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\(\Leftrightarrow x^2=8\cdot2=16\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
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Ta thấy \(10^{50}>10^{50}-3\)
\(\Rightarrow B=\frac{10^{50}}{10^{50}-3}>\frac{10^{50}+2}{10^{50}-3+2}=\frac{10^{50}+2}{10^{50}-1}=A\)
Vậy \(A< B\)
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35.6^3 = 7.5.6^3
7^5 = 7.7.7^3
7.5.6^3 < 7.7.7^3
Vậy 35.6^3 < 7^5
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a) \(2^{48}\) và \(8^{17}\)
= \(2^{48}\) và \(\left(2^3\right)^{17}\)
= \(2^{48}\) và \(2^{3.17}\)
= \(2^{48}\) và \(2^{51}\)
=> \(2^{48}\) \(< \) \(2^{51}\)
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a) \(142-\left[50-\left(2^3.10-2^3.5\right)\right]\)
= \(142-\left[50-\left(8.10-8.5\right)\right]\)
= \(142-\left[50-8.\left(10-5\right)\right]\)
= \(142-\left[50-8.5\right]\)
= \(142-\left[50-40\right]\)
= \(142-10\)
= \(132\)
b) \(375:\left\{32-\left[4+\left(5.3^2-42\right)\right]\right\}-14\)
= \(375:\left\{32-\left[4+\left(5.9-42\right)\right]\right\}-14\)
= \(375:\left\{32-\left[4+\left(45-42\right)\right]\right\}-14\)
= \(375:\left\{32-\left[4+3\right]\right\}-14\)
= \(375:\left\{32-7\right\}-14\)
= \(375:25-14\)
= \(15-14\)
= \(1\)
Chúc bạn học tốt !!!
a) 142 - [ 50 - ( 23 . 10 - 23 . 5 )]
= 142 - [ 50 - ( 8 . 10 - 8 . 5 )]
= 142 - [ 50 - ( 80 - 40 )]
= 142 - [ 50 - 40 ]
= 142 - 10
= 132
b) 375 : { 32 - [ 4 + ( 5 . 32 - 42 )]} - 14
= 375 : { 32 - [ 4 + ( 5 . 9 - 42 )]} - 14
= 375 : { 32 - [ 4 + ( 45 - 42 )]} - 14
= 375 : { 32 - [ 4 + 3 ]} - 14
= 375 : { 32 - 7 } - 14
= 375 : 25 - 14
= 15 - 14
= 1
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\(\frac{1}{2}+\frac{1}{5}+\frac{2}{9}\)
=\(\frac{45}{90}+\frac{18}{90}+\frac{20}{90}\)
=\(\frac{83}{90}\)
`A = 2 + 2^2 + ... + 2^50`
`2A = 2 . (2 + 2^2 + ... + 2^50)`
`2A = 2^2 + 2^3 + ... + 2^51`
`2A - A =(2^2 + 2^3 + ... + 2^51) - (2 + 2^2 + ... + 2^50)`
`A = 2^51 - 2`
Vậy: `A = 2^51 - 2`