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a) \(\frac{2^5\cdot2^{12}\cdot2^6}{2^{24}}=\frac{2^{23}}{2^{24}}=\frac{1}{2}\)
Các phần kia tương tự, à bạn đăng 1 2 câu hỏi 1 lần thôi, đăng nhiều quá ko ai trả lời đâu
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Thế à ! Vậy bạn hãy nhấp vào https://h.vn/hoi-dap/question/646555.html?pos=1792187 mà xem
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
d) \(\dfrac{8^4.3^6}{2^7.65}=\dfrac{\left(2^3\right)^4.3^6}{2^7.65}=\dfrac{2^{12}.3^6}{2^7.65}=\dfrac{2^7.2^5.3^6}{2^7.65}=\dfrac{2^5.3^6}{65}=\dfrac{23328}{65}\)
c) \(\left(\dfrac{3}{5}-\dfrac{3}{4}\right).\left(\dfrac{2}{6}-\dfrac{1}{5}\right)^2=\dfrac{3.4-3.5}{4.5}.\left(\dfrac{2.5-1.6}{6.5}\right)^2\\ =\dfrac{-3}{20}.\left(\dfrac{2}{15}\right)^2=\dfrac{-3}{20}.\dfrac{4}{225}\\ =\dfrac{-3}{4.5}.\dfrac{4}{75.3}=\dfrac{-1}{375}\)
a) \(\left(\frac{3}{7}+\frac{1}{2}\right)^2=\left(\frac{6}{14}+\frac{7}{14}\right)^2=\left(\frac{13}{14}\right)^2=\frac{169}{196}\)
b) \(\left(\frac{3}{4}-\frac{5}{6}\right)^2=\left(\frac{9}{12}-\frac{10}{12}\right)^2=\left(-\frac{1}{2}\right)^2=\frac{1}{4}\)
d) \(\left(-\frac{10}{3}\right)^5.\left(-\frac{6}{5}\right)^4=-\frac{2560}{3}\)
e) \(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(\frac{4}{5}-\frac{3}{4}\right)^2=\frac{17}{12}.\left(\frac{1}{20}\right)^2=\frac{17}{12}.\frac{1}{400}=\frac{17}{4800}\)
f) \(2:\left(\frac{1}{2}-\frac{2}{3}\right)^3=2:\left(-\frac{1}{6}\right)^3=2:-\frac{1}{216}=-432\)
\(a.\)
\(\left(\frac{2}{3}\right)^3=\frac{2^3}{3^3}=\frac{8}{27}\)
\(b.\)
\(\left(-2\frac{3}{4}\right)^2=\left(-\frac{11}{4}\right)^2=\frac{121}{16}\)
\(c.\)
\(\left(0,6\right)^4=0,1296\)
\(d.\)
\(\left(-\frac{1}{2}\right)^4=\frac{1^4}{2^4}=\frac{1}{16}\)
\(e.\)
\(\left(-\frac{1}{2}\right)^5=-\frac{1^5}{2^5}=-\frac{1}{32}\)
Câu 1:
a)\(\left(\frac{2}{3}\right)^2=\frac{4}{9}\) b)\(\left(-2\frac{3}{4}\right)^2=\left(-\frac{11}{4}\right)^2=\frac{121}{16}\)
c)\(\left(0,6\right)^4=\left(\frac{3}{5}\right)^4=\frac{81}{625}\) d)\(\left(-\frac{1}{2}\right)^4=\frac{1}{16}\)
e)\(\left(-\frac{1}{5}\right)^5=\frac{-1}{3125}\)
a: =>x-8/5=1/20-1/10=-1/20
=>x=-0,05+1,6=1,55
b: =>x-3/2=4/3 hoặc x-3/2=-4/3
=>x=17/6 hoặc x=1/6
c: =>\(\left|x-\dfrac{1}{3}\right|=\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{2}{3}=\dfrac{35}{12}\)
=>x-1/3=35/12 hoặc x-1/3=-35/12
=>x=39/12=13/4 hoặc x=-31/12
d: =>|x-5/8|=3/4
=>x-5/8=3/4 hoặc x-5/8=-3/4
=>x=11/8 hoặc x=-1/8
Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
\(\Leftrightarrow x:\frac{1}{45}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}:\frac{1}{45}=\frac{45}{2}\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
c) \(\frac{4-3x}{2x+5}=0\Leftrightarrow4-3x=0\)
\(\Leftrightarrow3x=4\Rightarrow x=\frac{4}{3}\)
d) \(\left(x-2\right).\left(x+\frac{2}{3}\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{3}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{3}{2}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
=> \(x:\frac{1}{45}=\frac{1}{2}\)
=> \(x=\frac{1}{2}.\frac{1}{45}\)
=> \(x=\frac{1}{90}\)
Vậy \(x=\frac{1}{90}.\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
=> \(\left\{{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2x=0+1=1\\2x=0-3=-3\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1:2\\x=\left(-3\right):2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{2};-\frac{3}{2}\right\}.\)
Mình chỉ làm được thế thôi nhé, mong bạn thông cảm.
Chúc bạn học tốt!
\(\left(\frac{2}{3}\right)^3=\frac{8}{27}\)
\(\left(-2\frac{3}{4}\right)^2=\frac{121}{16}\)
\(\left(0,6\right)^4=\frac{81}{625}\)
\(\left(-\frac{1}{2}\right)^4=\frac{1}{16}\)
\(\left(-\frac{1}{2}\right)^5=-\frac{1}{32}\)
\(\left(\frac{2}{3}\right)^3=\frac{2^3}{3^3}=\frac{8}{27}\)
\(\left(-2\frac{3}{4}\right)^2=\left(-\frac{11}{4}\right)^2=\frac{121}{16}\)
\(\left(0,6\right)^4=0,1296\)
\(\left(-\frac{1}{2}\right)^4=\frac{1}{16}\)
\(\left(-\frac{1}{2}\right)^5=\frac{1}{32}\)