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blue_cosy

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khongminh.tq55

1) [TEX]8^{x+1} + 8\frac{1}{2^{3x}}+3.2^{x+3}=125-24\frac{1}{2^x}[/TEX]
\Leftrightarrow[TEX]8.8^x+8\frac{1}{2^{3x}}+3.2^3.2^x=125-24\frac{1}{2^x}[/TEX]
\Leftrightarrow[TEX]8.8^x+8\frac{1}{8^x}+24.2^x+24\frac{1}{2^x}=125[/TEX]
đặt [TEX]t=2^x[/TEX]. ta có:
[TEX]8t^3+8\frac{1}{t^3}+24.t+24\frac{1}{t}=125[/TEX]
\Leftrightarrow[TEX]8(t^3+\frac{1}{t^3})+24(t+\frac{1}{t})=125[/TEX]
ta có: [TEX]t^3+\frac{1}{t^3}=(t+\frac{1}{t})^3-3(t-\frac{1}{t})[/TEX] nên:
[TEX]8(t^3+\frac{1}{t^3})+24(t+\frac{1}{t})=125 \Leftrightarrow8[(t+\frac{1}{t})^3-3(t-\frac{1}{t})]+24(t+\frac{1}{t})=125[/TEX]
\Leftrightarrow[TEX]8(t+\frac{1}{t})^3-24(t-\frac{1}{t})]+24(t+\frac{1}{t})=125[/TEX]
\Leftrightarrow[TEX]8(t+\frac{1}{t})^3=125[/TEX]
\Leftrightarrow[TEX]t+\frac{1}{t}=\frac{5}{2}[/TEX]
\Leftrightarrow[TEX]2t^2-5t+2=0[/TEX] \Leftrightarrow [TEX]t=2[/TEX] v [TEX]t=\frac{1}{2}[/TEX]
*[TEX]t=2[/TEX]\Leftrightarrow[TEX]2^x=2[/TEX]\Leftrightarrow[TEX]x=1[/TEX]
*[TEX]t=\frac{1}{2}[/TEX]\Leftrightarrow[TEX]2^x=\frac{1}{2}[/TEX]\Leftrightarrow[TEX]x=-1[/TEX]
:)):-j:-/^:)^:khi (75):
 
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