(2^x+5)^2=169
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A) \(\frac{7^{2003}+7^{2002}}{7^{2004}}=\frac{7^{2003}}{7^{2004}}+\frac{7^{2002}}{7^{2004}}=\frac{1}{7}+\frac{1}{7^2}=\frac{8}{49}\)
B) Xem lại đề bài
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( 169 - 42 ).( 169 - 52 ).( 169 - 62 )........( 169 - 142 ).( 169 - 152 )
= ( 132 - 42 ).( 132 - 52 ).( 132 - 62 )........( 132 - 132 ).( 132 - 142 ).( 132 - 152 )
= ( 132 - 42 ).( 132 - 52 ).( 132 - 62 ) ........0.( 132 - 142 ).( 132 - 152 )
= 0
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(169-42).(169-52)…(169-122).(169-132)
=(169-42).(169-52)…(169-122).(169-169)
=(169-42).(169-52)…(169-122).0
=0
(169 - 4^2) . (169 - 5^2) ... (169 - 13^2)
=(169-4^2).(169-5^2)...(169-169)
=0
![](https://rs.olm.vn/images/avt/0.png?1311)
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=(169-4^2).(169-5^20).....(169-169)
= 169-4^2).(169-5^20).......0
= 0
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (54 + 57) . (89 - 27) . (24 - 42)
= (54 + 57) . (89 - 27) . (16 - 16)
= (54 + 57) . (89 - 27) . 0
= 0
b) (72003 + 72002) : 72001
= 72003 : 72001 + 72002 : 72001
= 72 + 7
= 49 + 7
= 53
c) (169 - 42 ) × ( 169 - 52 ) ×......× ( 169 - 132)
= (169 - 42 ) × ( 169 - 52 ) ×......× ( 169 - 169)
= (169 - 42 ) × ( 169 - 52 ) ×......× 0
= 0
Tìm x biết
7 × ( x - 14 ) + 39 = 25 + 279 ÷ 9
=> 7 × ( x - 14 ) + 39 = 25 + 31
=> 7 × ( x - 14 ) + 39 = 56
=> 7 × (x - 14) = 56 - 39
=> 7 × (x - 14) = 17
=> x - 14 = 17 : 7
=> x - 14 = 17/7
=> x = 17/7 + 14
=> x = 115/7
\((5^4+4^7)\times(8^9-2^7)\times(2^4-4^2)\)
\(=(5^4+4^7)\times(8^9-2^7)\times0\)
\(=0\)
\(\left(2^x+5\right)^2=169\)
=>\(\left[{}\begin{matrix}2^x+5=13\\2^x+5=-13\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2^x=13-5=8\left(nhận\right)\\2^x=-13-5=-18\left(loại\right)\end{matrix}\right.\)
=>\(2^x=8\)
=>x=3
( 2x + 5 )2 = 169
(2x + 5 )2 = 132
2x + 5 = 13
2x = 13 - 5
2x = 8
2x = 23
x = 3
Vậy x = 3