\(\frac{2x^2-3x-2}{x^2-4}\)

\(=\frac{2x^2-4x+x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{2x\left(x-2\right)+\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{\left(2x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{2x+1}{x-2}\)

2x²-3x-2/x²-4=2x²-4x+x-2/x²-4=2x(x-2)+(x-2)/(x+2)(x-2)=(x-2)(2x-1)/(x+2)(x-2)=2x+1/x+2

\(ĐKXĐ:\)\(x\ne0,x\ne2\)

\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)

\(\Leftrightarrow\)\(\frac{x\left(x+2\right)-1\left(x-2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)

\(\Rightarrow\)\(x\left(x+2\right)-1\left(x-2\right)=2\)

\(\Leftrightarrow\)\(x^2+2x-x+2=2\)

\(\Leftrightarrow\)\(x^2+x+2-2=0\)

\(\Leftrightarrow\)\(x^2+x=0\)

\(\Leftrightarrow\)\(x\left(x+1\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\left(KTMĐKXĐ\right)\\x=1\left(TMĐKXĐ\right)\end{cases}}\)

Vậy.....

bài toán bị lỗi rồi bạn ơi

\(\frac{2x-1}{3}+\frac{5x+2}{2}=\frac{3x-5}{6}\)

\(\Leftrightarrow\)\(\frac{2\left(2x-1\right)+3\left(5x+2\right)}{6}=\frac{3x-5}{6}\)

\(\Rightarrow\)\(2\left(2x-1\right)+3\left(5x+2\right)=3x-5\)

\(\Leftrightarrow\)\(4x-2+15x+6=3x-5\)

\(\Leftrightarrow\)\(4x-2+15x+6-3x+5=0\)

\(\Leftrightarrow\)\(16x+9=0\)

\(\Leftrightarrow\)\(16x=-9\)

\(\Leftrightarrow\)\(x=\frac{-9}{16}\)

\(2016^{2016}\equiv3^{2016}\left(mod11\right)\)

\(3^{2016}=\left(3^5\right)^{403}.3=243^{403}.3\equiv1^{403}.3\left(mod11\right)\equiv3\left(mod11\right)\)

\(P=\left(\frac{1}{x}+\frac{1}{y}\right).\sqrt{1+x^2y^2}\)

\(\rightarrow P>2.\sqrt{\frac{1}{x}.\frac{1}{y}}.\sqrt{1+\left(xy\right)^2}\)

\(\rightarrow P>2.\sqrt{\frac{1}{xy}}.\sqrt{1+\left(xy\right)^2}\)

\(\rightarrow P>2\sqrt{\frac{1}{xy}+xy}\)

Đặt \(xy=t\)

\(\rightarrow P>2\sqrt{\frac{1}{t}+t}\)

Ta có :

\(1>x+y>2\sqrt{xy}\)

\(\rightarrow\sqrt{xy}< \frac{1}{2}\)

\(\rightarrow xy< \frac{1}{4}\)

\(\rightarrow t< \frac{1}{4}\)

Lại có :

\(\frac{1}{t}+t=\frac{15}{16t}+\left(\frac{1}{16}+t\right)\)

\(\rightarrow\frac{1}{t}+t>\frac{15}{16.\frac{1}{4}}+2\sqrt{\frac{1}{16}.t}\)

\(\rightarrow\frac{1}{t}+t>\frac{17}{4}\)

\(\rightarrow B>2.\sqrt{\frac{17}{4}}\)

\(\rightarrow B>\sqrt{17}\)

Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)

câu trả lời bằng hình