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Bài 2
\(a,\left(x-3\right)^2=9\Leftrightarrow\left(x-3\right)^2=3^2\Leftrightarrow x-3=3\Leftrightarrow x=6\)
\(b,\left(\frac{1}{2}+x\right)^2=16\Leftrightarrow\left(\frac{1}{2}+x\right)^2=4^2\Leftrightarrow\frac{1}{2}+x=4\Leftrightarrow x=\frac{7}{2}\)
a) \(\left(2x+3\right)^2=\frac{9}{144}\)
\(\Leftrightarrow\left(2x+3\right)^2=\left(\frac{1}{4}\right)^2=\left(-\frac{1}{4}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x+3=\frac{1}{4}\\2x+3=\frac{-1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{-11}{4}\\2x=\frac{-13}{4}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-11}{8}\\x=\frac{-13}{8}\end{cases}}}\)
Vậy ...
b) Ta có: \(\left(3x-1\right)^3=\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)
\(\Leftrightarrow3x-1=\frac{-2}{3}\Leftrightarrow3x=\frac{1}{3}\Leftrightarrow x=\frac{1}{9}\)
Vậy ....
c) \(x^{10}=25x^8\Leftrightarrow x^{10}:x^8=25\Leftrightarrow x^2=25\Leftrightarrow x=\left\{5;-5\right\}\)
Vậy ...
d) \(\frac{x^7}{81}=27\Leftrightarrow x^7=27.81=2187\)
Mà 37 = 2187 => x7 = 37 => x = 3
Vậy ....
e) \(\frac{x^8}{9}=729\Leftrightarrow x^8=729.9=6561\)
Mà 38 = (-3)8 = 6561
=> x8 = 38 = (-3)8
=> x = {-3;3}
Vậy ...
a) \(\frac{3^{17}.81^{11}}{27^{10}.9^{15}}=\frac{3^{17}.\left(3^4\right)^{11}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\frac{3^{17}.3^{44}}{3^{30}.3^{30}}=\frac{3^{61}}{3^{60}}=3\)
b) \(\frac{9^2.2^{11}}{16^2.6^3}=\frac{\left(3^2\right)^2.2^{11}}{\left(2^4\right)^2.\left(2.3\right)^3}=\frac{3^4.2^{11}}{2^8.2^3.3^3}=3\)
c) \(\frac{2^{10}.3^{31}+2^{40}.3^6}{2^{11}.3^{31}+2^{41}.3^6}=\frac{2^{10}.3^6.\left(3^{25}+2^{30}\right)}{2^{11}.3^6.\left(3^{25}+2^{30}\right)}=\frac{1}{2}\)
d) \(a.\left(-b\right).\left(-a\right)^2\left(-b\right)^3.\left(-a\right)^3.\left(-b\right)^4=-a^6b^8\)
a: \(=3^2\cdot3^3\cdot3^{-4}\cdot3^2=3^{2+3-4+2}=3^3\)
b: \(=2^2\cdot2^5:\left(2^3\cdot\dfrac{1}{2^4}\right)=2^7:\dfrac{1}{2}=2^8\)
c: \(=9\cdot32\cdot\dfrac{4}{9}=128=2^7\)
d: \(=\dfrac{1}{27}\cdot3^4=3^1\)
a)
\((3x-7)^5=0\Rightarrow 3x-7=0\Rightarrow x=\frac{7}{3}\)
b)
\(\frac{1}{4}-(2x-1)^2=0\)
\(\Leftrightarrow (2x-1)^2=\frac{1}{4}=(\frac{1}{2})^2=(-\frac{1}{2})^2\)
\(\Rightarrow \left[\begin{matrix} 2x-1=\frac{1}{2}\\ 2x-1=\frac{-1}{2}\end{matrix}\right.\Rightarrow \Rightarrow \left[\begin{matrix} x=\frac{3}{4}\\ x=\frac{1}{4}\end{matrix}\right.\)
c)
\(\frac{1}{16}-(5-x)^3=\frac{31}{64}\)
\(\Leftrightarrow (5-x)^3=\frac{1}{16}-\frac{31}{64}=\frac{-27}{64}=(\frac{-3}{4})^3\)
\(\Leftrightarrow 5-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{23}{4}\)
d)
\(2x=(3,8)^3:(-3,8)^2=(3,8)^3:(3,8)^2=3,8\)
\(\Rightarrow x=3,8:2=1,9\)
e)
\((\frac{27}{64})^9.x=(\frac{-3}{4})^{32}\)
\(\Leftrightarrow [(\frac{3}{4})^3]^9.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow (\frac{3}{4})^{27}.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow x=(\frac{3}{4})^{32}:(\frac{3}{4})^{27}=(\frac{3}{4})^5\)
f)
\(5^{(x+5)(x^2-4)}=1\)
\(\Leftrightarrow (x+5)(x^2-4)=0\)
\(\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2=4=2^2=(-2)^2\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=-5\\ x=\pm 2\end{matrix}\right.\)
g)
\((x-2,5)^2=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2\)
\(\Rightarrow \left[\begin{matrix} x-2,5=\frac{2}{3}\\ x-2,5=\frac{-2}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{19}{6}\\ x=\frac{11}{6}\end{matrix}\right.\)
h)
\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)
\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)
a) \(\frac{1}{81}\times\left(\frac{1}{3}\right)^{-2}\times9\times3^3\)
\(=\frac{3^7}{3^4}\)
\(=3^3\)
b) \(\left(2^5\times4\right)\div\left(2^3\times\frac{1}{16}\right)\)
\(=2^7\div\frac{2^3}{2^{\text{4}}}\)
\(=2^7\div\frac{1}{2}\)
=\(2^6\)