\(1,\)\(\frac{x+2}{x+3}+\frac{x-1}{x+1}=\frac{2}{x^2+4x+3}+1\)

\(\Rightarrow\frac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}=\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}\)

\(\Rightarrow\)\(x^2+3x+2+x^2-2x-3=2+x^2+4x+3\)

\(\Rightarrow x^2-3x-6=0\)

.....

\(\frac{x+1}{x-2}+\frac{2x-1}{x-1}=\frac{2}{x^2-3x+2}+\frac{11}{2}\)

\(\Rightarrow\frac{2\left(x+1\right)\left(x-1\right)}{2\left(x-2\right)\left(x-1\right)}+\frac{2\left(2x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)\(=\frac{4}{2\left(x-1\right)\left(x-2\right)}+\frac{22\left(x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)

\(\Rightarrow2x^2-2+4x^2-10x+4=4+22x^2-66x+44\)

.....

\(3,\)\(-2x^2+4x+3\)

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

\(=-2\left[\left(x^2-2x+1\right)-\frac{5}{2}\right]\)

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

Đa thức này lớn nhất =5 khi và chỉ khi \(\left(x-1\right)^2\)nhỏ nhất 

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

a,\(P=\frac{x^2+x}{x^2-2x+1}\div\left(\frac{x+1}{x}-\frac{1}{1-x}+\frac{2-x^2}{x^2-x}\right)\)

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

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

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

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

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

b,a,Để \(P\le1\Rightarrow\frac{x^2}{x-1}\le1\)

\(\Leftrightarrow\frac{x^2}{x-1}-1\le0\)

\(\Leftrightarrow\frac{x^2-x+1}{x-1}\le0\)

\(\Leftrightarrow x-1\le0\)

\(\Leftrightarrow x\le1\)

1.  A = -4 phần x+2

2.  2x^2 + x = 0 => x = 0 hoặc x = -1/2

    Với x = 0 thì A = -2

    Với x = -1/2 thì A = -8/3

3.   A = 1/2 =>  -4 phần x + 2  = 1/2

                  <=> -8 = x + 2 

                   <=> x = -10

4.   A nguyên dương => A > 0

                               => -4 phần x + 2 > 0

      Do -4 < 0 nên -4 phần x + 2 > 0 khi x + 2 < 0

                                                        => x < -2

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

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

b)\(2B=2x+5\Leftrightarrow2.\frac{x+1}{x}=2x+5\Leftrightarrow\frac{2x+2}{x}=2x+5\Leftrightarrow2x+2=2x^2+5x\)

\(\Leftrightarrow0=2x^2+3x-2\Leftrightarrow2x^2+4x-x-2=0\Leftrightarrow2x\left(x+2\right)-\left(x+2\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x+2\right)=0\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-2\end{cases}}\)

cảm ơn bạn nhé