BT5: \(C=\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}\)
\(C=\dfrac{2^{15}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}=\dfrac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}\)
\(C=\dfrac{2^{15}.3^8}{2^{15}.3^6}=\dfrac{3^8}{3^6}=3^2=9\)
BT4: Ta có: \(81\cdot27=3^4\cdot3^3=3^7\)
\(3^7:3^4=3^3\)
\(\Rightarrow3^7\ge3^n\ge3^3\)
\(\Rightarrow n\in\left\{7;6;5;4;3\right\}\)
`2,`
\(A=\dfrac{25^3\cdot2^{10}}{16^2\cdot625^2}\)
\(A=\dfrac{5^6\cdot2^{10}}{2^8\cdot5^6}\)
\(A=\dfrac{2^{10}}{2^8}\)
\(A=2^2\)
`3,`
\(B=\dfrac{5^{20}\cdot45^{10}}{75^{15}}\)
\(B=\dfrac{5^{20}\cdot15^{10}\cdot3^{10}}{15^{15}\cdot5^{15}}\)
\(B=\dfrac{1\cdot5^5\cdot3^{10}}{15^5}\)
\(B=\dfrac{5^5\cdot3^{10}}{5^5\cdot3^5}\)
\(B=\dfrac{3^{10}}{3^5}=3^5\)
`1,`
`a,`
`(3x-1)^2=4/25`
`=> (3x-1)^2 = (+-2/5)^2`
`=>` `TH1` ` 3x-1=2/5`
`=> 3x=2/5+1`
`=> 3x=7/5`
`=> x=7/5 \div 3`
`=> x=7/15`
`TH2:` `3x-1=-2/5`
`=> 3x=-2/5+1`
`=> 3x=3/5`
`=> x=3/5 \div 3`
`=> x=1/5`
`b,`
`(2x+3)^3+1/8=0`
`=> (2x+3)^3=-1/8`
`=> (2x+3)^3=(-1/2)^3`
`=> 2x+3=-1/2`
`=> 2x=-1/2-3`
`=> 2x=-7/2`
`=> x=-7/2 \div 2`
`=> x= -7/4`
`e,`
`11/12-|3/4+x|=1/6`
`=> |3/4+x| = 11/12-1/6`
`=> |3/4+x|=3/4`
`=>`\(\left[{}\begin{matrix}\dfrac{3}{4}+x=\dfrac{3}{4}\\\dfrac{3}{4}+x=-\dfrac{3}{4}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)
`d,`
`(x+5)/-25=-4/(x+5)`
`=> (x+5)^2 = (-25)*(-4)`
`=> (x+5)^2 = 100`
`=> (x+5)^2 = (+-10)^2`
`=>`\(\left[{}\begin{matrix}x+5=10\\x+5=-10\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=5\\x=-15\end{matrix}\right.\)
`4,`
`81*27 >= 3^n >= 3^7 \div 3^4`
\(3^4\cdot3^3\ge3^n\ge3^7\div3^4\)
\(3^7\ge3^n\ge3^3\)
`=> n={7;6; 5; 4;3}`
`5,`
\(C=\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}\)
\(C=\dfrac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}\)
\(C=\dfrac{3^8}{3^6}\)`=3^2`
