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\(\left|5\left(2x+3\right)\right|+\left|2\left(2x+3\right)\right|+\left|2x+3\right|=16\)
\(=8\left(2x+3\right)=16\)
\(\Rightarrow2x+3=2\)
\(\Rightarrow x=-\frac{1}{2}\)
a) 9 . 33 . 1/81 . 32
=32.33.1/34.32
=33
b) 4 . 25 : (23 . 1/16)
=22.25:(23.1/24)
=22.25:1/2
=22.24
=22+4
=26
c) 32 . 25 . (2/3)2
=32.25.22/32
=25.22
=25+2
=27
d) (1/3)2 . 1/3 . 92
=1/32.1/3.(32)2
=1/32.1/3.34
=1/32.33
=3
=31
a)
\(\Rightarrow3^x\left(3^2+3+1\right)=117\)
\(\Rightarrow3^x.13=117\)
\(\Rightarrow3^x=9\)
\(\Rightarrow3^x=3^2\)
=>x=2
b)
\(3^{2x+1}=3^{-4}\)
=> 2x+1= - 4
=>\(x=-\frac{5}{2}\)
c)
\(\left(x+2\right)^4=16\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x+2\right)^4=2^4\\\left(x+2\right)^4=\left(-2\right)^4\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+2=2\\x+2=-2\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\-4\end{array}\right.\)
Câu 1:
a) 2225 và 3150
Ta có:2225=(29)25=51225
3150=(36)25=72925
Vì 51225<72925
Suy ra: 2225<3150
Câu 2:
a)\(25^3:5^2=\left(5^2\right)^3:5^2=5^6:5^2=5^4\)
b)\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
c)\(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3+\frac{1}{4}:2=3+\frac{1}{8}=\frac{25}{8}\)
Câu 3:
a)\(9.3^3.\frac{1}{81}.3^2=3^2.3^3.3^2.\left(\frac{1}{3^4}\right)=3^7:3^4=3^3\)
b)\(4.2^5:\left(2^3.\frac{1}{16}\right)=2^2.2^5:\left(2^3.\frac{1}{2^4}\right)=2^7:\frac{1}{2}=2^8\)
c)\(3^2.2^5.\left(\frac{2}{3}\right)^2=288.\frac{4}{9}=2^7\)
d)\(\left(\frac{1}{3}\right)^3.\frac{1}{3}.9^2=\left(\frac{1}{3}\right)^4.\left(3^2\right)^2=3^4.\left(\frac{1}{3}\right)^4=3^4:3^4=1\)
3x*3x-2=81
\(\Rightarrow3^{x+x-2}=81\)
\(\Rightarrow3^{2x-2}=3^4\)
\(\Rightarrow2x-2=4\)
\(\Rightarrow2x=6\Rightarrow x=3\)
\(3^x.3^{x-2}=81\)
\(\Rightarrow3^x.3^x:3^2=81\)
\(\Rightarrow3^{2x}=729\)
\(\Rightarrow3^{2x}=3^6\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
Vậy x = 3