\(a,\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\dfrac{1}{2}\)
\(b,\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
\(c,\left(2x+3\right)^2=\dfrac{9}{121}\)
\(\Rightarrow\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
\(\Rightarrow2x+3=\dfrac{3}{11}\)
\(\Rightarrow2x=-\dfrac{30}{11}\)
\(\Rightarrow x=-\dfrac{15}{11}\)
\(d,\left(2x-1\right)^3=-\dfrac{8}{27}\)
\(\Rightarrow\left(2x-1\right)^3=\left(-\dfrac{2}{3}\right)^3\)
\(\Rightarrow2x-1=-\dfrac{2}{3}\)
\(\Rightarrow2x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{6}\)
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\Leftrightarrow x=\dfrac{-1}{4}\)
\(\left(2x+3\right)^2=\dfrac{9}{121}\Leftrightarrow\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\Leftrightarrow2x+3=\dfrac{3}{11}\Leftrightarrow x=\dfrac{-15}{11}\)
\(\left(2x-1\right)^3=-8\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\Leftrightarrow2x-1=-2\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\)
\(\text{Câu 1 :}\) \(\text{Tìm }x\)
\(\text{a) }\left(2x-1\right)^3=-8\\ \Leftrightarrow\left(2x-1\right)^3=-2^3\\ \Leftrightarrow2x-1=-2\\ \Leftrightarrow2x=-1\\ \Leftrightarrow x=-\dfrac{1}{2}\\ \text{Vậy }x=-\dfrac{1}{2}\)
\(\text{b) }\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\\ \Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\\ \Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\\ \Leftrightarrow x=\dfrac{1}{4}-\dfrac{1}{2}\\ \Leftrightarrow x=-\dfrac{1}{4}\\ \text{Vậy }x=-\dfrac{1}{4}\)
\(\text{Câu 2 : So Sánh}\)
\(\text{a) }99^{20}\text{ và }9999^{10}\\ \text{Ta có : }99^{20}=\left(99^2\right)^{10}=9801^{10}\\ \text{Mà }9801^{10}< 9999^{10}\\ \Rightarrow99^{20}< 9999^{10}\\ \\ \text{Vậy }99^{20}< 9999^{10}\)
\(\text{b) }3^{4000}\text{ và }9^{2000}\\ \text{Ta có : }3^{4000}=\left(3^2\right)^{2000}=9^{2000}\\ \text{ Mà }9^{2000}=9^{2000}\\ \Rightarrow3^{4000}=9^{2000}\\ \text{Vậy }3^{4000}=9^{2000}\)
\(\text{c) }2^{332}\text{ và }3^{223}\\ \text{Ta có : }2^{332}< 2^{333}\\ 3^{223}< 3^{222}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ (1) }\\ \text{Ta lại có : }2^{333}=\left(2^3\right)^{111}=8^{111}\\ 3^{222}=\left(3^2\right)^{111}=9^{111}\\ \\ \text{Mà }8^{111}< 9^{111}\\ \Rightarrow2^{333}< 3^{222}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\left(2\right)\\ \text{Từ }\left(1\right)\text{ và }\left(2\right)\text{ suy ra : }2^{332}< 2^{333}< 3^{222}< 3^{223}\\ \Rightarrow2^{332}< 3^{223}\\ \text{Vậy }2^{332}< 3^{223}\)
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Tại mình không để ý nên làm thiếu câu \(c\) và câu \(d\) bài \(1\) giờ mình bổ sung nhé :
\(\text{c) }\left(2x+3\right)^2=\dfrac{9}{121}\\ \Leftrightarrow\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\\ \Leftrightarrow2x+3=\dfrac{3}{11}\\ \Leftrightarrow2x=-\dfrac{30}{11}\\ \Leftrightarrow x=-\dfrac{15}{11}\\ \text{Vậy }x=-\dfrac{15}{11}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \)
\(\text{d) }\left(2x-1\right)^3=-\dfrac{8}{27}\\ \Leftrightarrow\left(2x-1\right)^3=\left(-\dfrac{2}{3}\right)^3\\ \Leftrightarrow2x-1=-\dfrac{2}{3}\\ \Leftrightarrow2x=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{6}\\ \text{Vậy }x=\dfrac{1}{6}\)
