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\(\frac{4^{20}(2^{20}+1)}{4^{15}(4^{10}+1)}=4^5=1024 \)
Bài 2
Ta có; \((x-2)^{2012}\ge0\)
\(|y^2-9|^{2014}\ge0\)
=>\((x-2)^{2012}+|y^2+9|^{2014}\ge0\)
Mà \((x-2)^{2012}+|y^2+9|^{2014}=0\)
=>x-2=0=>x=2
y^2-9=0=>y=+-3
45^10*5^20/75^15
=5^10*9^10*5^20/(5^2)^15
=5^10*5^20*9^10/5^30
=9^10
(0.8)^5/(0.4)^6
=(0.4)^5*2^5/(0.4)^6
=2^5/(0.4)
=32/(0.4)
=80
2^15*9^4/6^6*8^3
=2^15*(3^2)^4/2^6*3^6*(2^3)^3
=2^15*3^8/2^6*3^6*2^9
=3^2
=9
Tính :
a) \(\frac{8^{14}}{4^{12}}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^{12}}=\frac{2^{42}}{2^{24}}=2^{18}=262144.\)
b) \(\frac{120^3}{40^3}=\left(\frac{120}{40}\right)^3=3^3=27.\)
Tìm x:
b) \(x^2-0,25=0\)
\(\Rightarrow x^2=0+0,25\)
\(\Rightarrow x^2=0,25\)
\(\Rightarrow\left[{}\begin{matrix}x=0,5\\x=-0,5\end{matrix}\right.\)
Vậy \(x\in\left\{0,5;-0,5\right\}.\)
c) \(\frac{8}{2^x}=2\)
\(\Rightarrow2^x=8:2\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
Chúc bạn học tốt!
a, 2 mũ 17 phần 2 mũ 14
b,=30
mình chỉ làm được 2 câu thôi,chúc cậu học tốt!
a.
\(\frac{45^{10}\times5^{20}}{75^{15}}=\frac{\left(3^2\times5\right)^{10}\times5^{20}}{\left(3\times5^2\right)^{15}}=\frac{3^{20}\times5^{10}\times5^{20}}{3^{15}\times5^{30}}=3^5=243\)
b.
\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,8\right)^5}{\left(0,4\right)^5}\times\frac{1}{\left(0,4\right)}=\left(\frac{0,8}{0,4}\right)^5\times\frac{1}{\frac{4}{10}}=2^5\times\frac{5}{2}=2^4\times5=16\times5=80\)
c.
\(\frac{2^{15}\times9^4}{6^6\times8^3}=\frac{2^{15}\times\left(3^2\right)^4}{\left(2\times3\right)^6\times\left(2^3\right)^3}=\frac{2^{15}\times3^8}{2^6\times3^6\times2^9}=3^2=9\)
Chúc bạn học tốt ^^
1. sai dấu nhé
2.a, \(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(3^2.5\right)^{10}.5^{20}}{\left(5^2.3\right)^{15}}=\frac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5=243\)
b, \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(\frac{4}{5}\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\cdot2\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\right)^5\cdot2^5}{\left(\frac{2}{5}\right)^5\cdot\frac{2}{5}}=2^5\div\frac{2}{5}=32\cdot\frac{5}{2}=80\)
c, \(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^2}{2^{15}}=3^2=9\)
a/ 4510 . 520 / 7515 = (32.5)10.520/ (3.52)15 = 320.530/ 315.530 = 35=243
b/ 215. 94 / 66. 83 = 215.38/ 26.36.29 = 32 = 9
a)\(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(3^2.5\right)^{10}.5^{20}}{3^{15}.\left(5^2\right)^{15}}=\frac{3^{20}.5^{30}}{3^{15}.5^{30}}=3^5=243\)
b) \(\frac{2^{15}\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)
Bài 1:
a)
\(\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{(2^3)^{20}+(2^2)^{20}}{(2^2)^{25}+(2^6)^{5}}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}(2^{20}+1)}{2^{30}(2^{20}+1)}=2^{10}\)
b)
\(\frac{45^{10}.5^{20}}{75^{15}}=\frac{(3^2.5)^{10}.5^{20}}{(3.5^2)^{15}}=\frac{3^{20}5^{30}}{3^{15}.5^{30}}=\frac{3^{20}}{3^{15}}=3^5\)
Bài 2:
Ta thấy $(x-2)^{2012}=[(x-2)^{1006}]^2\geq 0$ với mọi $x\in\mathbb{R}$
$|b^2-9|^{2014|\geq 0$ với mọi $b\in\mathbb{R}$ (tính chất trị tuyệt đối)
Do đó để tổng của chúng bằng $0$ thì:
\((x-2)^{2012}=|b^2-9|^{2014}=0\)
\(\Leftrightarrow \left\{\begin{matrix} x-2=0\\ b^2-9=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=2\\ b=\pm 3\end{matrix}\right.\)
Vậy.......