\(P=\left(\dfrac{x}{x-2}+\dfrac{1}{x^2-4}\right):\dfrac{x+1}{x+2}\left(x\ne-1;x\ne\pm2\right)\)

a. \(=\left[\dfrac{x\left(x+2\right)}{x^2-4}+\dfrac{1}{x^2-4}\right].\dfrac{x+2}{x+1}\)

\(=\dfrac{x^2+2x+1}{x^2-4}.\dfrac{x+2}{x+1}\)

\(=\dfrac{\left(x+1\right)^2}{\left(x+2\right)\left(x-2\right)}.\dfrac{x+2}{x+1}\)

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

b. Thay x = 1/2 vào P, ta đc:

\(P=\dfrac{\dfrac{1}{2}+1}{\dfrac{1}{2}-2}=\dfrac{\dfrac{3}{2}}{-\dfrac{3}{2}}=-1\)

\(a.a\ne0;a\ne\pm3\)

\(b.P=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\left(1-\dfrac{6a-18}{a^2-9}\right)\)

\(=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}.\left[1-\dfrac{6\left(a-3\right)}{\left(a+3\right)\left(a-3\right)}\right]\)

\(=\dfrac{a+3}{2a}.\left(1-\dfrac{6}{a+3}\right)\)

\(=\dfrac{a+3}{2a}-\dfrac{a+3}{2a}.\dfrac{6}{a+3}\)

\(=\dfrac{a+3}{2a}-\dfrac{6}{2a}\)

\(=\dfrac{a-3}{2a}\)

c. Với: \(P=0\Leftrightarrow\dfrac{a-3}{2a}=0\)

\(\Leftrightarrow a-3=0\)

\(\Leftrightarrow a=3\left(loại\right)\)

Vậy ko có a thỏa mãn để P = 0

Với \(P=1\Leftrightarrow\dfrac{a-3}{2a}=1\)

\(\Leftrightarrow2a=a-3\)

\(\Leftrightarrow a=-3\) (loại)

Vậy ......

\(2x-\sqrt{x}=0\left(Đk:x\ge0\right)\)

\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\2\sqrt{x}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)

a. \(M=\dfrac{11x^2+11x}{x^2-1}\left(đk:x\ne\pm1\right)\)

\(=\dfrac{11x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{11x}{x-1}\)

b. \(\dfrac{11x}{x-1}=x.\dfrac{11}{x-1}\)

\(M\in Z\Leftrightarrow x.\dfrac{11}{x-1}\in Z\Leftrightarrow\dfrac{11}{x-1}\in Z\) (vì x nguyên)

\(\Leftrightarrow x-1\inƯ_{\left(11\right)}=\left\{\pm1;\pm11\right\}\)

+) \(x-1=1\Leftrightarrow x=2\)

+) \(x-1=-1\Leftrightarrow x=0\)

+) \(x-1=11\Leftrightarrow x=12\)

+) \(x-1=-11\Leftrightarrow x=-10\)

\(\dfrac{5xy-4y}{2x^2y^3}+\dfrac{3xy+4y}{2x^2y^3}=\dfrac{5xy-4y+3xy+4y}{2x^2y^3}=\dfrac{8xy}{2x^2y^3}=\dfrac{4}{xy^2}\)

a. Xét 2 tam giác vuông: \(\Delta OHM\) và \(\Delta OKM\) có:

\(\left\{{}\begin{matrix}OM.chung\\\widehat{HOM}=\widehat{KOM}\left(gt\right)\end{matrix}\right.\)

\(\Rightarrow\Delta OHM=\Delta OKM\) (cạnh huyền - góc nhọn)

\(\Rightarrow OH=OK\)

b. Tương tự câu a, suy ra: \(\Delta OHI=\Delta OKI\left(c.g.c\right)\)

\(\Rightarrow IH=IK\) hay I là trung điểm của HK

\(\left(2x^3-3x^2-11x+6\right):\left(x-3\right)\)

\(=\dfrac{2x^3-3x^2-11x+6}{x}-\dfrac{2x^3-3x^2-11x+6}{3}\)

\(=2x^2-3x-11+\dfrac{6}{x}-\dfrac{2}{3}x^3+x^2+\dfrac{11}{3}x-2\)

\(=-\dfrac{2}{3}x^3+3x^2+\dfrac{2}{3}x-13+\dfrac{6}{x}\)

a) \(x\left(2x-3\right)-6+4x=0\)

\(\Leftrightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)

b) \(x^3+2x^2+x=0\)

\(\Leftrightarrow x\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow x\left(x+1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

c) \(x^2-6x+8=0\)

\(\Leftrightarrow x^2-2x-4x+8=0\)

\(\Leftrightarrow x\left(x-2\right)-4\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

B. 48 km/h