\(\hept{\begin{cases}\frac{3}{5}x-\frac{2}{5}y+\frac{5}{3}x-y-x=1\\\frac{2}{3}x-y+2x-\frac{3}{2}y-y=1\end{cases}}\)<=>\(\hept{\begin{cases}\frac{19}{15}x-\frac{7}{5}y=1\\\frac{8}{3}x-\frac{7}{2}y=1\end{cases}}\)<=>x=3;y=2

\(\hept{\begin{cases}4x^2-5y-5=4x^2-12x+9\\21x+6=10y-5-3x\end{cases}}\)<=>\(\hept{\begin{cases}12x-5y=14\\24x-10y=-11\Leftrightarrow12x-5y=-\frac{11}{2}\end{cases}}\)

=>pt vô nghiệm

\(\hept{\begin{cases}\sqrt{2x}+2\sqrt{3y}=5\\3\sqrt{2x}-\sqrt{3y}=\frac{9}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{2x}+2\sqrt{3y}=5\\6\sqrt{2x}-2\sqrt{3y}=9\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{2x}+2\sqrt{3y}=5\\7\sqrt{2x}=14\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{2x}+2\sqrt{3y}=5\\\sqrt{2x}=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{2x}+2\sqrt{3y}=5\\2x=2^2=4\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{2.2}+2\sqrt{3y}=5\\x=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2+2\sqrt{3y}=5\\x=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2\sqrt{3y}=3\\x=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{3y}=\frac{3}{2}\\x=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}3y=\left(\frac{3}{2}\right)^2=\frac{9}{4}\\x=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=\frac{3}{4}\\x=2\end{cases}}\)

Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\frac{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Áp dụng:

\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)

\(=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}=1-\frac{1}{\sqrt{n+1}}< 1\left(đpcm\right)\)

Ta có: \(\hept{\begin{cases}a;b;c\ge0\\a+b+c=1\end{cases}}\Rightarrow0\le a;b;c\le1\Rightarrow\hept{\begin{cases}a^2\le a\\b^2\le b\\c^2\le c\end{cases}}\)

\(\sqrt{5a+4}+\sqrt{5b+4}+\sqrt{5c+4}\)

\(=\sqrt{a+4a+4}+\sqrt{b+4b+4}+\sqrt{c+4c+4}\)

\(\ge\sqrt{a^2+4a+4}+\sqrt{b^2+4b+4}+\sqrt{c^2+4c+4}=a+2+b+2+c+2=7\)

\("="\Leftrightarrow a;b;c\) là hoán vị của 0;0;1

xét \(\Delta ABH\)vg tại H có

AB² = BH² + AH²   ( Đ/Lí py - ta - go )

30²  = BH²  + 24²

BH² = 324

BH= 18 cm

xét \(\Delta\)ABC vg tại A có AH \(\perp\)BC

AB² = BH . BC ( hệ thức về cạnh và đường cao trong tg vg )

30² = 18 . BC

BC = 50 cm

#mã mã#

=3.4142135

    3 + \(2\sqrt{2}\) 

= \(\left(\sqrt{2}\right)^2+2\sqrt{2}+1\)

= \(\left(\sqrt{2}+1\right)^2\)

#mã mã#