1.

\(\left\{{}\begin{matrix}x>2\\\frac{5}{2}+3\le x+\frac{3}{2}x\\2x\le5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>2\\\frac{5}{2}x\ge\frac{11}{2}\\x\le\frac{5}{2}\end{matrix}\right.\) \(\Rightarrow\frac{11}{5}\le x\le\frac{5}{2}\)

\(\Rightarrow a+b=\frac{11}{5}+\frac{5}{2}=D\)

2.

\(\left\{{}\begin{matrix}6x-4x>7-\frac{5}{7}\\4x-2x< 25-\frac{3}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>\frac{22}{7}\\x< \frac{47}{4}\end{matrix}\right.\)

\(\Rightarrow\frac{22}{7}< x< \frac{47}{4}\Rightarrow x=\left\{4;5...;11\right\}\) có 8 giá trị

3.

\(\left\{{}\begin{matrix}5x-4x< 5+2\\x^2< x^2+4x+4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x< 7\\x>-1\end{matrix}\right.\)

\(\Rightarrow-1< x< 7\Rightarrow x=\left\{0;1;...;6\right\}\)

\(\Rightarrow\sum x=1+2+...+6=21\)

4.

\(\left\{{}\begin{matrix}x^2-2x+1\le8-4x+x^2\\x^3+6x^2+12x+8< x^3+6x^2+13x+9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x\le7\\x\ge-1\end{matrix}\right.\) \(\Rightarrow-1\le x\le\frac{7}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}x_{min}=-1\\x_{max}=3\end{matrix}\right.\) \(\Rightarrow S=2\)

5.

\(\left\{{}\begin{matrix}x>\frac{1}{2}\\x< m+2\end{matrix}\right.\)

Hệ đã cho có nghiệm khi và chỉ khi:

\(m+2>\frac{1}{2}\Rightarrow m>-\frac{3}{2}\)

1/ ĐKXĐ:...

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y-2}=4\\\frac{12}{x}+\frac{3}{y-2}=3\end{matrix}\right.\) \(\Rightarrow\frac{10}{x}=-1\Rightarrow x=-10\)

\(\frac{4}{-10}+\frac{1}{y-2}=1\Rightarrow\frac{1}{y-2}=\frac{7}{5}\Rightarrow y-2=\frac{5}{7}\Rightarrow y=\frac{19}{7}\)

2/ ĐKXĐ:...

Đặt \(\left\{{}\begin{matrix}\frac{1}{2x-y}=a\\\frac{1}{x+y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a-b=0\\3a-6b=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{9}\\b=\frac{2}{9}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{2x-y}=\frac{1}{9}\\\frac{1}{x+y}=\frac{2}{9}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x-y=9\\x+y=\frac{9}{2}\end{matrix}\right.\) \(\Rightarrow...\)

3/ \(\Leftrightarrow\left\{{}\begin{matrix}5x+10y=3x-1\\2x+4=3x-6y-15\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-x+6y=-19\end{matrix}\right.\) \(\Rightarrow...\)

4/ Bạn tự giải

a/ \(\Leftrightarrow\left\{{}\begin{matrix}3x-4y=11\\-x-10y=-15\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=5\\y=1\end{matrix}\right.\)

b/ \(\Leftrightarrow\left\{{}\begin{matrix}x+y=8\\\frac{2x}{3}+\frac{x}{4}-\frac{y}{6}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=8\\\frac{11}{12}x-\frac{y}{6}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=8\\11x-2y=12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{28}{13}\\y=\frac{76}{13}\end{matrix}\right.\)

ĐKXĐ: ...

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{4}{x+1}+\frac{1}{y}=\frac{4}{x}\\\frac{2}{x+1}+\frac{4}{x}=\frac{3}{y}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{12}{x+1}+\frac{3}{y}=\frac{12}{x}\\\frac{2}{x+1}+\frac{4}{x}=\frac{3}{y}\end{matrix}\right.\)

\(\Rightarrow\frac{14}{x+1}+\frac{4}{x}=\frac{12}{x}\Leftrightarrow\frac{14}{x+1}=\frac{8}{x}\)

Tới đây chắc bạn giải tiếp được

ĐKXĐ:...

a) \(\left\{{}\begin{matrix}\frac{x}{2}=\frac{y}{3}\\\frac{x+8}{y+4}=\frac{9}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{2y}{3}\\\frac{\frac{2y}{3}+8}{y+4}=\frac{9}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{-12}{19}\\x=\frac{-8}{19}\end{matrix}\right.\)

Vậy...

b) \(\left\{{}\begin{matrix}0,75x-3,2y=10\\x\sqrt{3}-y\sqrt{2}=4\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3,2y+10}{0,75}\\\frac{\left(3,2y+10\right)\sqrt{3}}{0,75}-y\sqrt{2}=4\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{\frac{16\sqrt{3}}{5}y+10\sqrt{3}-\frac{3\sqrt{2}}{4}y}{0,75}=4\sqrt{3}\\x=\frac{3,2y+10}{0,75}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y\left(\frac{16\sqrt{3}}{5}-\frac{3\sqrt{2}}{4}\right)+10\sqrt{3}=3\sqrt{3}\\x=\frac{3,2y+10}{0,75}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{-140\sqrt{3}}{64\sqrt{3}-15\sqrt{2}}\\x=\frac{\frac{-448\sqrt{3}}{64\sqrt{3}-15\sqrt{2}}+10}{0,75}\end{matrix}\right.\)

Nghiệm đẹp lắm.

c) \(\left\{{}\begin{matrix}\frac{2x+3}{y-1}=\frac{4x+1}{2y+1}\\\frac{x+2}{y-1}=\frac{x-4}{y+2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+3\right)\left(2y+1\right)-\left(y-1\right)\left(4x+1\right)=0\\\left(x+2\right)\left(y+2\right)-\left(y-1\right)\left(x-4\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x+5y+4=0\\3x+6y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2y\\-12y+5y+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{4}{7}\\x=\frac{-8}{7}\end{matrix}\right.\)

Vậy...