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a) \(\sqrt{\frac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\frac{41.289}{164}}=\sqrt{\frac{289}{4}}=\frac{17}{2}\)
b) tương tự ý a
c) \(\left(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\right)^2=7+4\sqrt{3}+7-4\sqrt{3}-2.\sqrt{7+4\sqrt{3}}.\sqrt{7-4\sqrt{3}}\)
\(=14-2\sqrt{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=14-2\sqrt{49-48}\)
\(=14-2.1=12\)
\(\Rightarrow\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}=\sqrt{12}=2\sqrt{3}\)
1) \(\frac{\sqrt{165^2-124^2}}{164}=\frac{\sqrt{\left(165-124\right)\left(165+124\right)}}{164}=\frac{\sqrt{41}\cdot\sqrt{289}}{164}=\frac{\sqrt{41}\cdot17}{164}=\frac{17}{4\sqrt{41}}\)
2) \(\frac{\sqrt{149^2-76^2}}{\sqrt{457^2-384^2}}=\frac{\sqrt{\left(149+76\right)\left(149-76\right)}}{\sqrt{\left(457+384\right)\left(457-384\right)}}=\frac{\sqrt{225}\cdot\sqrt{73}}{\sqrt{841}\cdot\sqrt{73}}=\frac{25}{29}\)
a: \(=\sqrt{\dfrac{25}{16}\cdot\dfrac{49}{9}\cdot\dfrac{1}{100}}=\dfrac{5}{4}\cdot\dfrac{7}{3}\cdot\dfrac{1}{10}=\dfrac{35}{120}=\dfrac{7}{24}\)
b: \(=\sqrt{1.44\cdot0.81}=1.2\cdot0.9=1.08\)
c: \(=\sqrt{\dfrac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\dfrac{1}{4}\cdot289}=\dfrac{17}{2}\)
d: \(=\sqrt{\dfrac{\left(149-76\right)\left(149+76\right)}{\left(457-384\right)\left(457+384\right)}}=\sqrt{\dfrac{225}{841}}=\dfrac{15}{29}\)
Với mọi n nguyên dương ta có:
\(\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=1\Rightarrow\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}\)
Với k nguyên dương thì
\(\frac{1}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k+1}+\sqrt{k}}\Rightarrow\frac{2}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k-1}+\sqrt{k}}+\frac{1}{\sqrt{k+1}+\sqrt{k}}=\sqrt{k}-\sqrt{k-1}+\sqrt{k+1}-\sqrt{k}\)
\(=\sqrt{k+1}-\sqrt{k-1}\)(*)
Đặt A = vế trái. Áp dụng (*) ta có:
\(\frac{2}{\sqrt{1}+\sqrt{2}}>\sqrt{3}-\sqrt{1}\)
\(\frac{2}{\sqrt{3}+\sqrt{4}}>\sqrt{5}-\sqrt{3}\)
...
\(\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-\sqrt{79}\)
Cộng tất cả lại
\(2A=\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{4}}+....+\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-1=8\Rightarrow A>4\left(đpcm\right)\)
3.
Theo bất đẳng thức cô si ta có:
\(\sqrt{b-1}=\sqrt{1.\left(b-1\right)}\le\frac{1+b-1}{2}=\frac{b}{2}\Rightarrow a.\sqrt{b-1}\le\frac{a.b}{2}\)
Tương tự \(\Rightarrow b.\sqrt{a-1}\le\frac{a.b}{2}\Rightarrow a.\sqrt{b-1}+b.\sqrt{a-1}\le a.b\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=2\)
a, \(\sqrt{8}+\sqrt{18}-\sqrt{\frac{1}{2}}=2\sqrt{2}+3\sqrt{2}-\frac{1}{2}\sqrt{2}\)
\(=\frac{9}{2}\sqrt{2}\)
b, \(\frac{3-\sqrt{3}}{\sqrt{3}}+\frac{2\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)\)
\(=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}}+\frac{2\sqrt{2}}{\sqrt{2}+1}-\sqrt{2}-\sqrt{3}\)
\(=\sqrt{3}-1+\frac{2\sqrt{2}}{\sqrt{2}+1}-\sqrt{2}-\sqrt{3}\)
\(=\frac{2\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+1\right)\) \(=\frac{2\sqrt{2}-\left(\sqrt{2}+1\right)^2}{\sqrt{2}+1}\)
\(=\frac{2\sqrt{2}-2-2\sqrt{2}-1}{\sqrt{2}+1}=-\frac{2+1}{\sqrt{2}+1}\)
c, PT xác định với mọi x nha!
\(\sqrt{x^2-2x+1}=3\) \(\Rightarrow x^2-2x+1=9\)
\(\Leftrightarrow x^2-2x-8=0\)
\(\Leftrightarrow\left(x^2-4x\right)+\left(2x-8\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}}\)
Vậy...
bạn tự kl
a,
\(\frac{\sqrt{6}\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}+\sqrt{\frac{\left(2-\sqrt{2}\right)^2}{\left(2+\sqrt{2}\right).\left(2-\sqrt{2}\right)}}\)
=\(\sqrt{2}+\frac{2-\sqrt{2}}{\sqrt{2}}\)
=\(\sqrt{2}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}}\)
=\(\sqrt{2}+\sqrt{2}-1\)
=\(2\sqrt{2}-1\)
còn tiếp
b=,\(\frac{6\sqrt{3}}{3}-\frac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}-\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}-\sqrt{3}}\)
=\(6-1+\sqrt{3}-\sqrt{6}\)
=\(5+\sqrt{3}+\sqrt{6}\)
a) \(\sqrt{\frac{165^2-124^2}{164}}=\sqrt{\frac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\frac{41.289}{164}}\)
\(=\sqrt{\frac{11849}{164}}=\sqrt{72,25}=8,5\)
b)\(\sqrt{\frac{149^2-76^2}{457^2-384^2}}=\sqrt{\frac{\left(149-76\right)\left(149+76\right)}{\left(457-384\right)\left(457+384\right)}}\) \(=\sqrt{\frac{73.225}{73.841}}=\sqrt{\frac{225}{841}}=\sqrt{\frac{15^2}{29^2}}=\frac{15}{29}\)
c)\(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\) \(=\sqrt{2^2+3+2.2.\sqrt{3}}-\sqrt{2^2+3-2.2.\sqrt{3}}\)
\(=\sqrt{2^2+2.2.\sqrt{3}+\sqrt{3}^2}-\sqrt{2^2-2.2.\sqrt{3}+\sqrt{3}^2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}=\left(2+\sqrt{3}\right)-\left(2-\sqrt{3}\right)\)
\(=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}\)