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b: Ta có: 7x-8=713
nên 7x=721
hay x=103
c: Ta có: x-36:18=12
nên x-2=12
hay x=14
d: Ta có: (x-36):18=12
nên x-36=216
hay x=252
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a) \(5x+x=39-3^{11}:3^9\)
\(\Leftrightarrow6x=39-3^2\)
\(\Leftrightarrow6x=30\)
\(\Leftrightarrow x=5\)
b) \(2^x:2^5=16\)
\(\Leftrightarrow2^x:2^5=2^4\)
\(\Leftrightarrow2^x=2^4.2^5\)
\(\Leftrightarrow2^x=2^9\)
\(\Leftrightarrow x=9\)
c) \(7x-x=5^{21}:5^{19}+3.2^2-7^0\)
\(\Leftrightarrow6x=5^2+3.4-1\)
\(\Leftrightarrow6x=36\)
\(\Leftrightarrow x=6\)
d) \(7x-2x=6^{17}:6^{15}+44:11\)
\(\Leftrightarrow5x=6^2+4\)
\(\Leftrightarrow5x=40\)
\(\Leftrightarrow x=8\)
a)⇔6x=39-32
⇔6x=30
⇔ x=5
b)2x:25=16
⇔2x=24.25
⇔ 2x=29
⇔ x=9
c)⇔6x=52+3.22-1
⇔ 6x= 36
⇔ x=6
d)⇔5x=62+4
⇔ 5x=40
⇔ x=8
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a: 3x=81
nên x=27
b: \(5\cdot4^x=80\)
\(\Leftrightarrow4^x=16\)
hay x=2
c: \(2^x=4^5:4^3\)
\(\Leftrightarrow2^x=2^4\)
hay x=4
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a: \(\Leftrightarrow6x=30\)
hay x=5
b: \(\Leftrightarrow6x=25+12-1=36\)
hay x=6
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a: \(\Leftrightarrow8x=108+12=120\)
hay x=15
b: \(\Leftrightarrow6x=60\)
hay x=10
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Câu 1
a) \(48=2^4.3\)
\(60=2^2.3.5\)
\(72=2^3.3^2\)
\(ƯCLN\left(48;60;72\right)=2^2.3=12\)
\(ƯC\left(48;60;72\right)=Ư\left(12\right)=\left\{1;2;3;4;6;12\right\}\)
b) \(42=2.3.7\)
\(55=5.11\)
\(91=7.13\)
\(ƯCLN\left(42;55;91\right)=1\)
\(ƯC\left(42;55;91\right)=\left\{1\right\}\)
c) \(48=2^4.3\)
\(72=2^3.3^2\)
\(ƯCLN\left(48;72\right)=2^3.3=24\)
\(ƯC\left(48;72\right)=Ư\left(24\right)=\left\{1;2;3;4;6;8;12;24\right\}\)
Câu 2:
120 ⋮ \(x\); 168 ⋮ \(x\); 216 ⋮ \(x\);
\(x\) \(\in\) ƯC(120; 168; 216)
120 = 23.3.5; 168 = 23.3.7; 216 = 23.33
ƯClN(120; 168; 216) = 23.3 = 24
\(x\) \(\in\) Ư(24) = {1; 2; 3; 4; 6; 8; 12; 24}
Vì \(x\) > 20 nên \(x\) = 24
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\(180:\left\{\left(-30\right)+\left[4^2.5-\left(14+3^{11}:3^9\right)\right]\right\}\)
\(=180:\left\{\left(-30\right)+\left[16.5-\left(14+3^{11}.\dfrac{1}{3^9}\right)\right]\right\}\)
\(=180:\left\{\left(-30\right)+\left[80-\left(14+3^2\right)\right]\right\}\)
\(=180:\left\{\left(-30\right)+\left[80-\left(14+9\right)\right]\right\}\)
\(=180:\left[\left(-30\right)+80-23\right]\)
\(=180:27\\ =\dfrac{20}{3}\)
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\(4^3\cdot4^{x-1}=64\)
\(\Leftrightarrow4^{x-1}=1\)
\(\Leftrightarrow x-1=0\)
hay x=1
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a)3\(^{11}\):3\(^9\)−147:7\(^2\)
=3\(^{11}\)\(^{-9}\)−147:49
=3\(^2\)−147:49
=9−3
=6
b)295−(31−22×5)w)295-(31-22×5)
=295−(31−2\(^2\).5)\(^2\)
=295−11\(^2\)
=295−121
=174
6x=39-32
6x=39-9
6x=30
x=5