Bài 3: Liên hệ giữa phép nhân và phép khai phương

Áp dụng bđt Bunyakovsky:

\(A^2=\left(\sqrt{x+3}+\sqrt{5-x}\right)^2\)

\(\le\left(1^2+1^2\right)\left(x+3+5-x\right)=16\)

\(\Leftrightarrow A\le4\)

\("="\Leftrightarrow x=1\)

\(A=\sqrt{x+3}+\sqrt{5-x}\)

\(\Leftrightarrow A^2=\left(\sqrt{x+3}+\sqrt{5-x}\right)^2\)

Áp dụng BĐT Bunhiacopsky ta được :

\(A^2\le\left(1^2+1^2\right)\left(x+3+5-x\right)\)

\(\Leftrightarrow A^2\le16\)

\(\Leftrightarrow A\le4\) (đpcm)

Dấu "=" xảy ra khi (x + 3) = (5 - x) \(\Leftrightarrow\) 2x = 2 \(\Leftrightarrow\) x = 1

b: \(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)

\(=4\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}\)

\(=-4\sqrt{5\sqrt{3}}\)

\(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}=\dfrac{\sqrt{12+2.2.2\sqrt{3}+4}-\sqrt{12-2.2.2\sqrt{3}+4}}{\sqrt{2}}=\dfrac{2\sqrt{3}+2-2\sqrt{3}+2}{\sqrt{2}}=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\)

\(\sqrt{8+4\sqrt{3}-\sqrt{8-4\sqrt{3}}}\)

=\(\sqrt{12+2.2.2\sqrt{3}+4}-\sqrt{12-2.2.2\sqrt{3}+4}\)

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

=\(\dfrac{4}{\sqrt{2}}=2\sqrt{2}\)

\(\sqrt{10+\sqrt{19}}+\sqrt{10-\sqrt{19}}\)

\(=\sqrt{10^2-\left(\sqrt{19}\right)^2}\)

\(=\sqrt{100-19}\)

= \(\sqrt{81}\)

\(=9\)

\(\sqrt{10+\sqrt{19}}+\sqrt{10-\sqrt{19}}\)

=\(\sqrt{10^2-\left(\sqrt{19}\right)^2}\)

=\(\sqrt{100-19}\)

=\(\sqrt{81}\)

= 9 (đpcm)

\(a.A=\left(\sqrt{5-2\sqrt{6}}+\sqrt{2}\right)\sqrt{3}=\left(\sqrt{3-2\sqrt{3}.\sqrt{2}+2}+\sqrt{2}\right)\sqrt{3}=3\) \(b.B=\sqrt{4+2\sqrt{3}}+\sqrt{5+2\sqrt{6}}+\sqrt{2}=\sqrt{3+2\sqrt{3}+1}+\sqrt{3+2.\sqrt{3}.\sqrt{2}+2}+\sqrt{2}=\sqrt{3}+1+\sqrt{3}+\sqrt{2}+\sqrt{2}=2\sqrt{3}+2\sqrt{2}+1\) \(c.2+\sqrt{17-4\sqrt{9+4\sqrt{5}}}=2+\sqrt{17-4\sqrt{5+2.2\sqrt{5}+4}}=2+\sqrt{17-4\left(\sqrt{5}+2\right)}=2+\sqrt{5-2.2\sqrt{5}+4}=2+\sqrt{5}-2=\sqrt{5}\)

a.

\(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}\\ =\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\\ =\sqrt{81-17}\\ =\sqrt{64}\\=8\)

\(a.VT=\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{81-17}=8=VP\)

\(b.\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}=3\sqrt{3}-\sqrt{2}\) ( thiếu đề )

\(VT=\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}=\dfrac{1}{3-2\sqrt{3}.\sqrt{2}+2}+\dfrac{2}{3+2\sqrt{3}.\sqrt{2}+2}=\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2}{\sqrt{3}+\sqrt{2}}=\sqrt{3}+\sqrt{2}+2\sqrt{3}-2\sqrt{2}=3\sqrt{3}-\sqrt{2}=VP\)

Lời giải:

Ta có:

\(5,5+3\sqrt{2}=5,5+2\sqrt{\frac{9}{2}}=\frac{9}{2}+1+2\sqrt{\frac{9}{2}.1}=(\sqrt{\frac{9}{2}}+1)^2\)

\(\Rightarrow \sqrt{5,5+3\sqrt{2}}=\sqrt{\frac{9}{2}}+1\)

Tương tự:\(\sqrt{5,5-3\sqrt{2}}=\sqrt{\frac{9}{2}}-1\)

Do đó:

\(\frac{\sqrt{5,5+3\sqrt{2}}-\sqrt{5,5-3\sqrt{2}}}{6\sqrt{2}}=\frac{\sqrt{\frac{9}{2}}+1-(\sqrt{\frac{9}{2}}-1)}{6\sqrt{2}}\)

\(=\frac{2}{6\sqrt{2}}=\frac{1}{3\sqrt{2}}\)

\(\dfrac{\sqrt{5,5+3\sqrt{2}}-\sqrt{5,5-3\sqrt{2}}}{6\sqrt{2}}=\dfrac{\sqrt{2}\cdot\sqrt{5,5+3\sqrt{2}}-\sqrt{2}\cdot\sqrt{5,5-3\sqrt{2}}}{12}=\dfrac{\sqrt{11+2\cdot3\sqrt{2}}-\sqrt{11-2\cdot3\sqrt{2}}}{12}=\dfrac{\sqrt{9+2\cdot3\cdot\sqrt{2}+2}-\sqrt{9-2\cdot3\cdot\sqrt{2}+2}}{12}=\dfrac{\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}}{12}=\dfrac{3+\sqrt{2}-3+\sqrt{2}}{12}=\dfrac{2\sqrt{2}}{12}=\dfrac{\sqrt{2}}{6}\)

\(\dfrac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}=\dfrac{\sqrt{a}\left(\sqrt{ab}+1\right)-\sqrt{b}\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}=\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{ab}-1}\)

P/s : Mình sửa đề .