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Câu 2: A = \(^{1+2+2^2+2^{ }^3+...+2^{2017}}\)
2A = \(2+2^2+2^3+...+2^{2018}\)
Suy ra 2A - A =\(2^{2018}-1\) Do đó A < B
1. Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=t\Rightarrow a=2016t,b=2017t,c=2018t\)
\(\left(a-c\right)^3=\left(2016t-2018t\right)^3=\left(-2t\right)^3=-8t^3\)
\(8\left(a-b\right)^2\left(b-c\right)=8\left(2016t-2017t\right)^2\left(2017t-2018t\right)=8.\left(-t\right)^2.\left(-t\right)=-8t^3\)
Vậy \(\left(a-c\right)^3=8\left(a-b\right)^2\left(b-c\right)\)
=>\(-B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2012}\right)\)
=\(\frac{1}{2}.\frac{2}{3}...\frac{2011}{2012}=\frac{1}{2012}\)
a) Có \(A=\frac{\sqrt{x}+2}{\sqrt{x}-2}=\frac{\sqrt{x}-2+4}{\sqrt{x}-2}=\frac{\sqrt{x}-2}{\sqrt{x}-2}+\frac{4}{\sqrt{x}-2}=1+\frac{4}{\sqrt{x}-2}\)
Để A đạt giá trị nguyên thì: \(\sqrt{x}-2\in U\left(4\right)\)
TH1: \(\sqrt{x}-2=1\Rightarrow x=9\)
TH2: \(\sqrt{x}-2=-1\Rightarrow x=1\)
TH3: \(\sqrt{x}-2=2\Rightarrow x=16\)
TH4: \(\sqrt{x}-2=-2\Rightarrow x=0\)
TH5: \(\sqrt{x}-2=4\Rightarrow x=36\)
TH6: \(\sqrt{x}-2=-4\Rightarrow\) k tồn tại x
Vậy:...
a) \(\left(x-2\right)^3=-27\)
\(\Rightarrow\left(x-2\right)^3=\left(-3\right)^3\)
\(\Rightarrow x-2=-3\)
\(\Rightarrow x=-1\)
Vậy \(x=-1\)
b) \(\left(2x+1\right)^4=81\)
\(\Rightarrow\left(2x+1\right)^4=3^4=\left(-3\right)^4\)
\(\left\{{}\begin{matrix}\left(2x+1\right)^4=3^4\Rightarrow2x+1=3\Rightarrow x=1\\\left(2x+1\right)^4=\left(-3\right)^4\Rightarrow2x+1=-3\Rightarrow x=-2\end{matrix}\right.\)
Vậy \(x=1;x=-2\)
c) Bạn xem lại đề bài nhé!
d) \(\left(5x-2\right)^{10}=\left(5x-2\right)^{100}\)
\(\Rightarrow\left(5x-2\right)^{10}-\left(5x-2\right)^{100}=0\)
\(\Rightarrow\left(5x-2\right)^{10}.\left[1-\left(5x-2\right)^{90}\right]=0\)
+) TH1: \(\left(5x-2\right)^{10}=0\)
\(\Rightarrow5x-2=0\)
\(\Rightarrow x=\dfrac{2}{5}\)
+) TH2: \(1-\left(5x-2\right)^{90}=0\)
\(\Rightarrow\left(5x-2\right)^{90}=1\)
\(\Rightarrow\left(5x-2\right)^{90}=1^{90}=\left(-1\right)^{90}\)
\(\Rightarrow\left\{{}\begin{matrix}\left(5x-2\right)^{90}=1^{90}\Rightarrow5x-2=1\Rightarrow x=\dfrac{3}{5}\\\left(5x-2\right)^{90}=\left(-1\right)^{90}\Rightarrow5x-2=-1\Rightarrow x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{5};\dfrac{2}{5};\dfrac{3}{5}\right\}\)
Câu 2a đánh thiếu đề rồi : I x+1I + I x+2I + I x+3 I = x
2c)
Ta có: \(25-y^2\le25\Rightarrow8\left(x-2012\right)^2\le25\)
\(\Rightarrow\left(x-2012\right)^2\le3\)
\(\Rightarrow\left[\begin{matrix}\left(x-2012\right)^2=0\\\left(x-2012\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x-2012=0\\\left[\begin{matrix}x-2012=1\\x-2012=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=2012\\\left[\begin{matrix}x=2013\\x=2011\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}y=5\\\left[\begin{matrix}y=\sqrt{17}\\y=\sqrt{17}\end{matrix}\right.\end{matrix}\right.\)(loại)
Vậy x=2012,y=5
gọi biểu thức 2^4+2^8+....+2^2016 là A ta có
A=2^4+2^8+.....+2^2016
8A=2^4+2^8+.....+2^2010
8A-A=2-2^2010
7A=1+2-2^2010
A = 1 + 24 + 28 + ...... + 22012 + 22016
24A = 24 + 28 + 212 + ..... + 22016 + 22020
24A - A = (24 + 28 + 212 + ..... + 22016 + 22020) - (1 + 24 + 28 + ...... + 22012 + 22016)
15A = 22010 - 1
\(\Rightarrow A=\frac{2^{2010}-1}{15}\)
B = 1 + 22 + 24 + ....... + 22016 + 22018
22B = 22 + 24 + 26 + ........ + 22018 + 22020
22B - B = (22 + 24 + 26 + ........ + 22018 + 22020) - (1 + 22 + 24 + ....... + 22016 + 22018)
3B = 22010 - 1
\(\Rightarrow B=\frac{2^{2010}-1}{3}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{2^{2010}-1}{15}}{\frac{2^{2010}-1}{3}}=\frac{\left(2^{2010}-1\right).\frac{1}{15}}{\left(2^{2010}-1\right).\frac{1}{3}}=\frac{\frac{1}{15}}{\frac{1}{3}}=\frac{\frac{1}{3}.\frac{1}{5}}{\frac{1}{3}}=\frac{1}{5}\)
Đáng ra phải là 22020 chứ bạn