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a) 35 - 102 : 52 + 23 . 7
= 3^5 - 2^2 +2^3 .7
= 2^2.(3^5 + 2.7)
=4.(243 +14)
= 4.257=1028
b) 24 - 5 . 23 - 2 . 33
= 2^3 (2-5)-2. 27
=8. (-3) - 54
= -24 -54
=-78
c) 36 . 57 - 36 . 47 + 63 . 62 - 63 . 52
=36 (57-47) + 63 (62-62)
=36.10+63.10
=10.(36+63)
=10.99
=990
d) 24 . 5 - { 140 - [ 868 - 12 ( 3087 : 72 + 40 ) ] }
=16 . 5 - { 140 - [ 868 - 12 ( 3087 : 49 + 1) ] }
=80 - { 140 - [ 868 - 12 ( 63+ 1) ] }
=80 - { 140 - [ 868 - 12.64 ] }
=80 - { 140 - [ 868 - 768 ] }
=80 - { 140 - 100}
=80 - 40 =40
e) 723 . 542
____________
108
(23. 32 ) 3 .(2.33)2
___________________
22.33
bạn xem có đúng ko nha
29. 36 . 22.36
___________________
22.33
211. 312
___________
22.33
= 28. 39
=256.19683
=5038848
f) 310 . 11 + 310 . 5
__________________
39 . 24
310 . (11 + 5)
__________________
39 . 24
310 . 16
_______
39 . 24
310 . 24
_______
39 . 24
=3
\(a)2x^2-98=0\)
\(2x^2=0+98\)
\(2x^2=98\)
\(x^2=98:2\)
\(x^2=49\)
\(\rightarrow x^2=7^2\)
\(\rightarrow x=7\)
Vậy x = 7
Đặt \(A=5+5^3+5^5+....+5^{47}+5^{49}\)
\(\Rightarrow5^2A=5^3+5^5+5^7+.....+5^{49}+5^{51}\)
\(\Rightarrow5^2A-A=\left(5^3+5^5+5^7+....+5^{49}+5^{51}\right)-\left(3+3^3+3^5+....+5^{47}+5^{49}\right)\)
\(\Rightarrow24A=5^{51}-5\)
\(\Rightarrow A=\dfrac{5^{51}-5}{24}\)
Vậy ............................................................
1)a) \(\left(3x-7\right)^5=32\Rightarrow\left(3x-7\right)^5=2^5\)
\(\Rightarrow3x-7=2\Rightarrow3x=9\Rightarrow x=3\)
Vậy \(x=3\)
b) \(\left(4x-1\right)^3=-27.125\)
\(\Rightarrow\left(4x-1\right)^3=-3^3.5^3=-15^3\)
\(\Rightarrow4x-1=-15\Rightarrow4x=-14\Rightarrow x=-3,5\)
Vậy \(x=-3,5\)
c) \(3^{4x+4}=81^{x+3}\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\)
\(\Rightarrow4x=4x+8\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)^7=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^7-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^7.\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^7=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=-1\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)