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NN
25 tháng 8 2017
\(a,\left(7x-11\right)^3=2^5.5^2+200.\)
\(\left(7x+11\right)^3=32.25+200.\)
\(\left(7x+11\right)^3=800+200.\)
\(\left(7x-11\right)^3=1000.\)
\(\left(7x-11\right)^3=10^3.\)
\(\Rightarrow7x-11=10.\)
\(\Rightarrow x=\left(10+11\right):3=7\in Z.\)
Vậy.....
\(b,3^x+25=26.2^2+2.3^0.\)
\(3^x+25=26.4+2.\)
\(3^x+25=104+2.\)
\(3^x+25=106.\)
\(3^x=106-25.\)
\(3^x=81.\)
\(3^x=3^4\Rightarrow x=4\in Z.\)
Vậy.....
\(c,2^x+3.2=64.\)(có vấn đề).
\(d,5^{x+1}+5^x=750.\)
\(5^x.5^1+5^x+1=750.\)
\(5^x\left(5^1+1\right)=750.\)
\(5^x\left(5+1\right)=750.\)
\(5^x.6=750.\)
\(5^x=750:6.\)
\(5^x=125.\)
\(5^x=5^3\Rightarrow x=3\in Z.\)
Vậy.....
\(e,x^{15}=x.\)
\(\Rightarrow x\left(x^{14}-1\right)=0\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right..\)
\(f,\left(x-5\right)^4=\left(x-5\right)^6.\)
\(\Leftrightarrow\left(x-5\right)^4-\left(x-5^6\right)=0.\)
\(\Leftrightarrow\left(x-5\right)^4\left[1-\left(x-5\right)^2\right]=0.\)
\(\Leftrightarrow\left(x-5\right)^4\left(1-x+5\right)\left(1+x-5\right)=0.\)
\(\Leftrightarrow\left(x-5\right)^4\left(6-x\right)\left(x-4\right)=0.\)
\(\Leftrightarrow\left(x-5\right)^4=0\Rightarrow x-5=0\Rightarrow x=5\in Z.\)
\(6-x=0\Rightarrow x=6\in Z.\)
\(x-4=0\Rightarrow x=4\in Z.\)
Vậy.....
N
3 tháng 10 2018
\(x^{2018}-x^{18}=0\)
\(x^{18}.\left(x^{2018}-1\right)=0\)
\(=>\orbr{\begin{cases}x^{18}=0\\x^{2018}-1=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
NA
3 tháng 10 2018
b) 275 > 81x
<=> 315 > 34x
<=> 15 > 4x
<=> x < 15 /4
c) 1252+x > 258
<=> 53(2+x) > 516
<=> 3(2+x) > 16
<=> 6 + 3x > 16
<=> 3x > 10
<=> x > 10/3
d) 5x . 5x+1 . 5x+2 <= 100...0 ( 18 số 0 ) : 218
<=> 5x+x+1+x+2 <= 1018 : 218
<=> 53x+3 <= 518
<=> 3x+3 <= 18
<=> 3x <= 15
<=> x <= 5
( <= là bé hơn hoặc bằng )
ND
1
TS
30 tháng 9 2017
a) 127 : 67 =27.67:67
=27
b)275:813=(33)5:(34)3
=315: 312
=33
c)183:93=23.93:93
=23
d)1253:254=(53)3:(52)4
=59:58
=5
LH
9 tháng 10 2017
127:67=(12:6)7=27
275:813=(33)5:(34)3=315:312=33
183;93=(18:9)3=23
1253:254=(53)3:(52)4=59:58=51=5
LG
4 tháng 8 2017
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 ............................................................
LG
4 tháng 8 2017
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 \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
D
22 tháng 9 2017
\(2^5.4=2^5.2^2=2^7\);\(5^4.25=5^4.5^2=5^6\); \(16^3.2^3=\left(2^4\right)^3.2^3=2^{12}.2^3=2^{15}\)
\(625^5:25^7=\left(5^4\right)^5:\left(5^2\right)^7=5^{20}:5^{14}=5^6\); \(25^6:125^3=\left(5^2\right)^6:\left(5^3\right)^3=5^{12}:5^9=5^3\)
\(12^4.3^4=\left(2^2.3\right)^4.3^4=2^8.3^4.3^4=2^8.3^8=6^8\); \(9^6:3^2=\left(3^2\right)^6:3^2=3^{12}:3^2=3^{10}\)
\(2^3.2^4.2=2^8\)
PQ
0
29 tháng 5 2022
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
MS
2
T
2 tháng 10 2018
\(a.2^x=32\Leftrightarrow2^x=2^5\Leftrightarrow x=5\)
\(b.\left(x-5\right)^3=8\Leftrightarrow\left(x-5\right)^3=2^3\Leftrightarrow x-5=2\Leftrightarrow x=7\)
\(c.5^{x+1}=25\Leftrightarrow5^{x+1}=5^2\Leftrightarrow x+1=2\Leftrightarrow x=1\)
\(d.3\cdot\left(6x+3\right)=3^5\)
\(\Leftrightarrow6x+3=3^4\)
\(\Leftrightarrow6x=3^4-3=81-3=78\)
\(\Leftrightarrow x=78:613\)
525 x 5x-1 = 525
5x-1 = 525 : 525
5x+1 = 1
Vì bất kì luỹ thừa nào mũ 0 đều bằng 1
Nên => x = 1 (vì 1-1 = 0 mà 50 = 1 => x=1)
\(5^{25}\times5^{x+1}=5^{25}\)
\(5^{x+1}=5^{25}\div5^{25}\)
\(5^{x+1}=1\)
\(5^x\times5=5^0\)
\(5^x=5^0\div5\)
\(\Rightarrow5^x\in\varnothing\)
\(\Rightarrow x\in\varnothing\)
Vậy \(x\in\varnothing\)