( 0,125 )⁸ . 64⁴

\(=\left(\frac{1}{8}\right)^8.64^4\)

\(=\frac{1}{8^8}.\left(8^2\right)^4\)

\(=\frac{1}{8^8}.8^8=1\)

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\(\frac{2^{19}.27+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(=\frac{2^{19}.3^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)

\(=\frac{2^{19}.3^3+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)

\(=\frac{2^{19}.3^3.\left(1+3^6.5\right)}{2^{19}.3^3.\left(3^6+2.3.3^6\right)}\)

\(=\frac{1+3^6.5}{3^6+2.3.3^6}\)

\(=\frac{1+3^6.5}{3^6+2.3.3^6}=\frac{3646}{5103}\)

Biết làm rồi :>>>>>>

Ta có : D = 3 . | 2x - 1 | + | 6x - 4 |

=> D = | 3 | . | 2x - 1 | + | 6x - 4 |

=> D = | 6x - 3 | + | 6x - 4 |

Vì \(\hept{\begin{cases}\left|6x-3\right|\ge6x-3\forall x\\\left|6x-4\right|\ge-6x+4\forall x\end{cases}}\)

=> | 6x - 3 | + | 6x - 4 | ≥ ( 6x - 3 ) + ( -6x + 4 )

=> D ≥ 1

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}6x-3\ge0\\6x-4\le0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}6x\ge3\\6x\le4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\ge\frac{1}{2}\\x\le\frac{2}{3}\end{cases}}\Leftrightarrow\frac{1}{2}\le x\le\frac{2}{3}\)

Tặng Hoàng Sơn một k

ớp 5 làm đc 

\(( x^4 )^2 = \frac{x^{19}}{x^{10}}\)

=> x⁸ = x⁹

=> x⁹ - x⁸ = 0

=> x⁸ . ( x - 1 ) = 0

\(\Rightarrow\orbr{\begin{cases}x^8=0\\x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

x ^4x2 = x^19-10   x^8=x^9   x=1 

=1/2 nha bn 

em ko bt ạ

em mới lớp 6 thui