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5⁸.5² = 5⁸⁺²
= 5¹⁰
= 9765625
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4⁹ : 64²
= 4⁹ : (4³)²
= 4⁹ : 4⁶
= 4⁹⁻⁶
= 4³
= 64
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2²⁵ : 32⁴
= 2²⁵ : (2⁵)4
= 2²⁵ : 2²⁰
= 2²⁵⁻²⁰
= 2⁵
= 32
------------
125³ : 25⁴
= (5³)³ : (5²)⁴
= 5⁹ : 5⁸
= 5⁹⁻⁸
= 5
\(\frac{22}{9}-\left(x+\frac{1}{2}\right)^2=\frac{7}{3}\)
\(\left(x+\frac{1}{2}\right)^2=\frac{22}{9}-\frac{7}{3}\)
\(\left(x+\frac{1}{2}\right)^2=\frac{1}{9}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{3}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}-\frac{1}{2}\)
\(x=-\frac{1}{6}\)
\(\frac{22}{9}-\left(x+\frac{1}{2}\right)^2=\frac{7}{3}\)
\(\left(x+\frac{1}{2}\right)^2=\frac{22}{9}=\frac{7}{3}\)\
\(\left(x+\frac{1}{2}\right)^2=\frac{22}{9}-\frac{21}{9}\)
\(\left(x+\frac{1}{2}\right)^2=\frac{1}{9}\)
\(\Rightarrow x+\frac{1}{2}=\pm\frac{1}{3}\)
TH1:\(x+\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}-\frac{1}{2}\)
\(x=-\frac{1}{6}\)
TH2:\(x+\frac{1}{2}=-\frac{1}{3}\)
\(x=-\frac{1}{3}-\frac{1}{2}\)
\(x=-\frac{5}{6}\)
Vậy \(x\in\left\{-\frac{1}{6};-\frac{5}{6}\right\}\)
\(-\left(-220\right)-78-220-\left(-78\right)-23=220-78-220+78-23\)
\(=\left(220-220\right)+\left(-78+78\right)-23\)
\(=0-23\)
\(=-23\)