nhanh giúp mình với ạ mình cảm ơn
bài 2:
a: Ta có: \(\left(x+5\right)^3=-64\)
\(\Leftrightarrow x+5=-4\)
hay x=-9
b: Ta có: \(\left(2x-3\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
Bài 1:
a.
\((2^3)^4-(2^6)^2=2^{3.4}-2^{6.2}=2^{12}-2^{12}=0\)
b.
\(32^5:8^3=(2^5)^5:(2^3)^3=2^{25}:2^9=2^{25-9}=2^{16}\)
c.
\(81^4.27^6=(3^4)^4.(3^3)^6=3^{16}.3^{18}=3^{16+18}=3^{34}\)
d.
\(x^{12}:(x^3)^4=x^{12}: x^{12}=1\)
Bài 2:
a, b: Đã có người làm
c.
\((\frac{1}{2})^{2x-1}=\frac{1}{8}=(\frac{1}{2})^3\)
\(\Leftrightarrow 2x-1=3\Leftrightarrow x=2\)
d.
\((\frac{-1}{3})^{x-3}=\frac{1}{81}=(\frac{-1}{3})^4\)
$\Leftrightarrow x-3=4$
$\Leftrightarrow x=7$
e.
\((x-3)^{10}=(x-3)^{30}\)
\(\Leftrightarrow (x-3)^{10}[(x-3)^{20}-1]=0\)
\(\Rightarrow \left[\begin{matrix} (x-3)^{10}=0\\ (x-3)^{20}=1\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x-3=0\\ x-3=\pm 1\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=3\\ x=4\\ x=2\end{matrix}\right.\)
f.
Vì \((x+1,5)^8\geq 0, \forall x\in\mathbb{R}\); \((2,7-y)^{12}\geq 0, \forall y\in\mathbb{R}\)
Do đó để tổng của chúng $=0$ thì:
$(x+1,5)^8=(2,7-y)^{12}=0$
$\Leftrightarrow x=-1,5; y=2,7$
Bài 3:
a.
\(A=\frac{2^{13}.3^7}{2^{15}.3^2.(3^2)^2}=\frac{2^{13}.3^7}{2^{15}.3^6}=\frac{3}{2^2}=\frac{3}{4}\)
b.
\(B=\frac{1+3^4+3^8+3^{12}}{(1+3^4+3^8+3^{12})+(3^2+3^6+3^{10}+3^{14})}\)
\(=\frac{1+3^4+3^8+3^{12}}{(1+3^4+3^8+3^{12})(1+3^2)}=\frac{1}{3^2+1}=\frac{1}{10}\)
c.
\(C=27.\frac{-2^5}{3^5}.\frac{5^4}{2^4}.\frac{2}{5^3}=\frac{-2^6.3^3.5^4}{2^4.3^5.5^3}=-\frac{2^2.5}{3^2}=\frac{-20}{9}\)
\(D=\frac{8^5.5^8-2^5.10^9}{2^{16}.5^7+20^8}=\frac{(2^3)^5.5^8-2^5(2.5)^9}{2^{16}.5^7+(2^2.5)^8}\)
\(=\frac{2^{15}.5^8-2^{14}.5^9}{2^{16}.5^7+2^{16}.5^8}=\frac{2^{14}.5^8(2-5)}{2^{16}.5^7(1+5)}\)
\(=\frac{-2^{14}.3.5^8}{2^{17}.3.5^7}=\frac{-5}{2^3}=\frac{-5}{8}\)
