Ta có

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

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

Ta có: \(x^2+2x+2x\sqrt{x+3}=9-\sqrt{x+3}\)       \(\left(ĐK:x\ge-3\right)\)

    \(\Leftrightarrow\left(x^2+2x\sqrt{x+3}+x+3\right)+x+\sqrt{x+3}=12\)

    \(\Leftrightarrow\left(x+\sqrt{x+3}\right)^2+\left(x+\sqrt{x+3}\right)-12=0\)

    \(\Leftrightarrow\left(x+\sqrt{x+3}\right)\left(x+\sqrt{x+3}+1\right)-12=0\)

Đặt \(a=x+\sqrt{x+3}\)\(\Leftrightarrow\)\(a+1=x+\sqrt{x+3}+1\)     

Ta lại có: \(a.\left(a+1\right)-12=0\)

         \(\Leftrightarrow a^2+a-12=0\)

         \(\Leftrightarrow a^2-3a+4a-12=0\)

         \(\Leftrightarrow a\left(a-3\right)+4\left(a-3\right)=0\)

         \(\Leftrightarrow\left(a+4\right)\left(a-3\right)=0\)

         \(\Leftrightarrow\orbr{\begin{cases}a+4=0\\a-3=0\end{cases}}\)

+ \(a+4=0\)\(\Leftrightarrow\)\(x+\sqrt{x+3}+4=0\)

                            \(\Leftrightarrow\)\(x+4=-\sqrt{x+3}\)

                            \(\Leftrightarrow\)\(\left(x+4\right)^2=\left(-\sqrt{x+3}\right)^2\)

                            \(\Leftrightarrow\)\(x^2+8x+16=x+3\)

                            \(\Leftrightarrow\)\(x^2+7x+13=0\)

                            \(\Leftrightarrow\)\(\left(x^2+7x+\frac{49}{4}\right)+\frac{3}{4}=0\)

                            \(\Leftrightarrow\)\(\left(x+\frac{7}{2}\right)^2+\frac{3}{4}=0\)

   Vì \(\left(x+\frac{7}{2}\right)^2+\frac{3}{4}>0\forall x\)mà \(\left(x+\frac{7}{2}\right)^2+\frac{3}{4}=0\)

         \(\Rightarrow\)Phương trình \(\left(x+\frac{7}{2}\right)^2+\frac{3}{4}=0\)vô nghiệm

+ \(a-3=0\)\(\Leftrightarrow\)\(x+\sqrt{x+3}-4=0\)

                            \(\Leftrightarrow\)\(x-3=-\sqrt{x+3}\)

                            \(\Leftrightarrow\)\(\left(x-3\right)^2=\left(-\sqrt{x+3}\right)^2\)

                            \(\Leftrightarrow\)\(x^2-6x+9=x+3\)

                            \(\Leftrightarrow\)\(x^2-7x+6=0\)

                            \(\Leftrightarrow\)\(\left(x^2-x\right)-\left(6x-6\right)=0\)

                            \(\Leftrightarrow\)\(x.\left(x-1\right)-6.\left(x-1\right)=0\)

                            \(\Leftrightarrow\)\(\left(x-6\right).\left(x-1\right)=0\)

                            \(\Leftrightarrow\)\(\orbr{\begin{cases}x-6=0\\x-1=0\end{cases}}\)

                            \(\Leftrightarrow\)\(\orbr{\begin{cases}x=6\left(TM\right)\\x=1\left(TM\right)\end{cases}}\)

Vậy \(S=\left\{1;6\right\}\)

Tính nhanh:3.8.46+2.3.5.12+19.4.6

a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne-3\\x\ne3\end{cases}}\)

\(A=\left(\frac{1}{3}+\frac{3}{x^2-3x}\right):\left(\frac{x^2}{27-3x^2}+\frac{1}{x+3}\right)\)\(=\left[\frac{1}{3}+\frac{3}{x\left(x-3\right)}\right]:\left(\frac{-x^2}{3x^2-27}+\frac{1}{x+3}\right)\)

\(=\left[\frac{x\left(x-3\right)}{3x\left(x-3\right)}+\frac{9}{3x\left(x-3\right)}\right]:\left[\frac{-x^2}{3\left(x^2-9\right)}+\frac{1}{x+3}\right]\)

