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\(a)3^{x+1}-3^x=162\)
\(\Leftrightarrow3^x\cdot3-3^x=162\)
\(\Leftrightarrow3^x\left(3-1\right)=162\)
\(\Leftrightarrow3x\cdot2=162\)
\(\Leftrightarrow3x=162:2\)
\(\Leftrightarrow3x=81\)
\(\Leftrightarrow x=81:3\)
\(\Leftrightarrow x=27\)
Vậy x=27
\(b)\left(1-x\right)^3=216\)
\(\Leftrightarrow\left(1-x\right)^3=6^3\)
\(\Leftrightarrow x-1=6\)
\(\Leftrightarrow x=6+1\)
\(\Leftrightarrow x=7\)
Vậy x=7
\(c)5^{x+1}-2\cdot5^x=375\)
\(\Leftrightarrow5^x\cdot5-2\cdot5^x=375\)
\(\Leftrightarrow5^x\cdot\left(5-2\right)=375\)
\(\Leftrightarrow5^x\cdot3=375\)
\(\Leftrightarrow5^x=375:3\)
\(\Leftrightarrow5^x=125\)
\(\Leftrightarrow5^x=5^3\)
\(\Leftrightarrow x=3\)
Vậy x=3
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32010- ( 32009 + 32008 + ... + 3 + 1 )
Đặt A = 1 + 3 + ... + 32009
=> 3A = 3 + 32 + ... + 32010
=> 3A - A = 32010 - 1
Nên 32010 - ( 32010 - 1 ) = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5^{x+1}-2.5^x=375\Rightarrow5^x.5-2.5^x=375\Rightarrow3.5^x=375\Rightarrow5^x=125=5^3\Rightarrow x=3\)
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\(2^3+3.\left(\frac{1}{9}\right)^0-2.4+\left[\left(-2\right)^2:\frac{1}{2}\right].8\)
\(=8+3.1-8+\left(4:\frac{1}{2}\right).8\)
\(=\left(8-8\right)+3+8.8\)
\(=3+64=67\)
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\(A=\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+.....+\left(-3\right)^{2004}\)
\(-3A=\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{2005}\)
\(-3A-A=\left(-3^{2005}-\left(-3^0\right)\right)\)
\(A=\frac{\left(-3^{2005}-\left(-3^0\right)\right)}{-4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)3x4-5x3y+6x2-10xy+2
=(3x4-5x3y)+(6x2-10xy)+2
=x3(3x-5y)+2x(3x-5y)+2
=x3.0+2x.0+2
=0+0+2
=2
2) x5-2010x4+2010x3-2010x2+2010x-2020
=x5-(2009+1)x4+(2009+1)x3-(2009+1)x2+(2009+1)x-2009-11
=x5-(x+1)x4+(x+1)x3-(x+1)x2+(x+1)x-x-11
=x5-x5-x4+x4+x3-x3-x2+x2+x-x-11
=-11
3²\0,375² = (3\0375)² = 8² = 64
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