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\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(=\frac{2^{19}.\left(3^3\right)^3+15.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(=\frac{2^{19}.3^9+15.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^9+15.2^{18}.3^8}{2^{19}.3^9+2^{20}.3^{10}}\)
\(=\frac{2^{18}.3^8\left(2.3+15\right)}{2^{19}.3^9\left(1+2.3\right)}\)
\(=\frac{6+15}{2.3\left(1+6\right)}\)
\(=\frac{21}{6.7}\)
\(=\frac{21}{42}\)
\(=\frac{1}{2}\)
Có P =\(\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\dfrac{2^{19}.\left(3^3\right)^3+5.3.\left(3^2\right)^4}{\left(2.3\right)^9+\left(3.2^2\right)^{10}}\)=\(\dfrac{2^{19}.3^9+5.3.2^{18}.3^8}{3^9.2^9.2^{10}+3^{10}.\left(2^2\right)^{10}}=\dfrac{2^{19}.3^9+5.2^{18}.3^9}{3^9.2^{19}+3^{10}.2^{20}}=\dfrac{2^{18}.3^9.\left(2+5\right)}{3^9.2^{19}.\left(1+3.2\right)}=\dfrac{2^{18}.3^9.7}{3^9.2^{19}.7}\)
=\(\dfrac{1}{2}\)
\(=\frac{2^{19}3^9+3\cdot5\cdot2^{18}\cdot3^8}{2^9\cdot3^9\cdot2^{10}+4^{10}\cdot3^{10}}=\frac{2^{19}\cdot3^9+5\cdot2^{18}\cdot3^9}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}=\frac{2^{18}\cdot3^9\cdot\left(2+5\right)}{2^{19}\cdot3^9\left(1+6\right)}=\frac{1}{2}\)
\(A=\frac{2^{19}.\left(2^3\right)^3+15.\left(2^2\right)^9.\left(3^2\right)^4}{2^9.3^9.2^{10}+\left(2^2.3\right)^{10}}=\frac{2^{19}.3^9+15.2^{18}.3^8}{2^{19}.3^9+2^{20}.3^{10}}=\frac{2^{18}.3^8.\left(2.3+15\right)}{2^{19}.3^9.\left(1+2.3\right)}\)
\(=\frac{2^{18}.3^8.21}{2^{19}.3^9.7}=\frac{21}{2.3.7}=\frac{1}{2}\)
2,Đặt \(\frac{x}{3}=\frac{y}{6}\)\(=k\)
Ta có x=3k; y=6k
Vì x+y=90 nên:3k+6k=90
\(\Leftrightarrow\)k(3+6)=90
9k=90
k=90:9=10
Suy ra k=10\(\hept{\begin{cases}x=3.10=30\\y=6.10=60\end{cases}}\)
3,
Đặt \(\frac{x}{3}=\frac{y}{6}\)\(=k\)
Ta có x=3k; y=6k
Vì 4x-y=42 nên:4.3k-6k=42
\(\Leftrightarrow\) 12k-6k=42
6k=42
k=42:6=7
Suy ra k=7\(\hept{\begin{cases}x=3.7=21\\y=6.7=42\end{cases}}\)
4,
Đặt \(\frac{x}{3}=\frac{y}{6}\)\(=k\)
Ta có x=3k; y=6k
Vì xy=162 nên:3k.6k=162
\(\Leftrightarrow\)k2.18=162
k2=162:18
k2=9
k=\(\pm\)3
Với k=3\(\hept{\begin{cases}x=3.3=9\\y=6.3=18\end{cases}}\)
Với k=-3\(\hept{\begin{cases}x=3.\left(-3\right)=-9\\y=6.\left(-3\right)=-18\end{cases}}\)
5,
Đặt \(\frac{x}{3}=\frac{y}{6}\)\(=k\)
Ta có x=3k; y=6k
Vì 2x2-y2=-8 nên:2.(3k)2-(6k)2=-8
\(\Leftrightarrow\)2.9k2-36k2=-8
18k2-36k2=-8
-18k2=-8
k2=-8/-18=4/9
k=\(\pm\)\(\frac{2}{3}\)
Với k=\(\frac{2}{3}\)\(\hept{\begin{cases}x=\frac{2}{3}.3=2\\y=\frac{2}{3}.6=4\end{cases}}\)
Với k=\(\frac{-2}{3}\)\(\hept{\begin{cases}x=\frac{-2}{3}.3=-2\\y=\frac{-2}{3}.6=-4\end{cases}}\)
6,
Đặt \(\frac{x}{3}=\frac{y}{6}\)\(=k\)
Ta có x=3k; y=6k
Vì x-y=9 nên:3k-6k=9
\(\Leftrightarrow\) -3k=9
k=9:(-3)
k=-3
Suy ra\(\hept{\begin{cases}x=-3.3=-9\\y=-3.6=-18\end{cases}}\)