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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
T
24 tháng 7 2018
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/ \(4.5^2-2.3^2\)
\(=2^2.5^2-2.3^2\)
\(=2^2.25-2^2.14\)
\(=2^2.\left(25-14\right)\)
\(=2^2.11\)
\(=4.11=44\)
XS
9 tháng 7 2018
Bài 1 tự làm!
Bài 2:
a, \(\left(3x-4\right)\left(x-1\right)^3=0\Rightarrow\left[{}\begin{matrix}3x-4=0\\\left(x-1\right)^3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=1\end{matrix}\right.\)
b, \(2^{2x-1}:4=8^3\Rightarrow2^{2x-1}:2^2=2^9\)
\(\Rightarrow2x-1-2=9\Rightarrow2x-3=9\Rightarrow2x-12\Rightarrow x=6\)
c, Đề chưa rõ
d, \(\left(x+2\right)^5=2^{10}\Rightarrow\left(x+2\right)^5=4^5\Rightarrow x+2=4\Rightarrow x=2\)
e, \(\left(3x-2^4\right).7^3=2.7^4\Rightarrow3x-2^4=2.7^4:7^3\Rightarrow3x-16=2.7=14\)
\(\Rightarrow3x=14+16=30\Rightarrow x=\dfrac{30}{3}=10\)
f, \(\left(x+1\right)^2=\left(x+1\right)^0\Rightarrow\left(x+1\right)^2=1\) (vì x0 = 1)
\(\Rightarrow x+1=1\Rightarrow x=0\)
E
30 tháng 9 2015
a)(x-6)2 . (x-6) =64
(x-6)3=64
(x-6)3=43
=>x-6=3
x=9
b)(x+5)3 : (x+5) =144
(x+5)2=144
(x+5)2=122
=>x+5\(\in\left\{12;-12\right\}\)
\(\Rightarrow x\in\left\{7;-17\right\}\)
c)3x+42=196 : (194 . 19)-2.12036
3x+42=196 :195-2
3x+42=19-2
3x+42=17
3x+16=17
3x=1
=>x=0
d)(19x + 2.52) =52 - 42
(19x + 2.52) =25-16
(19x + 2.52) =9
19x + 50 =9
19x = -41
x=\(\frac{-41}{19}\)
tính hợp lí(nếu có thể)
(52004 - 52006):(52005.5)
=(52004 - 52006):52006
=(52004 - 52006).\(\frac{1}{5^{2006}}\)
=\(\frac{5^{2004}}{5^{2006}}\) - 1
=\(\frac{1}{5^2}\) -1
=\(\frac{1}{25}-1=\frac{-24}{25}\)
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.....
NN
2
5 tháng 10 2017
ý a)\(\left(x-5\right)^4=\)\(\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^4=\)\(\left(x-5\right)^4\left(x-5\right)\)
\(\Rightarrow1=x-5\)
\(\Rightarrow x=4\)
ý b)\(x^{2006}=x^2\)
vì 2006 ; 2 là số chẵn
\(\Rightarrow x=1;-1\)
27 tháng 3 2018
a)
\(\dfrac{7}{5}+\dfrac{5}{6}:5-\dfrac{3}{8}\cdot\left(-3\right)\\ =\dfrac{7}{5}+\dfrac{1}{6}+\dfrac{9}{8}\\ =\dfrac{168+20+135}{120}\\ =\dfrac{323}{120}\)
a)Ta có:04=06
14=16
(-1)4=(-1)6
=>x-5 E {-1;0;1}
=>x E {4;5;6}
b)Ta có: 02006=02
12006=12
(-1)2006=(-1)2
=>x E {-1;'0;1}
a) (x-5)4 = (x-5)6
=>(x-5)4 - (x-5)6=0
=>(x-5)4[1-(x-5)2]=0
=>(x-5)4[1-x2+10x-25]=0
=>(x-5)4[6x-24+4x-x2]=0
=>(x-5)4[6(x-4)-x(x-4)]=0
=>(x-5)4(6-x)(x-4)=0
=>(x-5)4=0 hoặc 6-x=0 hoặc x-=0
=>x=5 hoặc x=6 hoặc x=4
b)x2006=x2
=>x2006-x2=0
=>x2(x2004-1)=0
=>x2=0 hoặc x2004-1=0
=>x=0 hoặc x=1 hoặc -1