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a: ĐKXĐ: 2x-10>=0
=>2x>=10
=>x>=5
b: \(\sqrt{A^2B}=\sqrt{A^2}\cdot\sqrt{B}=\left|A\right|\cdot\sqrt{B}\)
\(\sqrt{72}=\sqrt{36\cdot2}=6\sqrt{2}\)
c: \(A=\sqrt{16}+\sqrt{81}=4+9=13\)
\(B=\sqrt{\dfrac{\left(15\sqrt{5}+5\sqrt{200}-3\sqrt{450}\right)}{\sqrt{10}}}\)
\(=\sqrt{\dfrac{15}{\sqrt{2}}+5\sqrt{20}-3\sqrt{45}}\)
\(=\sqrt{\dfrac{15\sqrt{2}+2\sqrt{5}}{2}}=\sqrt{\dfrac{30\sqrt{2}+4\sqrt{5}}{4}}\)
\(=\dfrac{\sqrt{30\sqrt{2}+4\sqrt{5}}}{2}\)
\(C=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\left(2+\sqrt{3}\right)\)
\(=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}-\left(2+\sqrt{3}\right)+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
\(=2+\sqrt{3}-2-\sqrt{3}+\sqrt{2}=\sqrt{2}\)
đặt \(a=5+2\sqrt{6}\).ta sẽ chứng minh với dạng tổng quát \(\left[a^n\right]\)là 1 số tự nhiên lẻ.
ta có: \(a^n=\left(5+2\sqrt{6}\right)^n=x+y\sqrt{6}\)(x,y là các số tự nhiên) (*)
đặt \(b=5-2\sqrt{6}\Rightarrow b^n=x-y\sqrt{6}\)
\(\Rightarrow a^n+b^n=2x\)
mà \(0< b=5-2\sqrt{6}< 1\)
\(\Rightarrow0< b^n< 1\)
\(\Rightarrow2x-1< a^n=2x-b^n< 2x\)
nên \(\left[a^n\right]=2x-1\)lẻ vì x nguyên.
p/s:(*) : thử \(\left(5+2\sqrt{6}\right)^2,\left(5+2\sqrt{6}\right)^3\)đều có dạng \(A+B\sqrt{6}\)
a) \(\sqrt{200}-\sqrt{32}+\sqrt{72}\)
\(=\sqrt{10^2\cdot2}-\sqrt{4^2\cdot2}+\sqrt{6^2\cdot2}\)
\(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}\)
\(=\left(10-4+6\right)\sqrt{2}\)
\(=12\sqrt{2}\)
b) \(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\dfrac{1}{5}}\)
\(=4\cdot2\sqrt{5}-3\cdot5\sqrt{5}+5\cdot3\sqrt{5}-3\sqrt{5}\)
\(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}\)
\(=\left(8-15+15-3\right)\sqrt{5}\)
\(=5\sqrt{5}\)
c) \(\left(2\sqrt{8}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\sqrt{20}-2\sqrt{2}\right)\)
\(=\left(2\cdot2\sqrt{2}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\cdot2\sqrt{5}-2\sqrt{2}\right)\)
\(=\left(3\sqrt{5}-3\sqrt{2}\right)\left(72-10\sqrt{5}-2\sqrt{2}\right)\)
\(=5\sqrt{2}-\dfrac{3}{2}\cdot4\sqrt{2}-\dfrac{1}{3}\cdot6\sqrt{2}+8=-3\sqrt{2}+8\)
\(=3\sqrt{2}-3\sqrt{2}+2\sqrt{2}+6\sqrt{2}=8\sqrt{2}\)
a) \(\Rightarrow\dfrac{x^2}{\sqrt{5}}=\sqrt{20}\Rightarrow x^2=\sqrt{20.5}=\sqrt{100}=10\)
\(\Rightarrow x=\pm\sqrt{10}\)
b)ĐKXĐ: \(x\ge0\)
\(\Rightarrow3\sqrt{2x}+\sqrt{2}-6\sqrt{2}+4=0\)
\(\Rightarrow3\sqrt{2x}=5\sqrt{2}-4\)
\(\Rightarrow18x=50+16-40\sqrt{2}\)
\(\Rightarrow x=\dfrac{66-40\sqrt{2}}{18}\)
\(a,\Leftrightarrow\dfrac{x^2}{\sqrt{5}}=\sqrt{20}=2\sqrt{5}\Leftrightarrow x^2=2\sqrt{5}\cdot\sqrt{5}=10\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{10}\\x=-\sqrt{10}\end{matrix}\right.\)
\(b,ĐK:x\ge0\\ PT\Leftrightarrow3\sqrt{2x}+\dfrac{1}{7}\cdot7\sqrt{2}-6\sqrt{2}+4=0\\ \Leftrightarrow3\sqrt{2x}=5\sqrt{2}-4\\ \Leftrightarrow\sqrt{2x}=\dfrac{5\sqrt{2}-4}{3}\\ \Leftrightarrow2x=\dfrac{66-40\sqrt{2}}{9}\\ \Leftrightarrow x=\dfrac{66-40\sqrt{2}}{18}=\dfrac{33-20\sqrt{2}}{9}\left(tm\right)\)
a)\(\left(\sqrt{10}-\sqrt{15}+3\sqrt{3}\right)\sqrt{5}-\sqrt{72}\)
\(=\sqrt{15}-\sqrt{15}+15-6\sqrt{2}\)
\(15-6\sqrt{2}\)
b)\(\dfrac{\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right)}{8\sqrt{10}}\)
\(=\dfrac{\left(15.5\sqrt{2}+5.10\sqrt{2}-3.15\sqrt{2}\right)}{8\sqrt{10}}\)
\(=\dfrac{\left(75\sqrt{2}+50\sqrt{2}-45\sqrt{2}\right)}{8\sqrt{10}}\)
\(=\dfrac{80\sqrt{2}}{8\sqrt{10}}=\dfrac{10\sqrt{2}}{\sqrt{10}}=\sqrt{20}=2\sqrt{5}\)
Ta có :
\(a^2=72+\sqrt{72+\sqrt{72+\sqrt{72+.......}}}\)
\(\Leftrightarrow a^2=72+a\Leftrightarrow a^2-a-72=0\Leftrightarrow\left(a-9\right)\left(a+8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a=9\\a=-8\end{cases}}\)
Mà a > 0 nên a = 9 \(\Rightarrow\left[a\right]=9\)