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NV
7 tháng 5 2020

\(\left(\sqrt{2}.\sqrt{2}x+\sqrt{7}.\frac{\sqrt{7}}{y}\right)^2\le\left(2+7\right)\left(2x^2+\frac{7}{y^2}\right)\)

\(\Rightarrow\sqrt{2x^2+\frac{7}{y^2}}\ge\frac{1}{3}\left(2x+\frac{7}{y}\right)\)

\(\Rightarrow VT\ge\frac{1}{3}\left[2\left(a+b+c\right)+7\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]\)

\(VT\ge\frac{1}{3}\left(6+\frac{63}{a+b+c}\right)=\frac{1}{3}\left(6+\frac{63}{3}\right)=9\)

Dấu "=" xảy ra khi \(a=b=c=1\)

27 tháng 10 2021

TK: Cho các số thực dương a, b, c thỏa mãn a + b+ c = 3. Chứng minh rằng: \(\sqrt{2a^2+\frac{7}{b^2}}+\sqrt{2b^2+\frac{7}{... - Hoc24

20 tháng 9 2021

a) \(=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}-1-\sqrt{5}-1=-2\)

b) \(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(1+\sqrt{3}\right)^2}=2+\sqrt{3}-1-\sqrt{3}=1\)

c) \(=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}+1+\sqrt{7}-1=2\sqrt{7}\)

d) \(=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{5}+\sqrt{2}-\sqrt{2}+1=\sqrt{5}+1\)

a: \(=3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)-3\sqrt{6}\)

=3căn 6-6-3căn 6=-6

b: \(=\dfrac{a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\)

\(=\dfrac{a+\sqrt{ab}-a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}=\dfrac{2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)

 

17 tháng 8 2017

c. Ta có: C+E=\(\sqrt{45+\sqrt{2009}}+\sqrt{45-\sqrt{2009}}=\sqrt{\left(\sqrt{\dfrac{49}{2}}+\sqrt{\dfrac{41}{2}}\right)^2}+\sqrt{\left(\sqrt{\dfrac{49}{2}}-\sqrt{\dfrac{41}{2}}\right)^2}=\dfrac{7}{\sqrt{2}}+\dfrac{\sqrt{41}}{\sqrt{2}}+\dfrac{7}{\sqrt{2}}-\dfrac{\sqrt{41}}{\sqrt{2}}=\dfrac{2.7}{\sqrt{2}}=7\sqrt{2}\)

=> đpcm.

a) Sửa đề: \(A=\sqrt{8+2\sqrt{7}}-\sqrt{7}\)

Ta có: \(A=\sqrt{8+2\sqrt{7}}-\sqrt{7}\)

\(=\sqrt{7+2\cdot\sqrt{7}\cdot1+1}-\sqrt{7}\)

\(=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{7}\)

\(=\left|\sqrt{7}+1\right|-\sqrt{7}\)

\(=\sqrt{7}+1-\sqrt{7}\)

=1

b) Ta có: \(B=\sqrt{7+4\sqrt{3}}-2\sqrt{3}\)

\(=\sqrt{4+2\cdot2\cdot\sqrt{3}+3}-2\sqrt{3}\)

\(=\sqrt{\left(2+\sqrt{3}\right)^2}-2\sqrt{3}\)

\(=\left|2+\sqrt{3}\right|-2\sqrt{3}\)

\(=2+\sqrt{3}-2\sqrt{3}\)

\(=2-\sqrt{3}\)

c) Ta có: \(C=\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}\)

\(=\sqrt{13-2\cdot\sqrt{13}\cdot1+1}+\sqrt{13+2\cdot\sqrt{13}\cdot1+1}\)

\(=\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}\)

\(=\left|\sqrt{13}-1\right|+\left|\sqrt{13}+1\right|\)

\(=\sqrt{13}-1+\sqrt{13}+1\)

\(=2\sqrt{13}\)

d) Ta có: \(D=\sqrt{22-2\sqrt{21}}-\sqrt{22+2\sqrt{21}}\)

\(=\sqrt{21-2\cdot\sqrt{21}\cdot1+1}-\sqrt{21+2\cdot\sqrt{21}\cdot1+1}\)

\(=\sqrt{\left(\sqrt{21}-1\right)^2}-\sqrt{\left(\sqrt{21}+1\right)^2}\)

\(=\left|\sqrt{21}-1\right|-\left|\sqrt{21}+1\right|\)

\(=\sqrt{21}-1-\left(\sqrt{21}+1\right)\)

\(=\sqrt{21}-1-\sqrt{21}-1\)

=-2

29 tháng 10 2023

14:

\(A=\sqrt{-4x^2+4x+7}\)

\(=\sqrt{-\left(4x^2-4x-7\right)}\)

\(=\sqrt{-\left(4x^2-4x+1-8\right)}\)

\(=\sqrt{-\left(2x-1\right)^2+8}< =\sqrt{8}=2\sqrt{2}\)

Dấu = xảy ra khi 2x-1=0

=>\(x=\dfrac{1}{2}\)

13:

\(a+b+c>=\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)

=>\(2a+2b+2c-2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ac}>=0\)

=>\(\left(a-2\sqrt{ab}+b\right)+\left(b-2\sqrt{bc}+c\right)+\left(a-2\sqrt{ac}+c\right)>=0\)

=>\(\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{a}-\sqrt{c}\right)^2>=0\)(luôn đúng)

1 tháng 12 2016
  • \(A=\sqrt{11-2\sqrt{10}}=\sqrt{\left(\sqrt{10}-1\right)^2}=\sqrt{10}-1\)
  • \(B=\left(\sqrt{28}-2\sqrt{4}+\sqrt{7}\right).\sqrt{7}+7\sqrt{7}=\left(2\sqrt{7}-2\sqrt{4}+\sqrt{7}\right).\sqrt{7}+7\sqrt{7}\)

\(=\left(3\sqrt{7}-4\right).\sqrt{7}+7\sqrt{7}=3\sqrt{7}+3\sqrt{7}=6\sqrt{7}\)

  • \(C=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\frac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)

\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

  • \(D=0,2.\sqrt{10^2.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=2\sqrt{3}+2\left(\sqrt{3}-\sqrt{5}\right)=4\sqrt{3}-2\sqrt{5}\)
12 tháng 10 2021

c: Ta có: \(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)

\(=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\sqrt{10}\)