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Tìm x\(\in\)n
a,x3-23=25-(316:314+28:216)
b,5x-2-32=24-(68:66-62)
c,(x2-1)4=81
d,3x+42=196:(193.192)-3.1
a,3+2x-1=24-[42-(22-2)]
=>3+2x-1=24-(16-2)
=>3+2x-1=24-14
=>3+2x-1=10
=>2x-1=10-3
=>2x-1=7
=>sai đề
a) x3 - 23 = 25 - ( 316 : 314 + 28 : 2 6 )
x3 - 23 = 25 -( 32 + 22 )
x3 - 23 = 25- ( 9 + 4 )
x3 - 23= 25 - 13
x3 - 8 = 32 - 13
x3 - 8 = 19
x3 = 19 + 8
x3 = 27
x3 = 33
=> x = 3
Vậy x =3
b) ( x2 - 1)4 = 81
( x2 - 1 )4 = 34
=> x2 - 1 = 3
x2 = 3 + 1
x2 = 4
x2 = 22
=> x = 2
Vậy x= 2
c) 5x-2 - 32 = 24 - (68 : 66 - 62 )
5x-2 - 32 = 16 - ( 62 - 62)
5x-2 - 9 = 16 - 0
5x-2 - 9= 16
5x - 2 = 16 + 9
5x - 2 = 25
5x - 2 = 52
=> x - 2 = 2
x = 2 + 2
x = 4
Vậy x = 4
d) 3x + 42 = 196 : ( 193 . 192 ) 3.12005
3x + 16 = 196 : 195 . 3 . 1
3x + 16 = 19 . 3
3x + 16 = 57
3x = 57 - 16
3x = 41
Hình như phép này đề bài sai rồi thì phải
a) 27^16 : 9^10
Ta có: (3.9)^16 : 9^10
= 3^16.9^16: 9^10
= 3^16. 9^6
= 3^16.(3^2)^6
=3^16.3^12
=3^28
1) \(2^{x+1}\cdot2^{2014}=2^{2015}\)\(\Leftrightarrow2^{2014x+2014}=2^{2015}\)\(\Leftrightarrow2014x+2014=2015\)\(\Leftrightarrow x=\frac{1}{2014}\)
2) \(7x-2x=\frac{6^{17}}{6^{15}}+\frac{44}{11}\)\(\Leftrightarrow5x=6^2+4=36+4=40\)\(\Leftrightarrow x=\frac{40}{5}=8\)
3) \(3^x=9\)\(\Leftrightarrow3^x=3^2\)\(\Leftrightarrow x=2\)
4) \(7x-x=\frac{5^{21}}{5^{19}}+3\cdot2^2-7^0\)\(\Leftrightarrow6x=5^2+3\cdot4-1=25+12-1=36\)\(\Leftrightarrow x=6\)
5) \(4^x=64\)\(\Leftrightarrow4^x=4^3\)\(\Leftrightarrow x=3\)
6) \(9^{x-1}=9\)\(\Leftrightarrow x-1=1\)\(\Leftrightarrow x=0\)
7) \(\frac{2^x}{2^5}=1\)\(\Leftrightarrow2^{x-5}=2^0\)\(\Leftrightarrow x-5=0\)\(\Leftrightarrow x=5\)
8) \(\left(5x-9\right)^3=216\)\(\Leftrightarrow\left(5x-9\right)^3=6^3\)\(\Leftrightarrow5x-9=6\)\(\Leftrightarrow5x=15\)\(\Leftrightarrow x=3\)
9) \(5\cdot3^{7x-11}=135\)\(\Leftrightarrow5.3^{7x-11}=5.3^3\)\(\Leftrightarrow3^{7x-11}=3^3\)\(\Leftrightarrow7x-11=3\)\(\Leftrightarrow7x=14\Leftrightarrow x=2\)
10) \(2.3^x=19\cdot3^8-81^2\)\(\Leftrightarrow2.3^x=19\cdot3^8-3^8=18.3^8=2.3^{11}\)\(\Leftrightarrow3^x=3^{11}\Leftrightarrow x=11\)
Đây là cách làm của mình. Bạn có thể chỉnh sửa tuỳ ý theo cách làm của bạn nhé ^^
Học tốt ^3^
Bài 1 :
a/ \(a^3.a^9=a^{3+9}=a^{12}\)
b/\(\left(a^5\right)^7=a^{5.7}=a^{35}\)
c/ \(\left(a^6\right).4.a^{12}=a^{24}.a^{12}.4=a^{24+12}.4=a^{36}.4\)
d/ \(\left(2^3\right)^5.\left(2^3\right)^3=2^{15}.2^9=2^{15+9}=2^{24}\)
e/ \(5^6:5^3+3^3.3^2\)
\(=5^3+3^5=125+243=368\)
i/ \(4.5^2-2.3^2\)
\(=2^2.5^2-2.3^2\)
\(=2^2.25-2^2.14\)
\(=2^2.\left(25-14\right)\)
\(=2^2.11\)
\(=4.11=44\)
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
3 phần trên đễ quá mik ko làm mik chỉ làm phàn 4 thôi nhé
4) ta có: (x-3)^x+2=(x-3)^x+6
=>(x-3)^x*(x-3)^2=(x-3)^x*(x-3)^6
=>(x-3)^x=(x-3)^x*(x-3)^4
=>(x-3)^x*(x-3)^4-(x-3)^x*1=0
=>(x-3)^x*((x-3)^4-1)=0
=>(x-3)^x=0 hoặc (x-3)^4-1=0
còn lại cậu tự làm nha nó đẽ mà
1: =>\(5^{x-2}-9=2^4-\left(6^2-6^2\right)\)
=>\(5^{x-2}=16+9=25\)
=>x-2=2
=>x=4
2: \(\Leftrightarrow3^x+16=19^6:19^5-3=19-3=16\)
=>3^x=0
=>x=0
3: \(\Leftrightarrow2^x+2^x\cdot16=272\)
=>2^x*17=272
=>2^x=16
=>x=4
4: \(\Leftrightarrow2^{x-1}+3=24-\left(4^2-2^2+1\right)=24-\left(16-4+1\right)\)
=>\(2^{x-1}+3=24-16+4-1=8+4-1=12-1=11\)
=>2^x-1=8
=>x-1=3
=>x=4