P = \(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

= \(1-\frac{1}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{1}{x-2}\)

= \(1-\left(\frac{1}{x+3}+\frac{1}{x-2}+\frac{5}{\left(x-2\right)\left(x+3\right)}\right)\)

\(=1-\left(\frac{x+3+x-2+5}{\left(x-2\right)\left(x+3\right)}\right)=1-\frac{2x+6}{\left(x-2\right)\left(x+3\right)}=1-\frac{2\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)

\(=1-\frac{2}{x-2}\)

Khi đó P = \(1-\frac{2}{x-2}\)

Khi P = -3/4

=> \(1-\frac{2}{x-2}=-\frac{3}{4}\)

=> \(\frac{2}{x-2}=\frac{7}{4}\)

=> 7(x - 2) = 2.4

=> 7(x - 2) = 8

=> x - 2 = 8/7

=> x = \(\frac{22}{7}\)

Vậy khi x = 22/7 thì P = -3/4

\(P=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

\(=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}+\frac{1}{2-x}\)

\(=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{1}{x-2}\)

\(=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)

\(=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-12-x}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)

Neu P = -3/4 thi : 

\(\frac{x-4}{x-2}=-\frac{3}{4}\Leftrightarrow4x-16=-3x+6\Leftrightarrow7x=22\Leftrightarrow x=\frac{22}{7}\)

\(x^2-2xy+2y^2-x+8=\left(x-y-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2+\frac{15}{2}\ge\frac{15}{2}\)

Dấu "=" xảy ra khi \(x=1,y=\frac{1}{2}\)

Bài làm

\(P=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

a) ĐKXĐ : \(\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)

\(=\frac{x+2}{x+3}-\frac{5}{x^2+3x-2x-6}-\frac{1}{x-2}\)

\(=\frac{x+2}{x+3}-\frac{5}{x\left(x+3\right)-2\left(x+3\right)}-\frac{1}{x-2}\)

\(=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x^2-4x+3x-12}{\left(x+3\right)\left(x-2\right)}\)

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

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

b) x² - 9 = 0 <=> ( x - 3 )( x + 3 ) = 0

<=> \(\orbr{\begin{cases}x=3\left(nhan\right)\\x=-3\left(loai\right)\end{cases}}\)

x = 3 => \(P=\frac{3-4}{3-2}=-1\)

c) \(P=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\)

Để P đạt giá trị nguyên => \(\frac{2}{x-2}\)nguyên

=> \(2⋮x-2\)

=> \(x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Vậy ...

a, \(x\ne-1;3\)

b, Ta có : \(P=\frac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=1\)

\(\Leftrightarrow\frac{3x\left(x+1\right)}{\left(x+1\right)\left(2x-6\right)}=1\Leftrightarrow\frac{3x}{2\left(x-3\right)}=1\)

\(\Leftrightarrow3x=2x-6\Leftrightarrow x=-6\)

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

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

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

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

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

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

Bài làm

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

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

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

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

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