chứng minh:
\(\frac{\left(5\sqrt{3}+\sqrt{50}\right).\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
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chứng minh:
\(\frac{\left(5\sqrt{3}+\sqrt{50}\right).\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
\(\frac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(2-2\sqrt{2}.\sqrt{3}+3\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}=\frac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{5\left(\sqrt{3}-\sqrt{2}\right)}=\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2=3-2=1\)
\(VT=\frac{\left(5\sqrt{3}+5\sqrt{2}\right).\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)
\(=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)^2.\left(5-2\sqrt{6}\right)}{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5\sqrt{3}-5\sqrt{2}\right)}\)\(=\frac{\left(75+50\sqrt{6}+50\right).\left(5-2\sqrt{6}\right)}{75-50}\)
\(=\frac{25\left(5+2\sqrt{6}\right).\left(5-2\sqrt{6}\right)}{25}=5^2-\left(2\sqrt{6}\right)^2\)\(=25-24=1=VP\)
bn chép lại đề nhé
\(=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)
\(=\frac{\left(75+50\sqrt{6}+50\right)\left(\sqrt{3}-\sqrt{2}\right)}{75-50}\)
\(A=\dfrac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\\ =\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}\\ =\dfrac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{\sqrt{3}-\sqrt{2}}\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\\ =3-2\\ =1\)
Vậy \(A=1\)
\(\frac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
\(=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)
\(=\frac{5\left(\sqrt{3}+\sqrt{2}\right)\left(3-2.\sqrt{3}.\sqrt{2}+2\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\frac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{5\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=1\)
DÀI QUÁ MK KO GHI ĐƯỢC NÊN VIẾT KQ LUÔN NHA !!!
ĐẲNG THỨC ĐÓ = 1 NHA Hatsune Miku !
\(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)
\(\Leftrightarrow\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-\sqrt{1}=4\)
\(\Leftrightarrow\sqrt{25}-\sqrt{1}=4\Leftrightarrow5-1=4\)(đúng)
Vậy \(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)(đpcm)
\(M=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{11-6\sqrt{2}}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{2-6\sqrt{2}+9}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{\left(3-\sqrt{2}\right)^2}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+3-\sqrt{2}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{6}}\)
\(=\sqrt{16+32\sqrt{6}}\)
22) \(\frac{1}{\sqrt{5}+\sqrt{2}}+\frac{1}{\sqrt{5}-\sqrt{2}}\)
\(=\frac{\left(\sqrt{5}-\sqrt{2}\right)+\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}\)
\(=\frac{2\sqrt{5}}{\sqrt{5^2}-\sqrt{2^2}}\)
\(=\frac{2\sqrt{5}}{5-2}=\frac{2\sqrt{5}}{3}\)
cm > hay < ?