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a) \(\left(3x-2^4\right).7^3=2.7^4\)\(\Leftrightarrow3x-2^4=2.7^4:7^3\)
\(\Leftrightarrow3x-16=2.7\)\(\Leftrightarrow3x-16=14\)\(\Leftrightarrow3x=30\)
\(\Leftrightarrow x=10\)
Vậy \(x=10\)
b) \(3x+4x=\left|-75\right|+23\)\(\Leftrightarrow7x=75+23\)
\(\Leftrightarrow7x=98\)\(\Leftrightarrow x=14\)
Vậy \(x=14\)
a) \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)
=> \(3x\cdot7^3-2^4\cdot7^3=2\cdot7\cdot7^3\)
=> \(3x\cdot7^3=14\cdot7^3+16\cdot7^3\)
=> \(3x\cdot7^3=\left(14+16\right)\cdot7^3\)
=> \(3x\cdot7^3=30\cdot7^3\)
=> \(3x=30\)(bỏ hai vế 73)
=> \(x=10\)
Vậy x = 10
b) \(3x+4x=\left|-75\right|+23\)
=> \(7x=75+23\)
=> \(7x=98\)
=> \(x=14\)
Vậy x = 14
Bạn chú ý đến thừa số cuối cùng
24=16
42=16
Do đó 24-42=0
Vậy cả tích bằng 0
3.42+(57:56)-(2.24)
=3.42+57-6-24+1
=3.42+51-25
=(3.42)+5-32
=48+5-32
=53-32
=21
\(1+2+2^2+2^3+2^4+...+2^{22}+2^{23}\Leftrightarrow\left(1+2\right)+2^2\left(1+2\right)+...+2^{22}\left(1+2\right)\)
\(\Rightarrow3+2^2\cdot3+...2^{22}\cdot3\Leftrightarrow3\cdot\left(2^0+2^1+...+2^{22}\right)⋮3\left(đpcm\right)\)
\(\Rightarrow3\cdot\frac{\left(2^0+2^1+...+2^{22}\right)}{7}\Leftrightarrow3\cdot7\left(2^0+2^1+2^2\right)⋮3,7\left(đpcm\right)\)
(-7)2+(-49).[ -15+(-7)4:73 ]+(-1)2014
=49+(-49).[-15+ 7]+1
=49+(-49).(-8)+1
=49+392+1
=441+1
=442
A = 12 : {390 : [500 - (125 + 35.7)]}
A = 12 : {390 : [500 - (125 + 245)]}
A = 12 : {390 : [500 - 370]}
A = 12 : {390 : 130}
A = 12 : 3 = 4
B = 12000 - (1500 . 2 + 1800.3 + 1800.2 : 3)
B = 12000 - (3000 + 5400 + 1200)
B = 12000 - 9600 = 2400
C = \(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\frac{11\cdot3^{22}\cdot3^7-\left(3^2\right)^{15}}{2^2\cdot3^{28}}=\frac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\frac{11\cdot3\cdot3^{28}-3^2\cdot3^{28}}{2^2\cdot3^{28}}\)
\(=\frac{33\cdot3^{28}-9\cdot3^{28}}{4\cdot3^{28}}=\frac{\left(33-9\right)\cdot3^{28}}{4\cdot3^{28}}=\frac{24}{4}=6\)
a)\(S=1+5+5^2+...+5^{10}\)
\(5S=5+5^2+5^3+...+5^{11}\)
\(5S-S\)hay 4S\(=5^{11}-1\)
\(\Rightarrow S=\left(5^{11}-1\right):4\)
b)\(S=1+7+7^2+...+7^{10}\)
\(7S=7+7^2+7^3+...+7^{11}\)
\(7S-S\)hay 6S\(=7^{11}-1\)
\(\Rightarrow S=\left(7^{11}-1\right):6\)
Học tốt nha!!!
\(7I=7+7^2+7^3+7^4+...+7^{102}\)
\(6I=7I-I=7^{102}-1\Rightarrow I=\frac{7^{102}-1}{6}\)