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Ta có:
\(\frac{12}{\sqrt{13}}+\frac{13}{\sqrt{12}}=\frac{12\sqrt{13}}{13}+\frac{13\sqrt{12}}{12}=\frac{13\sqrt{13}-\sqrt{13}}{13}+\frac{12\sqrt{12}+\sqrt{12}}{12}\)\(=\sqrt{12}+\sqrt{13}+\frac{1}{\sqrt{12}}-\frac{1}{\sqrt{13}}>\sqrt{12}+\sqrt{13}\)
Với n > 0 Ta có:
\(\frac{1}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}+\sqrt{n}}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}\)
\(=\sqrt{n+1}+\sqrt{n}\)
\(\Rightarrow\frac{1}{\sqrt{16}-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+...+\frac{1}{\sqrt{10}-\sqrt{9}}\)
\(=\sqrt{16}+\sqrt{15}-\sqrt{15}-\sqrt{14}+...+\sqrt{10}+\sqrt{9}\)
\(\sqrt{16}+\sqrt{9}=3+4=7\)
C = \(\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}}-\sqrt{12+2\sqrt{11}}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
C = \(\left(\sqrt{12+2\sqrt{\left(\sqrt{13}+1\right)^2}}-\sqrt{\left(\sqrt{11}+1\right)^2}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
C = \(\left(\sqrt{14+2\sqrt{13}}-\left(\sqrt{11}+1\right)\right)\left(\sqrt{11}+\sqrt{13}\right)\)
C = \(\left(\sqrt{\left(\sqrt{13}+1\right)^2}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)
C = \(\left(\sqrt{13}+1-\sqrt{11}-1\right)\left(\sqrt{13}+\sqrt{11}\right)\)
C \(\left(\sqrt{13}-\sqrt{11}\right)\left(\sqrt{13}+\sqrt{11}\right)\) = \(13-11\) = \(2\)
cho P = \(\frac{\sqrt{x}+2}{\sqrt{x}+1}\) , Tìm GTLN của P
Đặt \(\sqrt{12}=a;\sqrt{13}=b\)
Theo đề, ta có:
\(\dfrac{a^2}{b}+\dfrac{b^2}{a}>a+b\)
\(\Leftrightarrow a^2+b^2-a^2-2ab-b^2>0\)
\(\Leftrightarrow2ab< 0\)(đúng)