\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{5\sqrt{3}}{2}=\dfrac{5\sqrt{3}}{2}-9\sqrt{3}=\dfrac{5\sqrt{3}-18\sqrt{3}}{2}=\dfrac{-13\sqrt{3}}{2}\)

\(=\dfrac{1}{2}.4\sqrt{3}-2.5\sqrt{3}-\sqrt{3}+5.\dfrac{\sqrt{3}}{2}\)

\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{5\sqrt{3}}{2}\)

\(=-9\sqrt{3}+\dfrac{5\sqrt{3}}{2}=\dfrac{-18\sqrt{3}+5\sqrt{3}}{2}=-\dfrac{13\sqrt{3}}{2}\)

Giải hpt:

Đặt: \(\left[{}\begin{matrix}\sqrt{x-1}=a\\y+1=b\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}3a-2b=-1\\5a-9b=-13\end{matrix}\right.< =>\left\{{}\begin{matrix}15a-10b=-5\\15a-27b=-39\end{matrix}\right.< =>\left\{{}\begin{matrix}b=2\\15a-27\cdot2=-39\end{matrix}\right.< =>\left\{{}\begin{matrix}b=2\\a=1\end{matrix}\right.\)

Thay: \(\left[{}\begin{matrix}\sqrt{x-1}=1\\y+1=2\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

\(A=\sqrt{2a\left(b+1\right)}+\sqrt{2b\left(c+1\right)}+\sqrt{2c\left(a+1\right)}\)

\(A=\dfrac{1}{\sqrt{2}}\sqrt{4a\left(b+1\right)}+\dfrac{1}{\sqrt{2}}\sqrt{4b\left(c+1\right)}+\dfrac{1}{\sqrt{2}}\sqrt{4c\left(a+1\right)}\)

\(A\le\dfrac{1}{2\sqrt{2}}\left(4a+b+1\right)+\dfrac{1}{2\sqrt{2}}\left(4b+c+1\right)+\dfrac{1}{2\sqrt{2}}\left(4c+a+1\right)\)

\(A\le\dfrac{1}{2\sqrt{2}}\left[5\left(a+b+c\right)+3\right]=2\sqrt{2}\)

\(A_{max}=2\sqrt{2}\) khi \(a=b=c=\dfrac{1}{3}\)

\(a,ĐK:x+y\ne0;x\ne y\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+y}+\dfrac{4}{x-y}=\dfrac{14}{3}\left(1\right)\\\dfrac{3}{x+y}+\dfrac{4}{x-y}=5\left(2\right)\end{matrix}\right.\\ \left(2\right)-\left(1\right)=\dfrac{1}{x+y}=\dfrac{1}{3}\\ \Leftrightarrow x+y=3\\ \Leftrightarrow x=3-y\\ \text{Thay vào }\left(1\right)\Leftrightarrow\dfrac{2}{3}+\dfrac{4}{3-2y}=\dfrac{14}{3}\\ \Leftrightarrow\dfrac{4}{3-2y}=4\\ \Leftrightarrow3-2y=1\\ \Leftrightarrow y=1\Leftrightarrow x=2\)

Vậy hệ có nghiệm \(\left(x;y\right)=\left(2;1\right)\)

\(b,ĐK:y\ne-\dfrac{1}{2};x-2y\ne0\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{x-2y}+\dfrac{y}{1+2y}=3\left(1\right)\\\dfrac{6}{x-2y}-\dfrac{8}{1+2y}=-2\left(2\right)\end{matrix}\right.\\ \left(1\right)-\left(2\right)=\dfrac{y+8}{2y+1}=5\\ \Leftrightarrow y+8=10y+5\Leftrightarrow y=\dfrac{1}{3}\\ \text{Thay vào }\left(1\right)\Leftrightarrow\dfrac{6}{x-\dfrac{2}{3}}+\dfrac{\dfrac{1}{3}}{\dfrac{5}{3}}=3\\ \Leftrightarrow\dfrac{6}{x-\dfrac{2}{3}}=\dfrac{14}{5}\\ \Leftrightarrow x-\dfrac{2}{3}=\dfrac{15}{7}\Leftrightarrow x=\dfrac{59}{21}\)

Vậy hệ có nghiệm \(\left(x;y\right)=\left(\dfrac{59}{21};\dfrac{1}{3}\right)\)

\(c,HPT\Leftrightarrow\left\{{}\begin{matrix}xy-5x+3y-15=xy\\2xy+30x-9y-135=2xy\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-5x+3y=15\\10x-3y=45\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=12\\y=25\end{matrix}\right.\)

\(12,ĐK:x,y\ne0\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{2}{y}=4\\\dfrac{6}{x}-\dfrac{2}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{10}{x}=5\\\dfrac{2}{x}+\dfrac{1}{y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\\dfrac{1}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\left(tm\right)\)

\(13,\Leftrightarrow\left\{{}\begin{matrix}3\left(x+1\right)+2\left(x+2y\right)=4\\8\left(x+1\right)-2\left(x+2y\right)=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11\left(x+1\right)=22\\3\left(x+1\right)+2\left(x+2y\right)=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\6+2+4y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(14,ĐK:x+y\ne0;y\ne1\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x+y}+\dfrac{1}{y-1}=5\\\dfrac{4}{x+y}-\dfrac{8}{y-1}=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+y}-\dfrac{2}{y-1}=-1\\\dfrac{9}{y-1}=9\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+2}=1\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2=1\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\left(tm\right)\)

\(15,ĐK:x\ge-1\\ HPT\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+\sqrt{x+1}=4\\2\left(x+y\right)-6\sqrt{x+1}=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\sqrt{x+1}=14\\2\left(x+y\right)+\sqrt{x+1}=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3\left(tm\right)\\6+2y+2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\left(tm\right)\)

\(16,ĐK:x\ne1;y\ne-2\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x}{x-1}-\dfrac{2}{y+2}=4\\\dfrac{4x}{x-1}+\dfrac{2}{y+2}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7x}{x-1}=14\\\dfrac{2x}{x-1}+\dfrac{1}{y+2}=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\\dfrac{1}{y+2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\left(tm\right)\)

\(17,ĐK:x\ge0;y\ge1\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+2\sqrt{y-1}=5\\8\sqrt{x}-2\sqrt{y-1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}=9\\\sqrt{x}+2\sqrt{y-1}=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\\sqrt{y-1}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)

\(18,\Leftrightarrow\left\{{}\begin{matrix}8x-2\left|y+2\right|=6\\x+2\left|y+2\right|=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=9\\x+2\left|y+2\right|=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\\left|y+2\right|=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\\left[{}\begin{matrix}y=-1\\y=-3\end{matrix}\right.\end{matrix}\right.\\ 20,ĐK:y\ne1\\ HPT\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{3}{y-1}=5\\12x-\dfrac{3}{y-1}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}14x=14\\2x+\dfrac{3}{y-1}=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\\dfrac{3}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\left(tm\right)\)

\(21,ĐK:x\ne-1\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{9}{x+1}-6y=-3\\\dfrac{10}{x+1}+6y=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{19}{x+1}=19\\\dfrac{3}{x+1}-2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\3-2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\left(tm\right)\)

a: \(BC=\sqrt{AB^2+AC^2}=10\left(cm\right)\)

b: Xét ΔABI có

AH là đường cao

AH là đường trung tuyến

Do đó: ΔABI cân tại A

hay AB=AI