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`@` `\text {Ans}`
`\downarrow`
`a)`
\(2^{13}\cdot4^5\cdot8^2\)
`=`\(2^{13}\cdot\left(2^2\right)^5\cdot\left(2^3\right)^2\)
`=`\(2^{13}\cdot2^{10}\cdot2^6\)
`=`\(2^{29}\)
`b)`
\(100^6. 1000^4. 1000000\)\(33\)
`=`\(10^{12}\cdot10^{12}\cdot10^{198}=10^{222}\)
`c)`
\(24^5. 6^5. 8^8\)
`=`\(2^{15}\cdot3^5\cdot2^5\cdot3^5\cdot2^{24}\)
`=`\(2^{15+5+24}\cdot3^{5+5}\)
`=`\(2^{44}\cdot3^{10}\)
a, \(2^{13}\cdot4^5\cdot8^2=2^{13}\cdot2^{10}\cdot2^6=2^{19}.\)
b, \(100^6\cdot1000^4\cdot1000000^{33}=10^{12}\cdot10^{12}\cdot10^{198}=10^{222}.\)
c, \(24^5\cdot6^5\cdot8^8=2^{15}\cdot3^5\cdot2^5\cdot3^5\cdot2^{24}=2^{44}\cdot3^{10}.\)
\(a,3^4.3^5=3^{4+5}=3^9\\ b,12^8:12=12^{8-1}=12^7\\ c,4.8^6.2.8^3=\left(4.2\right).\left(8^6.8^3\right)=8.8^{6+3}=8.8^9=8^{1+9}=8^{10}\)
\(a.3^4.3^5=3^9\)
\(b.12^8:12=12^7\)
\(c.4.8^6.2.8^3\)
\(=\left(4.2\right)8^6.8^3\)
\(=8.8^6.8^3=8^{10}\)
a) \(3\cdot3^4\cdot3^5=3^{10}\)
b) \(7^3:7^2:7=7^{3-2-1}=7^0\)
c) \(\left(x^4\right)^3=x^{12}\)
`@` `\text {Ans}`
`\downarrow`
`g)`
`5*25*125`
`= 5*5^2 * 5^3`
`=`\(5^{1+2+3}=5^6\)
`h)`
`10*100*1000`
`= 10* 10^2 * 10^3`
`=`\(10^{1+2+3}\)
`= 10^6`
_____
`@` CT: \(a^m\cdot a^n=a^{m+n}\)
g) \(5\cdot25\cdot125\)
\(=5\cdot5^2\cdot5^3\)
\(=5^6\)
h) \(10\cdot100\cdot1000\)
\(=10\cdot10^2\cdot10^3\)
\(=10^6\)
2:
a: n=5^4
=>n=625
b: n^3=125
=>n^3=5^3
=>n=5
c: 11^n=1331
=>11^n=11^3
=>n=3
A. \(256:2^5\cdot4\)
\(=2^6:2^5\cdot2^2\)
\(=2^{6-5+2}\)
\(=2^3\)
B. \(\left(4x^2\right)^8:\left(4x^2\right)^4:\left(4x^2\right)^2\)
\(=\left(4x^2\right)^{8-4-2}\)
\(=\left(4x^2\right)^2\)
\(=\left[\left(2x\right)^2\right]^2\)
\(=\left(2x\right)^4\)
C. \(y^{50}:\left(y^5\right)^3:\left(y^2\right)^{10}\)
\(=y^{50}:y^{15}:y^{20}\)
\(=y^{50-15-20}\)
\(=y^{15}\)
c)
Có Q = P.(x+1) = \(\frac{-3.\left(x+1\right)}{\sqrt{x}-2}\)
\(\Leftrightarrow-Q=\frac{3x+3}{\sqrt{x}-2}=\frac{3x-6\sqrt{x}+6\sqrt{x}+3}{\sqrt{x}-2}=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)+6\left(\sqrt{x}-2\right)+15}{\sqrt{x}-2}\)
\(\Leftrightarrow-Q=3\sqrt{x}+6+\frac{15}{\sqrt{x}-2}\)
\(\Leftrightarrow-Q=3\left(\sqrt{x}-2\right)+\frac{15}{\sqrt{x}-2}+12\)
\(\Leftrightarrow-Q\ge2\sqrt{3\left(\sqrt{x}-2\right).\frac{15}{\sqrt{x}-2}}+12=2\sqrt{45}+12\)
\(\Leftrightarrow-Q\ge6\sqrt{5}+12\)
\(\Leftrightarrow Q\le-6\sqrt{5}-12\)
Dấu " = " Xảy ra khi :
\(3.\left(\sqrt{x}-2\right)=\frac{15}{\sqrt{x}-2}\Leftrightarrow\left(\sqrt{x}-2\right)^2=5\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=\sqrt{5}\\\sqrt{x}-2=-\sqrt{5}\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2+\sqrt{5}\Leftrightarrow x=9+4\sqrt{5}\\\sqrt{x}=2-\sqrt{5}\left(Sai\right)\end{cases}}}\)
Vậy Max Q = \(-6\sqrt{5}-12\)khi x = \(9+4\sqrt{5}\)
a ) 9^7 . 9^8 = 97+8 =915
b ) 8^6 : 8^4 = 86-4 = 82
c ) 5^3 . 8^3 = ( 5.8 )3 = 40^3
d ) 24^5 : 6^5 = ( 24:6 )5 = 4^5