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a: \(\left[\left(10-x\right)\cdot2+51\right]:3-2=3\)
=>\(\left[2\left(10-x\right)+51\right]:3=5\)
=>\(\left[2\left(10-x\right)+51\right]=15\)
=>\(2\left(10-x\right)=15-51=-36\)
=>10-x=-36/2=-18
=>\(x=10-\left(-18\right)=10+18=28\)
b: \(\left(x-12\right)-15=20-\left(17+x\right)\)
=>\(x-12-15=20-17-x\)
=>\(x-27=3-x\)
=>\(2x=30\)
=>\(x=\dfrac{30}{2}=15\)
c: \(720-\left[41-\left(2x-5\right)\right]=2^3\cdot5\)
=>\(720-\left[41-2x+5\right]=8\cdot5=40\)
=>\(\left[41-2x+5\right]=720-40=680\)
=>-2x+46=680
=>-2x=680-46=634
=>\(x=\dfrac{634}{-2}=-317\)
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(2¹⁷ + 15³).(3⁴⁵ - 6⁵).(2⁴ - 4²)
= (2¹⁷ + 15³).(3⁴⁵ - 6⁵).(16 - 16)
= (2¹⁷ + 15³).(3⁴⁵ - 6⁵).0
= 0
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Thế này à:
a, \(\frac{2^{13}+25}{2^{10}}\)
b, \(\frac{21^2.14.125}{35^5.6}\)
c,\(\frac{45^3.20^4.18^2}{180^5}\)
d, \(\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
2 mũ 13 ta chia cho 2 mũ 10 ta có là 2 mũ 3 chuyển thành 8 rồi cộng cho 25 ta có là 33
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2.4^x - 1 = 128
=> 4^x - 1 = 64
=> x - 1 = 3
=> x = 4
b, 17.4^2 + 83.16
= 17.16 + 83.16
= 16(17 + 83)
= 16.100
= 1600
c, a^5 . 10^10.a^15
= a^20.10^10
d, 3{65 - [5(4 + 2.3) - 15] : 7}
= 3{65 - [5(4 + 6) - 15] : 7}
= 3{65 - [5.10 - 15] : 7}
= 3{65 - [50 - 15] : 7}
= 3{65 - 35 : 7}
= 3{65 - 5}
= 3.60
= 180
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\(a,2^4.38-2^4.37=2^4.\left(38-37\right)=2^4.1=16\\ b,4^2.444446-4^3.111111\\ =4^2.\left(444446-4.111111\right)\\ =4^2.2=16.2=32\\ c,\left(2^9.3+2^9.5\right):2^{12}\\ =2^9.\left(3+5\right):2^{12}=2^9.8:2^{12}=2^9.2^3:2^{12}=2^{9+3-12}=2^0=1\\ d,13^2-\left(5^2.4+2^4.15\right)=13^2-5.4.\left(5+4.3\right)\\ =169-20.17\\ =169-340=-171\)
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a) 2^4 * 38 - 2^4 * 37 = 16 * 38 - 16 * 37 = 608 - 592 = 16
b) 4^2 * 444446 - 4^3 * 111111 = 16 * 444446 - 64 * 111111 = 7107136 - 7106944 = 192
c) (2^9 * 3 + 2^9 * 5) / 2^12 = (512 * 3 + 512 * 5) / 4096 = (1536 + 2560) / 4096 = 4096 / 4096 = 1
d) 13^2 - (5^2 * 4 + 2^2 * 15) = 169 - (25 * 4 + 4 * 15) = 169 - (100 + 60) = 169 - 160 = 9
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Bài 1:
a) Ta có: \(\left(2x-1\right)^{20}=\left(2x-1\right)^{18}\)
\(\Leftrightarrow\left(2x-1\right)^{20}-\left(2x-1\right)^{18}=0\)
\(\Leftrightarrow\left(2x-1\right)^{18}\left[\left(2x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-1\right)^{18}\cdot\left(2x-2\right)\cdot2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
b) Ta có: \(\left(2x-3\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
c) Ta có: \(\left(x-5\right)^2=\left(1-3x\right)^2\)
\(\Leftrightarrow\left(x-5\right)^2-\left(3x-1\right)^2=0\)
\(\Leftrightarrow\left(x-5-3x+1\right)\left(x-5+3x-1\right)=0\)
\(\Leftrightarrow\left(-2x-4\right)\left(4x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)
Bài 2:
a) \(15^{20}-15^{19}=15^{19}\left(15-1\right)=15^{19}\cdot14⋮14\)
b) \(3^{20}+3^{21}+3^{22}=3^{20}\left(1+3+3^2\right)=3^{20}\cdot13⋮13\)
c) \(3+3^2+3^3+...+3^{2007}\)
\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)
\(=13\left(3+...+3^{2005}\right)⋮13\)
( 2x - 3 ) . 15 = 5\(^{^2}\). 3
( 2x - 3 ) . 15 = 25 . 3
( 2x - 3 ) . 15 = 75
( 2x - 3 ) = 75 : 15
( 2x - 3 ) = 5
2x = 5 + 3
2x = 8
x = 8 : 2
x = 4
\(\left(2x-3\right)\cdot15=5^2\cdot3\)
\(\left(2x-3\right)\cdot15=25\cdot3\)
\(2x-3=75:15\)
\(2x-3=5\)
\(2x=3+5=8\)
\(x=8:2=4\)