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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
=\(\frac{1}{25}-1=\frac{-24}{25}\)
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
a,71.2-6.(2x+5)=10^5:10^3
142-6.(2x+5)=10^2
142-6.(2x+5)=100
6.(2x+5)=142-100
6.(2x+5)=42
2x+5=42:6
2x+5=7
2x=7-5
2x=2
x=1
Vậy x=1
a)5^8
b)x^5
c)6^4
d)6
\(5^3\cdot5^5=5^8\)
\(x^4\cdot x=x^5\)
\(6^5:6^2=6^3\)
\(36^3:6^5=6^6:6^5=6\)