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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:
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}=3\)
\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
a: \(BD\cdot CE\cdot BC\)
\(=\dfrac{HB^2}{AB}\cdot\dfrac{HC^2}{AC}\cdot\dfrac{AB\cdot AC}{AH}\)
\(=\dfrac{AH^4}{AH}=AH^3\)
b: \(\dfrac{BD}{CE}=\dfrac{HB^2}{AB}:\dfrac{HC^2}{AC}=\dfrac{HB^2}{AB}\cdot\dfrac{AC}{HC^2}=\dfrac{AB^4}{AB}\cdot\dfrac{AC}{AC^4}=\dfrac{AB^3}{AC^3}\)
câu a) bn có thể vào câu hỏi tương tự xem, cái này làm vui thôi
Ta có: \(BN=\frac{BH^2}{AB};CM=\frac{CH^2}{AC};AB.AC=AH.BC;BH.CH=AH^2\)
\(\sqrt[3]{BC^2}=\sqrt[3]{BN^2}+\sqrt[3]{CM^2}\)
\(\Leftrightarrow\)\(BC^2=BN^2+CM^2+3\sqrt[3]{\left(BN.CM\right)^2}\left(\sqrt[3]{BN^2}+\sqrt[3]{CM^2}\right)\)
\(\Leftrightarrow\)\(BC^2=BH^2-NH^2+CH^2-MH^2+3\sqrt[3]{\left(\frac{\left(BH.CH\right)^2}{AB.AB}\right)^2}.\sqrt[3]{BC^2}\)
\(\Leftrightarrow\)\(BC^2=\left(BH^2+CH^2\right)-\left(NH^2+MH^2\right)+3\sqrt[3]{\left(\frac{AH^4}{AH.BC}\right)^2}.\sqrt[3]{BC^2}\)
\(\Leftrightarrow\)\(BC^2=\left(BH+CH\right)^2-2BH.CH-\left(NH^2+MH^2\right)+3\sqrt[3]{\frac{AH^6}{BC^2}}.\sqrt[3]{BC^2}\)
\(\Leftrightarrow\)\(BC^2=BC^2-2AH^2-AH^2+3AH^2\) ( do \(NH^2=AM^2\) )
\(\Leftrightarrow\)\(BC^2=BC^2\) ( luôn đúng )
\(\Rightarrow\)\(\sqrt[3]{BC^2}=\sqrt[3]{BN^2}+\sqrt[3]{CM^2}\) đúng
b) bằng một cách nào đó \(\Delta NBH\) đã đồng dạng với \(\Delta ABC\) ( có góc B chung ) \(\Rightarrow\)\(\frac{BN}{AB}=\frac{BH}{BC}\)
Tương tự: \(\Delta MHC~\Delta ABC\) ( có góc C chung ) \(\Rightarrow\)\(\frac{CM}{AC}=\frac{CH}{BC}\)
\(\Rightarrow\)\(\frac{BN}{AB}+\frac{CM}{AC}=\frac{BH+CH}{BC}=1\)
\(\Leftrightarrow\)\(BN.AC+CM.AB=AB.AB\)
\(\Leftrightarrow\)\(BN\sqrt{AC^2}+CM\sqrt{AB^2}=AB.AC\)
\(\Leftrightarrow\)\(BN\sqrt{CH.BC}+CM\sqrt{BH.BC}=AH.BC\)
\(\Leftrightarrow\)\(BN\sqrt{CH}+CM\sqrt{BH}=AH\sqrt{BC}\) ( chia 2 vế cho \(\sqrt{BC}\ne0\) ) đpcm
c.
\(\left(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right)^2=2010\)
\(\leftrightarrow\) \(x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+1+x^2+y^2+x^2y^2=2010\)
\(\leftrightarrow\)\(x^2+x^2y^2+2x\sqrt{1+y^2}.y\sqrt{1+x^2}+y^2+x^2y^2=2009\)
\(\leftrightarrow\) \(\left(x\sqrt{1+y^2}+y\sqrt{1+x^2}\right)^2=2009\)
\(\leftrightarrow\) \(x\sqrt{1+y^2}+y\sqrt{1+x^2}=\sqrt{2009}\)