\(=\frac{x^2-3x+9}{3x\left(x-3\right)}:[\frac{-x^2}{3\left(x-3\right)\left(x+3\right)}+\frac{3\left(x-3\right)}{3\left(x-3\right)\left(x+3\right)}]\)

\(=\frac{x^2-3x+9}{3x\left(x-3\right)}:\frac{-x^2+3x-9}{3\left(x-3\right)\left(x+3\right)}\)\(=\frac{x^2-3x+9}{3x\left(x-3\right)}.\frac{3\left(x-3\right)\left(x+3\right)}{-\left(x^2-3x+9\right)}=\frac{x+3}{-x}=\frac{-x-3}{x}=-1-\frac{3}{x}\)

b) \(A< -1\)\(\Leftrightarrow-1-\frac{3}{x}< -1\)\(\Leftrightarrow\frac{-3}{x}< 0\)

mà \(-3< 0\)\(\Rightarrow x>0\)và \(x\ne3\)

Vậy \(A< -1\Leftrightarrow\hept{\begin{cases}x>0\\x\ne3\end{cases}}\)

c) Vì \(-1\inℤ\)\(\Rightarrow\)Để A nguyên thì \(\frac{3}{x}\inℤ\)\(\Rightarrow3⋮x\)

\(\Rightarrow x\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

So sánh với ĐKXĐ \(\Rightarrow x=\pm3\)loại

Vậy A nguyên \(\Leftrightarrow x=\pm1\)

\(\frac{6x^3\left(2y+1\right)}{5y}\cdot\frac{15}{2x^3\left(2y+1\right)}=\frac{9}{y}\)

\(\frac{3}{x^2-1}:\frac{6x}{2x^3\left(2y+1\right)}=\frac{3}{x^2-1}\cdot\frac{2x^3\left(2y+1\right)}{6x}=\frac{x^2\left(2y+1\right)}{x^2-1}\)

hok tốt.

\(\frac{6x^3\left(2y+1\right)}{5y}\cdot\frac{15}{2x^3\left(2y+1\right)}\)

\(=\frac{6x^3\left(2y+1\right)}{5y}\cdot\left[\frac{15}{2x^3\left(2y+1\right)}\right]\)

\(=\frac{180x^3y+90x^3}{20x^3y^2+10x^3y}\)

\(=\frac{180y+90}{20y^2+10y}\)

\(=\frac{18y+9}{2y^2+y}\)

\(=\frac{9\left(2y+1\right)}{y\left(2y+1\right)}\)

\(=\frac{9}{y}\)

\(\frac{y}{3x}+\frac{2y}{3x}=\frac{y+2y}{3x+3x}=\frac{3y}{3x}=\frac{y}{x}\)

\(\frac{4x-1}{3x^2y}-\frac{7x+1}{3x^2y}=\frac{4x-1-\left(7x+1\right)}{3x^2y}=\frac{-3x-2}{3x^2y}\)

\(\frac{6x-1}{3x^2y}+\frac{4x-1}{3x^2y}=\frac{6x-1+4x-1}{3x^2y}=\frac{10x-2}{3x^2y}\)

=\(\frac{30x^2y^3}{8x^2y^2}\)

Ta có :

\(\frac{6x^2y^2}{8xy^5}=\frac{3x}{4y^3}\)

\(\frac{x^2-xy}{5xy-5y^2}=\frac{x\left(x-y\right)}{5y\left(x-y\right)}=\frac{x}{5y}\)

Hok tốt !

a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n+1\right)}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}\)

b) \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\right)\)

         \(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

\(=\frac{1}{4}-\frac{1}{2\left(n+1\right)\left(n+2\right)}\)

Ta có : \(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\)

\(=\left[xy\left(x+y\right)+xyz\right]+\left[yz\left(y+z\right)+xyz\right]+xz\left(x+z\right)\)

\(=xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+z\right)\)

\(=y\left(x+y+z\right)\left(x+z\right)+xz\left(x+z\right)\)

\(=\left(x+z\right)\left(xy+y^2+yz+xz\right)\)

\(=\left(x+z\right)\left(x+y\right)\left(y+z\right)\)