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\(\left(3\times4^2\times7^2\right)\div\left(3^2\times3^{20}\right)\)
\(=\left(3\times28^2\right)\div3^{22}\)
\(=\frac{3\times28^2}{3^{22}}=\frac{28^2}{3^{21}}\)
.........
bạn có ra đề nhầm ko? đến máy tính cx ko ra
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a, Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
=> \(\frac{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}=1\)
=> đpcm
Study well ! >_<
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a) Ta có:
\(2^{300}=2^{3\cdot100}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=3^{2\cdot100}=\left(3^2\right)^{100}=9^{100}\)
Mà: \(8< 9\)
\(\Rightarrow8^{100}< 9^{100}\)
\(\Rightarrow2^{300}< 3^{200}\)
b) Ta có:
\(3^{500}=3^{5\cdot100}=\left(3^5\right)^{100}=243^{100}\)
\(7^{300}=7^{3\cdot100}=\left(7^3\right)^{100}=343^{100}\)
Mà: \(243< 343\)
\(\Rightarrow243^{100}< 343^{100}\)
\(\Rightarrow3^{500}< 7^{300}\)
c) Ta có:
\(8^5=\left(2^3\right)^5=2^{3\cdot5}=2^{15}=2\cdot2^{15}\)
\(3\cdot4^7=3\cdot\left(2^2\right)^7=3\cdot2^{2\cdot7}=3\cdot2^{14}\)
Mà: \(2< 3\)
\(\Rightarrow2\cdot2^{14}< 3\cdot2^{14}\)
\(\Rightarrow8^5< 3\cdot4^7\)
d) Ta có:
\(202^{303}=202^{3\cdot101}=\left(202^3\right)^{101}=8242408^{101}\)
\(303^{202}=303^{2\cdot101}=\left(303^2\right)^{101}=91809^{101}\)
Mà: \(8242408>91809\)
\(\Rightarrow8242408^{101}>91809^{101}\)
\(\Rightarrow202^{303}>303^{202}\)
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Câu 1
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{99}=\frac{49}{100}\)
cho mình nha bạn
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Bài 1:
a) \(4^{x+2}+4^x=68\)
\(\Rightarrow4^x\cdot\left(4^2+1\right)=68\)
\(\Rightarrow4^x\cdot17=68\)
\(\Rightarrow4^x=\dfrac{68}{17}\)
\(\Rightarrow4^x=4\)
\(\Rightarrow4^x=4^1\)
\(\Rightarrow x=1\)
b) \(5\cdot2^{x+4}-3\cdot2^x=308\)
\(\Rightarrow2^x\cdot\left(5\cdot2^4-3\right)=308\)
\(\Rightarrow2^x\cdot\left(5\cdot16-3\right)=308\)
\(\Rightarrow2^x\cdot77=308\)
\(\Rightarrow2^x=\dfrac{308}{77}\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\)
\(\Rightarrow x=2\)
c) \(4\cdot3^{x+1}+7\cdot3^x=513\)
\(\Rightarrow3^x\cdot\left(4\cdot3+7\right)=513\)
\(\Rightarrow3^x\cdot19=513\)
\(\Rightarrow3^x=\dfrac{513}{19}\)
\(\Rightarrow3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
d) \(5^{x+4}-5^x=3120\)
\(\Rightarrow5^x\cdot\left(5^4-1\right)=3120\)
\(\Rightarrow5^x\cdot\left(625-1\right)=3120\)
\(\Rightarrow5^x\cdot624=3120\)
\(\Rightarrow5^x\cdot\dfrac{3120}{624}\)
\(\Rightarrow5^x=5\)
\(\Rightarrow5^x=5^1\)
\(\Rightarrow x=1\)
f) \(3\cdot4^{2x+1}-16^x=2816\)
\(\Rightarrow3\cdot4^{2x+1}-\left(4^2\right)^x=2816\)
\(\Rightarrow3\cdot4^{2x+1}-4^{2x}=2816\)
\(\Rightarrow4^{2x}\cdot\left(3\cdot4-1\right)=2816\)
\(\Rightarrow4^{2x}\cdot11=2816\)
\(\Rightarrow4^{2x}=\dfrac{2816}{11}\)
\(\Rightarrow4^{2x}=256\)
\(\Rightarrow\left(2^2\right)^{2x}=2^8\)
\(\Rightarrow2^{4x}=2^8\)
\(\Rightarrow4x=8\)
\(\Rightarrow x=2\)
Bài 2:
\(2^x+124=5^y\)
\(\Rightarrow5^y-2^x=124\)
\(\Rightarrow5^y-2^x=125-1\)
\(\Rightarrow\left\{{}\begin{matrix}5^y=125\\2^x=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5^y=5^3\\2^x=2^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=3\\x=0\end{matrix}\right.\)
Vậy: ....
\(2^3.4^2\)
\(=2^3.\left(2^2\right)^2\)
\(=2^3.2^4\)
\(=2^{3+4}\)
\(=2^7\)
\(=128\)
_Chúc bạn học tốt_
\(2^3\cdot4^2=48\)
Em năm nay mới lên lớp 6 ! Em biết đc chút chút !