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a, (-0,2)2 \(\times\) 5 - \(\dfrac{2^{13}\times27^3}{4^6\times9^5}\)
= 0,04 \(\times\) 5 - \(\dfrac{2^{13}\times3^9}{2^{12}\times3^{10}}\)
= 0,2 - \(\dfrac{2}{3}\)
= \(\dfrac{2}{10}\) - \(\dfrac{2}{3}\)
= - \(\dfrac{7}{15}\)
b, \(\dfrac{5^6+2^2.25^3+2^3.125^2}{26.5^6}\)
= \(\dfrac{5^6+4.5^6+8.5^6}{26.5^6}\)
= \(\dfrac{5^6.\left(1+4+8\right)}{26.5^6}\)
= \(\dfrac{1}{2}\)
\(\frac{1}{9}\cdot27^x=3^x\)
\(3^x:27^x=\frac{1}{9}\)
\(\left(3:27\right)^x=\frac{1}{9}\)
\(\left(\frac{1}{9}\right)^x=\frac{1}{9}\)
\(\Rightarrow x=1\)
1/2x^5-3/4x^5+x^5y
thay x=1 và y=-1 vào biểu thức trên ,ta có:
1/2.1^5-3/4.1^5+1^5.(-1)
=1/2.1-3/4.1+1.(-1)
=1.[1/2-3/4+(-1)]
=1.[2/4-3/4-1]
=1.[-1/4-1]
=1.(-5/4)
=-5/4
a) Vì \(-45< -16\) nên \(\left(-\dfrac{45}{17}\right)^{15}< \left(\dfrac{-16}{17}\right)^{15}\)
b) Vì \(21< 23\) nên \(\left(-\dfrac{8}{9}\right)^{21}< \left(-\dfrac{8}{9}\right)^{23}\)
c) \(27^{40}=3^{3^{40}}=3^{120}\)
\(64^{60}=8^{2^{60}}=8^{120}\)
Vì \(3< 8\) nên \(3^{120}< 8^{120}\) hay \(27^{40}< 64^{60}\)
con ai kooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
b) (5/2-3x)=25/9
3x = 5/2-25/9
3x =-5/18
x =-5/18:3
x=-5/54
\(e.\left(x-1\right)^5=-32\)
\(\left(x-1\right)^5=\left(-2\right)^5\)
\(x-1=-2\)
\(x\) \(=-2+1\)
\(x\) \(=-1\)
Vậy \(x=-1\)
x13 = 27.x16
=> x13 - 27x16 = 0
=> x13(1 - 27x3) = 0
=> \(\orbr{\begin{cases}x^{13}=0\\1-27x^3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\27x^3=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x^3=\frac{1}{27}\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)
c) \(\left(\frac{1}{2}\right)^{2x-1}=\frac{1}{8}\)
=> \(\left(\frac{1}{2}\right)^{2x-1}=\left(\frac{1}{2}\right)^3\)
=> \(2x-1=3\)
=> \(2x=3+1\)
=> \(2x=4\)
=> \(x=4:2=2\)
\(b,\text{ }x^{13}=27\cdot x^{16}\)
\(x^{16}\text{ : }x^{13}=27\)
\(x^3=3^3\)
\(x=3\)