x=2+y

B=2.(2+y)(2+y)+y

=2(2+y)^2+y

=2(4+4y+y^2)+y

=8+8y+2y^2+y^2=8+8y+3y^2

\(A=\left(x-y\right)\left(x^2+xy+y^2\right)=2\left(x^2-2xy+y^2\right)+6xy=2\left(x-y\right)^2+6x\left(x-2\right)\)

\(=6\left(x^2-2x\right)+8=6\left(x-1\right)^2+2\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x=1\\y=-1\end{cases}}\)

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

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

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

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

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

\(=\frac{2.2x+4x^2}{\left(1+x\right)}.\frac{x}{4\left(x^3-3\right)}\)

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

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

\(=\frac{x}{1}.\frac{x}{\left(x^3-3\right)}\)

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

a = 48

b = \(\frac{1}{5}\)

c = 2

d \(\approx0,3904413457\)

cách giải nx bạn ơi 

\(a,ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm1\end{cases}}\)

Sao phân số thứ 2 là \(\frac{1-2}{1+x}\) .Bạn chép đề thật chuẩn mới trả lời đúng nhé

Ta có: \(A=\frac{3-2x}{3x-1}\)

=> \(3A=\frac{-\left(6x-9\right)}{3x-1}=\frac{-2\left(3x-1\right)+7}{3x-1}=-2+\frac{7}{3x-1}\)

Để A đạt GTLN <=> 3A đạt GTLN

<=> \(\frac{7}{3x-1}\)đạt GTLN 

<=> \(3x-1\)đạt GTNN

Do x \(\in\)Z <=> 3x - 1 = 2  <=> 3x = 3 <=> x = 1

Thay x = 1 vào 3A, ta có: \(-2+\frac{7}{3.1-1}=-2+7=5\)

        => \(A=5:3=\frac{5}{3}\)

Vậy x = 1 (x \(\in\)Z) thì A = \(\frac{3-2x}{3x-1}\)đạt Max

Cách khác nè. Ko bt đg ko

\(A=\frac{3-2x}{3x-1}\Leftrightarrow-A=\frac{2x-3}{3x-1}\Leftrightarrow-3A=\frac{6x-9}{3x-1}\)

\(\Leftrightarrow-3A=\frac{2\left(3x-1\right)-7}{3x-1}=2-\frac{7}{3x-1}\)

-3A đạt GTNN\(\Leftrightarrow\frac{7}{3x-1}\)đạt GTLN\(\Leftrightarrow3x-1\)đạt GTNN\(\Leftrightarrow3x-1=2\Leftrightarrow x=1\)(Vì x thuộc Z)

Khi đó \(\Leftrightarrow-3A==2-\frac{7}{3.1-1}=\frac{-3}{2}\)

Vậy GTLN của A là \(\frac{1}{2}\Leftrightarrow x=1\)

Sai thì mấy anh chị góp ý