x=32\\ 4x=32\\ x=8\\ b,3x+6x=81\\ 9x=81\\ x=9\\ c,\dfrac{1}{2}x+\dfrac{1}{3}x+4x=348\\ \left(\dfrac{1}{2}+\dfrac{1}{3}+4\right)x=348\\ \dfrac{29}{6}x=348\\ x=72\)

\(a,=23^4\\ b,=2^3.6^3\\ c,=2^4.3.5^4\\ d,=a^3+b^4\)

đk a ≠ 1  ; a ≥ 0

\(=\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{1-\sqrt{a}}\right)\\ =\left(1+\sqrt{a}\right)\left(1+\dfrac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\right)\\ =\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)

A= { x ϵ N /  x là các số lẻ < 21 }