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Bài 1 :
\(A=\frac{1}{3}-\frac{3}{4}-\frac{\left(-3\right)}{5}+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow A=\frac{3}{9}-\frac{3}{4}+\frac{9}{15}+\frac{1}{72}-\frac{2}{9}-\frac{2}{72}+\frac{1}{15}\)
\(\Rightarrow A=\left(\frac{3}{9}-\frac{2}{9}\right)+\left(\frac{9}{15}+\frac{1}{15}\right)+\left(\frac{1}{72}+\frac{-2}{72}\right)-\frac{3}{4}\)
\(\Rightarrow A=\frac{1}{9}+\frac{2}{3}+\frac{-1}{72}-\frac{3}{4}=\frac{8}{72}+\frac{48}{72}+\frac{-1}{72}-\frac{54}{72}\)
\(\Rightarrow A=\frac{1}{72}\)
Vậy : \(A=\frac{1}{72}\)
Bài 2:
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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.
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Bài 1:
1. \(x:-\left(-\frac{1}{2}\right)=-\frac{1}{2}\)
⇒ \(x:\frac{1}{2}=-\frac{1}{2}\)
⇒ \(x=\left(-\frac{1}{2}\right).\frac{1}{2}\)
⇒ \(x=-\frac{1}{4}\)
Vậy \(x=-\frac{1}{4}.\)
3. \(\frac{16}{2^n}=2\)
⇒ \(2^n=16:2\)
⇒ \(2^n=8\)
⇒ \(2^n=2^3\)
⇒ \(n=3\)
Vậy \(n=3.\)
4. \(\frac{-3^n}{81}=-27\)
⇒ \(\left(-3\right)^n=\left(-27\right).81\)
⇒ \(\left(-3\right)^n=-2187\)
⇒ \(\left(-3\right)^n=\left(-3\right)^7\)
⇒ \(n=7\)
Vậy \(n=7.\)
Chúc bạn học tốt!
Lời giải:
\(A-B=\frac{4}{2019^2}-\frac{4}{2019^4}\)
Dễ thấy $0< 2019^2< 2019^4\Rightarrow \frac{4}{2019^2}> \frac{4}{2019^4}$
$\Rightarrow A-B=\frac{4}{2019^2}-\frac{4}{2019^4}>0$
$\Rightarrow A>B$
thầy ơi vì sao \(A-B=\frac{4}{2019^2}-\frac{4}{2019^4}\)
\(\left(\dfrac{1}{38}-1\right).\left(\dfrac{1}{37}-1\right).\left(\dfrac{1}{36}-1\right)....\left(\dfrac{1}{2}-1\right)\)
\(=\left(\dfrac{1}{38}-\dfrac{38}{38}\right).\left(\dfrac{1}{37}-\dfrac{37}{37}\right).\left(\dfrac{1}{36}-\dfrac{36}{36}\right)....\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\)
\(=\left(\dfrac{-37}{38}\right).\left(\dfrac{-36}{37}\right).\left(\dfrac{-35}{36}\right)...\left(\dfrac{-1}{2}\right)\)
\(=\left(\dfrac{-1}{38}\right).\left(\dfrac{-1}{1}\right).\left(\dfrac{-1}{1}\right).....\left(-\dfrac{1}{1}\right)\)
\(=\left(\dfrac{-1}{38}\right).\left(-1\right).\left(-1\right).....\left(-1\right)\)
\(=\dfrac{\left(-1\right).\left(-1\right).....\left(-1\right)}{38}\)
\(=\dfrac{\left(-1\right)^{38}}{38}\)
\(=\dfrac{1}{38}\)