cho 3 số a,b,c t/m a+b+c=0 và -1<a\(\le b\le c\)<1. c/m a2+b2+c2<2
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TL:
\(\sqrt{8-3\sqrt{7}}-\sqrt{8+3\sqrt{7}}\)
\(=\frac{8-3\sqrt{7}-8-3\sqrt{7}}{\sqrt{8-3\sqrt{7}}+\sqrt{8+3\sqrt{7}}}\)
\(=\frac{-6\sqrt{7}}{\sqrt{8-3\sqrt{7}}+\sqrt{8+3\sqrt{7}}}\)
Cho \(A=\sqrt{8-3\sqrt{7}}-\sqrt{8+3\sqrt{7}}\)
CACH 1 : \(\Rightarrow A\sqrt{2}=\sqrt{16-6\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)
\(\Rightarrow A\sqrt{2}=\sqrt{9-2.3.\sqrt{7}+7}-\sqrt{9+2.3.\sqrt{7}+7}\)
\(\Rightarrow A\sqrt{2}=\sqrt{\left(3-\sqrt{7}\right)^2}-\sqrt{\left(3+\sqrt{7}\right)^2}\)
\(\Rightarrow A\sqrt{2}=|3-\sqrt{7}|-|3+\sqrt{7}|\)
\(\Rightarrow A\sqrt{2}=3-\sqrt{7}-3-\sqrt{7}=-2\sqrt{7}=-\sqrt{28}\)
\(\Rightarrow A=-\sqrt{14}\)
CACH 2 : \(A^2=8-3\sqrt{7}+8+3\sqrt{7}-2.\sqrt{8^2-\left(3\sqrt{7}\right)^2}\)
\(\Rightarrow A^2=16-2\sqrt{64-63}=16-2=14\)
\(\Rightarrow A=\sqrt{14}\) hoặc \(A=-\sqrt{14}\)
Mà \(8+3\sqrt{7}>8-3\sqrt{7}\) \(\Rightarrow\sqrt{8+3\sqrt{7}}>\sqrt{8-3\sqrt{7}}\)
Vây A âm \(\Rightarrow A=-\sqrt{14}\)
a,ĐKXĐ:\(\orbr{\begin{cases}x\ge2\\x\le-2\end{cases}}\)
\(\sqrt{x-2}.\sqrt{x+2}-\sqrt{x-2}=0\)
\(\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
\(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=4-\sqrt{x}-\sqrt{y}\left(đk:x;y>0\right)\)
\(\Leftrightarrow\frac{1}{\sqrt{x}}+\sqrt{x}+\frac{1}{\sqrt{y}}+\sqrt{y}=4\)
Do x,y là các số thực dương nên sử dụng BĐT AM-GM cho 2 số không âm ta có :
\(\frac{1}{\sqrt{x}}+\sqrt{x}\ge2\sqrt{\frac{1}{\sqrt{x}}.\sqrt{x}}=2\)
\(\frac{1}{\sqrt{y}}+\sqrt{y}\ge2\sqrt{\frac{1}{\sqrt{y}}.\sqrt{y}}=2\)
Cộng theo vế các bất đẳng thức cùng chiều ta được :
\(\frac{1}{\sqrt{x}}+\sqrt{x}+\frac{1}{\sqrt{y}}+\sqrt{y}\ge2+2=4\)
Dấu = xảy ra khi và chỉ khi \(\hept{\begin{cases}\frac{1}{\sqrt{x}}=\sqrt{x}\Leftrightarrow x=1\\\frac{1}{\sqrt{y}}=\sqrt{y}\Leftrightarrow y=1\end{cases}\Leftrightarrow}x=y=1\)
Vậy nghiệm của phương trình trên là \(x=y=1\)
\(a,\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}\)
\(=|\sqrt{2}-3|.\sqrt{9+6\sqrt{2}+2}\)
\(=(3-\sqrt{2}).\left(\sqrt{\left(3+\sqrt{2}\right)^2}\right)\)
\(=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)\)
\(=9-2=7\)
\(b,\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\frac{1}{3-\sqrt{3}}}\)
\(=\left(3-\sqrt{3}\right).\frac{\sqrt{1}}{\sqrt{3-\sqrt{3}}}\)
\(=\frac{3-\sqrt{3}}{\sqrt{3-\sqrt{3}}}\)
\(=\sqrt{3-\sqrt{3}}\)
\(c,-\frac{2}{3}\sqrt{\frac{\left(a-b\right)^3.b^5}{c}}.\frac{9}{4}\sqrt{\frac{c^3}{2\left(a-b\right)}}.\sqrt{98b}\)
\(=-\frac{2}{3}.\frac{\sqrt{\left(a-b\right)^3.b^5}}{\sqrt{c}}.\frac{9}{4}.\frac{\sqrt{c^3}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)
\(=-\frac{2}{3}.\frac{\left(a-b\right)b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{9}{4}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)
\(=-\frac{2}{3}.\frac{9}{4}.7.\frac{\left(a-b\right).b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.\sqrt{2b}\)
\(=-\frac{21}{2}.\left(a-b\right).b^2\sqrt{b}.c.\sqrt{b}\)
\(=\frac{-21}{2}.\left(a-b\right).b^3.c\)
\(d,\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}.2\sqrt{2}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\sqrt{2}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+4\sqrt{2}\right).2\sqrt{6}\)
\(=2.6-18\sqrt{2}+16\sqrt{3}\)
\(=12-18\sqrt{2}+16\sqrt{3}\)
\(F=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2012}+\sqrt{2011}}\)
\(F=\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{\sqrt{2012}-\sqrt{2011}}{\left(\sqrt{2012}+\sqrt{2011}\right)\left(\sqrt{2012}-\sqrt{2011}\right)}\)
\(F=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{2012}-\sqrt{2011}}{2012-2011}\)
\(F=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2012}-\sqrt{2011}\)
\(F=\sqrt{2012}-\sqrt{1}\)
\(F=\sqrt{2012}-1\)
\(\sqrt{\left(5+2\sqrt{6}\right)}+\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{\left(2+2.\sqrt{2}.\sqrt{3}+3\right)}+\sqrt{3-2\sqrt{3}.\sqrt{5}+5}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}\)
\(=\sqrt{2}+\sqrt{3}+\sqrt{3}+\sqrt{5}\)
\(=\sqrt{2}+2\sqrt{3}+\sqrt{5}\)
Ta có: \(\sqrt{5}< \sqrt{16}\left(=4\right)\)
\(\Rightarrow\sqrt{5}-4< 0\)
\(mà2\sqrt{3}>0\)
\(\Rightarrow2\sqrt{3}>\sqrt{5}-4\)
Ta có: \(\sqrt{5}< \sqrt{12}\)và \(-4< 0\)
\(\Rightarrow\sqrt{5}-4< \sqrt{12}+0=\sqrt{12}\)
mà \(\sqrt{12}=\sqrt{4.3}=\sqrt{4}.\sqrt{3}=2\sqrt{3}\)
\(\Rightarrow2\sqrt{3}=\sqrt{5}-4\)