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Sử dụng bất đẳng thức AM - GM ta dễ thấy:
\(LHS=\sqrt{a-1+2\sqrt{a-2}}+\sqrt{a-1-2\sqrt{a-2}}\)
\(\ge2\sqrt{\left(a-1+2\sqrt{a-2}\right)\left(a-1-2\sqrt{a-2}\right)}\)
\(=2\sqrt{\left(a-1\right)^2-4\left(a-2\right)}=2\sqrt{a^2-6a+9}=2\sqrt{\left(a-3\right)^2}\ge2\)( vì a khác 3 )
Hoặc cách khác như thế này:
\(LHS=\sqrt{a-1+2\sqrt{a-2}}+\sqrt{a-1-2\sqrt{a-2}}\)
\(=\sqrt{\left[a-2+2\sqrt{a+2}+1\right]}+\sqrt{\left[a-2-2\sqrt{a-2}+1\right]}\)
\(=\sqrt{\left(\sqrt{a-2}+1\right)^2}+\sqrt{\left(\sqrt{a-2}-1\right)^2}\)
\(=\left|\sqrt{a-2}+1\right|+\left|\sqrt{a-2}-1\right|\)
\(=\left|\sqrt{a-2}+1\right|+\left|1-\sqrt{a-2}\right|\ge\left|\sqrt{a-2}+1+1-\sqrt{a-2}\right|=2\)
Đẳng thức tự tìm nha
\(A=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
ĐK : \(\hept{\begin{cases}x,y>0\\x\ne y\end{cases}}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\frac{x+2\sqrt{xy}+y}{x-y}-\frac{x-2\sqrt{xy}+y}{x-y}\)
\(=\frac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y}{x-y}=\frac{4\sqrt{xy}}{x-y}\)
Với \(\hept{\begin{cases}x=7+2\sqrt{3}\\y=7-2\sqrt{3}\end{cases}}\)( tmđk )
=> \(A=\frac{4\sqrt{\left(7+2\sqrt{3}\right)\left(7-2\sqrt{3}\right)}}{7+2\sqrt{3}-\left(7-2\sqrt{3}\right)}\)
\(=\frac{4\sqrt{7^2-\left(2\sqrt{3}\right)^2}}{7+2\sqrt{3}-7+2\sqrt{3}}\)
\(=\frac{4\sqrt{49-12}}{4\sqrt{3}}\)
\(=\frac{4\sqrt{37}}{4\sqrt{3}}=\frac{\sqrt{37}}{\sqrt{3}}=\frac{\sqrt{37}\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{\sqrt{111}}{3}\)
Bài 1: Ta có: \(\sqrt{2020}-\sqrt{2019}=\frac{1}{\sqrt{2020}+\sqrt{2019}};\)\(\sqrt{2018}-\sqrt{2017}=\frac{1}{\sqrt{2018}+\sqrt{2017}}\)
Dễ thấy \(\sqrt{2020}+\sqrt{2019}>\sqrt{2018}+\sqrt{2017}\)nên \(\frac{1}{\sqrt{2020}+\sqrt{2019}}< \frac{1}{\sqrt{2018}+\sqrt{2017}}\)
Suy ra\(\sqrt{2020}-\sqrt{2019}< \sqrt{2018}-\sqrt{2017}\)
Bài 2: Xét biểu thức \(\sqrt{a^2+a^2\left(a+1\right)^2+\left(a+1\right)^2}=\sqrt{a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2}=\sqrt{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}=\sqrt{\left(a^2+a+1\right)^2}=a^2+a+1\)(Vì \(a^2+a+1>0\forall a\inℝ\))
Áp dụng công thức tổng quát trên, ta được: \(\sqrt{2019^2+2019^2.2020^2+2020^2}=2019^2+2019+1\)(là số tự nhiên) (đpcm)
2, a, \(a+\dfrac{1}{a}\ge2\)
\(\Leftrightarrow\dfrac{a^2+1}{a}\ge2\)
\(\Rightarrow a^2-2a+1\ge0\left(a>0\right)\)
\(\Leftrightarrow\left(a-1\right)^2\ge0\)( là đt đúng vs mọi a)
vậy...................
Câu 1:
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}=3\)
\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
\(\sqrt{x+2\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}\right)^2+2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-1}+1=2\\\sqrt{x-1}+1=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-1}=1\Leftrightarrow x-1=1\Leftrightarrow x=2\\\sqrt{x-1}=-3\left(vl\right)\end{cases}}\)
Vậy phương trình có tập nghiệm \(S=\left\{2\right\}\)
\(a+b\ge2\sqrt{ab}\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) ( đúng )
Áp dụng Bunhiacopski ta có:
\(S^2=\left(\sqrt{x-2}+\sqrt{y-3}\right)^2\le\left(1^2+1^2\right)\left(x-2+y-3\right)=2\left(x+y-5\right)=2\)
Dấu "=" bạn xét nốt
Bài 2:
a, Ta có
\(3\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}\)
= \(3\left|-2\right|+\left|-5\right|\)
=\(6+5\)
= 11
Vậy \(3\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}=11\)
b, Ta có
\(\sqrt{6+2\sqrt{5}}-\sqrt{5}\)
= \(\sqrt{5+2\sqrt{5}+1}-\sqrt{5}\)
= \(\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{5}\)
= \(\left|\sqrt{5}+1\right|-\sqrt{5}\)
= \(\sqrt{5}+1-\sqrt{5}=1\)
Vậy \(\sqrt{6+2\sqrt{5}}-\sqrt{5}=1\)
a. 2\(\sqrt{3.16}\)+\(\sqrt{3.9}\)+\(\sqrt{3}\)
=2.4.\(\sqrt{3}\)+3\(\sqrt{3}\)+\(\sqrt{3}\)
12\(\sqrt{3}\)