Ta có: \(B=\frac{x^2+1}{x^2-x+1}\)

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

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

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

Vì \(\frac{\left(x+1\right)^2}{3\left(x^2-x+1\right)}\ge0\forall x\)( Điều này các bạn tự CM nhé )

\(\Rightarrow\)\(B\ge\frac{2}{3}\)\(\Rightarrow\)\(B_{min}=\frac{2}{3}\)

Dấu "=" Xảy ra khi: \(x+1=0\)\(\Leftrightarrow\)\(x=-1\)

Vậy \(B_{min}=\frac{2}{3}\)\(\Leftrightarrow\)\(x=-1\)

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

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

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

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

\(=\frac{x-1}{x}:2=\frac{x-1}{2x}\)hay \(E=\frac{x-1}{2x}\)

mình làm ví dụ một câu thôi nhé ! 

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

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

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

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

<=> 4x( x + 1 )^2 = 4x ( x + 1 )^2 

<=> 4x ( x^2 + 2x + 1 ) = 4x ( x^2 + 2x + 1 )

<=> 4x^3 + 8x^2 + 4x - 4x^3 - 8x^2 - 4x = 0 

<=> 0 = 0 ( Vậy ta có đpcm ) 

Cảm ơn câu trả lời của bạn,mình quên không viết cap.Nhờ mọi người giải giúp mình bài 2 nhé!Bài 1 mình giải được rồi.Cảm ơn bạn lầu trên

Ta có: \(\left(x^2+1\right)^2+3x.\left(x^2+1\right)+2x^2=0\)

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

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

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

Vì  \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)mà \(\left(x^2+x+1\right)\left(x+1\right)^2=0\)

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

Vậy \(x=-1\)

Ta có: \(\left(2x+7\right)^2=\left(x+3\right)^2\)

    \(\Leftrightarrow4x^2+28x+49=x^2+6x+9\)

    \(\Leftrightarrow3x^2+22x+40=0\)

    \(\Leftrightarrow3\left(x^2+\frac{22}{3}x+\frac{40}{3}\right)=0\)

    \(\Leftrightarrow3\left(x^2+2.x.\frac{11}{3}+\frac{121}{9}-\frac{1}{9}\right)=0\)

    \(\Leftrightarrow\left(x+\frac{11}{3}\right)^2-\frac{1}{9}=0\)

    \(\Leftrightarrow\left(x+\frac{11}{3}\right)^2=\frac{1}{9}\)

    \(\Leftrightarrow x+\frac{11}{3}=\pm\frac{1}{3}\)

    \(\Leftrightarrow\orbr{\begin{cases}x+\frac{11}{3}=\frac{1}{3}\\x+\frac{11}{3}=-\frac{1}{3}\end{cases}}\)

    \(\Leftrightarrow\orbr{\begin{cases}x=-\frac{10}{3}\\x=-4\end{cases}}\)

Vậy \(x=-\frac{10}{3}\)hoặc \(x=-4\)

\(\left(2x+7\right)^2=\left(x+3\right)^2\)

\(\Leftrightarrow4x^2+28x+49=x^2+6x+9\)

\(\Leftrightarrow3x^2+22x+40=0\)giải delta ta có : 

\(x_1=-\frac{10}{3};x_2=-4\)

ĐK : n ∈ Z

n³ + 3n² - 4n

= n³ + 3n² + 2n - 6n

= n( n² + 3n + 2 ) - 6n

= n( n² + n + 2n + 2 ) - 6n

= n[ n( n + 1 ) + 2( n + 1 ) ] - 6n

= n( n + 1 )( n + 2 ) - 6n

Dễ dàng chứng minh n( n + 1 )( n + 2 ) ⋮ 6 và 6n ⋮ 6

=> n( n + 1 )( n + 2 ) - 6n ⋮ 6

hay n³ + 3n² - 4n ⋮ 6 ( đpcm )

ĐK : n ∈ Z

n³ + 3n² - 4n

= n³ + 3n² + 2n - 6n

= n( n² + 3n + 2 ) - 6n

= n( n² + n + 2n + 2 ) - 6n

= n[ n( n + 1 ) + 2( n + 1 ) ] - 6n

= n( n + 1 )( n + 2 ) - 6n

Dễ dàng chứng minh n( n + 1 )( n + 2 ) ⋮ 6 và 6n ⋮ 6

=> n( n + 1 )( n + 2 ) - 6n ⋮ 6

hay n³ + 3n² - 4n ⋮ 6 ( đpcm )

ta có

\(\left(7x-2x\right)\left(2x-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow5x.\left(2x-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x=0\\2x-1=0\end{cases}}\) hoặc \(x+3=0\)

hay ta có \(x\in\left\{0;\frac{1}{2};-3\right\}\)

đề bài ad cho sai