Tính :
A= 256 - \(\dfrac{256}{2}-\dfrac{256}{2^2}-\dfrac{256}{2^3}-....-\dfrac{256}{2^9}\)
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Câu 2 : Tính nhanh:
\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+...+\(\dfrac{1}{256}\)+\(\dfrac{1}{512}\)
Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}+\dfrac{1}{512}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}+\dfrac{1}{256}\)
\(\Rightarrow A=2A-A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}+\dfrac{1}{256}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{256}-\dfrac{1}{512}\)
\(\Rightarrow A=1-\dfrac{1}{512}=\dfrac{511}{512}\)
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Đặt ⇒2A=1+12+14+...+1128+1256⇒2A=1+12+14+...+1128+1256
⇒A=1−1512=511512
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\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{128}\)+\(\dfrac{1}{256}\)
Tính nhanh
\(\left(x-\dfrac{2}{15}\right)^3=\dfrac{8}{125};\left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\)
\(\left(x-\dfrac{2}{15}\right)^3=\dfrac{8}{125}\)
\(\left(x-\dfrac{2}{15}\right)^3=\left(\dfrac{2}{5}\right)^3\)
\(x-\dfrac{2}{15}=\dfrac{2}{5}\)
\(x=\dfrac{2}{5}+\dfrac{2}{15}\)
\(x=\dfrac{8}{15}\)
\(\left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\)
\(\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\)
\(2x+5=4\)
\(2x=-1\)
\(x=-0,5\)
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\(\left(x-\dfrac{2}{15}\right)^3=\dfrac{8}{125}\\ \Rightarrow\left(x-\dfrac{2}{15}\right)^3=\left(\dfrac{2}{5}\right)^3\\ \Rightarrow x-\dfrac{2}{15}=\dfrac{2}{5}\\ \Rightarrow x=\dfrac{2}{5}+\dfrac{2}{15}\\ \Rightarrow x=\dfrac{6}{15}+\dfrac{2}{15}\\ \Rightarrow x=\dfrac{8}{15}\\ \left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\\ \Rightarrow\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\\ \Rightarrow2x+5=4\\ \Rightarrow2x=4-5\\ \Rightarrow2x=-1\\ \Rightarrow x=-\dfrac{1}{2}\)
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Đơn giản hóa (\(\dfrac{1}{2}\))\(\dfrac{-1}{4}\)
A. -16 B. -\(\sqrt{2}\) C. -\(\dfrac{1}{16}\) D.\(\dfrac{1}{256}\) E.\(\sqrt{2}\)
(\(\dfrac{-3}{4}\))3x - 1 = \(\dfrac{256}{81}\)
Có thể chia ra 2 trường hợp
\(\left(\dfrac{-3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
\(\Rightarrow\left(\dfrac{-3}{4}\right)^{3x-1}=\left(\dfrac{4}{3}\right)^4\)
Xem lại đề
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\(\left(-\dfrac{3}{4}\right)3x-1=\dfrac{256}{81}\)
\(-\dfrac{9}{4}\cdot x=\dfrac{337}{81}\)
\(x=\dfrac{337}{81}\cdot-\dfrac{4}{9}\)
\(x=-\dfrac{1348}{729}\)
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Sửa đề:
(-3/4)³ˣ⁻¹ = 81/256
(-3/4)³ˣ⁻¹ = (-3/4)⁴
3x - 1 = 4
3x = 4 + 1
3x = 5
x = 5/3
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A = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\)+ \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\) + \(\dfrac{1}{256}\)
\(A=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{128}-\dfrac{1}{256}\right)\)
\(A=1-\dfrac{1}{256}\)
\(A=\dfrac{255}{256}\)
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\(\left(\dfrac{1}{2}\right)^x+\left(\dfrac{1}{2}\right)^{x+3}=\dfrac{9}{256}\)
tìm x thõa mãn
Lời giải:
\(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+3}=\frac{9}{256}\)
\(\Leftrightarrow \left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^x.\left(\frac{1}{2}\right)^3=\frac{9}{256}\)
\(\Leftrightarrow \left(\frac{1}{2}\right)^x.(1+\frac{1}{8})=\frac{9}{256}\Rightarrow \left(\frac{1}{2}\right)^x=\frac{1}{32}=\left(\frac{1}{2}\right)^5\)
\(\Rightarrow x=5\)
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Đơn giản hoá \(\left(\dfrac{1}{2}\right)^{\dfrac{-1}{4}}\)
A) -16 B) -\(\sqrt{2}\) C) -\(\dfrac{1}{16}\) D) -\(\dfrac{1}{256}\) E) \(\sqrt{2}\)
Lời giải:
\(\left(\frac{1}{2}\right)^{\frac{-1}{4}}=(2^{-1})^{\frac{-1}{4}}=2^{\frac{1}{4}}=\sqrt[4]{2}\)
Không đáp án nào đúng.
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tìm x:
\(a,5^x.\left(5^2\right)^3=625\)
\(b,\left(\dfrac{12}{15}\right)^x=\left(\dfrac{5}{4}\right)^{-2}-\left(\dfrac{-3}{5}\right)^4\)
\(c,\left(\dfrac{-3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
\(d,172x^2-7^9:98^3=2^{-3}\)
