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+) \(\frac{-3}{4}-x=\frac{2}{5}\)
=> \(-x=\frac{2}{5}+\frac{3}{4}\)
=> \(-x=\frac{23}{20}\)
=> \(x=\frac{-23}{20}\)
Vậy \(x=\frac{-23}{20}\)
+) \(\frac{-4}{x}=\frac{x}{-16}\)
=> \(x.x=-4.\left(-16\right)\)
=> \(x^2=64\)
=> \(x=\sqrt{64}=8\)
hoặc \(x=-\sqrt{64}=-8\)
Vậy \(x=8\) hoặc \(x=-8\)
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\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( 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) =
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a: \(\Leftrightarrow\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}+\dfrac{5}{48}=\dfrac{6}{48}=\dfrac{1}{8}\)
\(\Leftrightarrow-\dfrac{1}{12}< x< \dfrac{1}{8}\)
=>x=0
c: \(\Leftrightarrow x=\dfrac{-1}{2}\cdot\dfrac{1}{4}=\dfrac{-1}{8}\)
d: \(\Leftrightarrow x^8=x^7\)
=>x(x-1)=0
=>x=0(loại) hoặc x=1(nhận)
e: \(\Leftrightarrow3^x=\dfrac{3^{10}}{3^9}=3\)
hay x=1
f: =>x-1=20
hay x=21
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a, \(\frac{x-1}{x+5}=\frac{6}{7}\)
\(\Rightarrow\left(x-1\right).7=\left(x+5\right).6\)
\(\Rightarrow7x-7=6x+30\)
\(\Rightarrow7x-6x=7+30\)
\(\Rightarrow x=37\)
Vậy x=37
b, \(\left(2x-\frac{1}{2}\right)^4+\frac{11}{16}=\frac{3}{4}\)
\(\Rightarrow\left(2x-\frac{1}{2}\right)^4=\frac{1}{16}\)
\(\Rightarrow\left(2x-\frac{1}{2}\right)^4=\left(\frac{1}{2}\right)^4\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=\frac{1}{2}\\2x-\frac{1}{2}=-\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=1\\2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=0\end{cases}}\)
Vậy \(x=\frac{1}{2}\) hoặc \(x=0\)
\(\frac{x}{-4}=\frac{-16}{x}\)
=>x.x = -4 . (-16)
=>x2=64
=>x2=82
=>x=8
Vậy x=8
\(\frac{x}{-4}=\frac{-16}{x}\)
=> x2= -4.(-16)
=> x2= 64
=> x2= 82
\(\Rightarrow\orbr{\begin{cases}x=8\\x=-8\end{cases}}\)
Vậy \(x=\pm8\)