a) ĐKXĐ của phương trình P là: \(\left[{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)

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

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

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

\(=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\) \(=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\) (Vì \(x+3\ne0\))

Mà \(P=\frac{-3}{4}\) nên \(\frac{x-4}{x-2}=-\frac{3}{4}\)

\(\Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\)

\(\Leftrightarrow4x-16=6-3x\)

\(\Leftrightarrow4x+3x=6+16\)

\(\Leftrightarrow7x=22\) \(\Leftrightarrow x=\frac{22}{7}\) (thỏa mãn ĐKXĐ)

Vậy để \(P=-\frac{3}{4}\) thì \(x=\frac{22}{7}\)

c) Để \(P\) nguyên thì \(\frac{x-4}{x-2}\in Z\) \(\Rightarrow x-4⋮x-2\)

\(\Leftrightarrow x-4-\left(x-2\right)⋮x-2\) (Vì \(\left(x-2\right)⋮x-2\))

\(\Leftrightarrow x-4-x+2⋮x-2\)

\(\Leftrightarrow-2⋮x-2\)

\(\Rightarrow x-2\in\)Ư(\(-2\))\(=\left\{1;-1;2;-2\right\}\)

\(\Leftrightarrow x\in\left\{3;1;4;0\right\}\)

Vậy để \(P\) nguyên thì \(x\in\left\{3;1;4;0\right\}\)

d) Ta có: \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow x\in\left\{3;-3\right\}\)

Mà \(P=\frac{x-4}{x-2}\) (2) . Thay \(x=3\) vào (2) ta được:

\(P=\) \(\frac{3-4}{3-2}\) \(=-1\)

Thay \(x=-3\) vào (2) ta được :

\(P=\frac{-3-4}{-3-2}=\frac{7}{5}\)

Vậy giá trị của biểu thức \(P\) khi : \(x=3\) là \(-1\)

: \(x=-3\) là \(\frac{7}{5}\)

a, ĐKXĐ: \(x\notin\left\{-2;\pm3\right\}\)

\(B=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{x-4}{x-3}-\frac{x-1}{x+3}\right):\frac{x+3-1}{x+3}\\ =\frac{21+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{x+2}\\ =\frac{21+x^2-x-12-\left(x^2-4x+3\right)}{x-3}\cdot\frac{1}{x+2}\\ =\frac{x^2-x+9-x^2+4x-3}{\left(x-3\right)\left(x+2\right)}\\ =\frac{3x+6}{\left(x-3\right)\left(x+2\right)}\\ =\frac{3\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}=\frac{3}{x-3}\)

b, Ta có:

\(\left|2x+1\right|=5\Leftrightarrow\left\{{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-3\left(ktm\right)\end{matrix}\right.\)

Suy ra, với \(x=2\), ta được:

\(B=\frac{3}{2-3}=\frac{3}{-1}=-3\)

c, Để \(B=\frac{-3}{5}\) thì:

\(\frac{3}{x-3}=\frac{-3}{5}\\ \Leftrightarrow\frac{-3}{3-x}=\frac{-3}{5}\\ \Leftrightarrow3-x=5\Leftrightarrow x=-2\left(ktm\right)\)

Hay không có giá trị nào sao cho \(B=\frac{-3}{5}\).

d, Do 3>0 nên để B<0 thì: \(x-3< 0\Leftrightarrow x< 3\).

Kết hợp với ĐKXĐ, ta có điều kiện: \(\left\{{}\begin{matrix}x< 3\\x\notin\left\{-2;-3\right\}\end{matrix}\right.\)

Chúc bạn học tốt nha.

a) ĐKXĐ: x∉{3;-3}

Ta có: \(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)

\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right):\left(\frac{x+3}{x+3}-\frac{1}{x+3}\right)\)

\(=\frac{21+x^2-x-12-\left(x^2-4x+3\right)}{\left(x-3\right)\left(x+3\right)}:\frac{x+2}{x+3}\)

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

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

b) Ta có: |2x+1|=5

⇔\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Do x=-3 không thỏa mãn ĐKXĐ nên ta chỉ tính giá trị của B tại x=2

Thay x=2 vào biểu thức \(B=\frac{3}{x-3}\), ta được:

\(\frac{3}{2-3}=\frac{3}{-1}=-3\)

Vậy: -3 là giá trị của biểu thức \(B=\frac{3}{x-3}\) tại x=2

c) Ta có: \(B=\frac{-3}{5}\)

⇔\(\frac{3}{x-3}=\frac{-3}{5}\)

\(\Leftrightarrow x-3=\frac{5\cdot3}{-3}=\frac{15}{-3}=-5\)

hay x=-2(tm)

Vậy: Khi \(B=\frac{-3}{5}\) thì x=-2

d) Để B<0 thì \(\frac{3}{x-3}< 0\)

mà 3>0

nên x-3<0

hay x<3

Vậy: Khi x<3 và x≠-3 thì B<0

a) \(B=\left(\frac{9-3x}{\left(x-1\right)\left(x+5\right)}+\frac{\left(x+5\right)^2}{\left(x-1\right)\left(x+5\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x+5\right)\left(x-1\right)}\right)\)\(:\frac{7\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)

\(B=\frac{9-3x+x^2+10x+25-\left(x^2-1\right)}{\left(x-1\right)\left(x+5\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{7\left(x-2\right)}\)

\(B=\frac{7x+35}{\left(x-1\right)\left(x+5\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{7\left(x-2\right)}\)

\(B=\frac{7\left(x+5\right)\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+5\right)\cdot7\left(x-2\right)}=\frac{x+1}{x-2}\)

b) \(\left(x+5\right)^2-9x-45=0\)

\(\Leftrightarrow x^2+10x+25-9x-45=0\)

\(\Leftrightarrow x^2+x-20=0\)

\(\Leftrightarrow x^2-4x+5x-20=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-5\left(KTM\right)\\x=4\left(TM\right)\end{matrix}\right.\)

Với x = 4 ta có \(B=\frac{4+1}{4-2}=\frac{5}{2}\)

Y

c) \(B=\frac{x+1}{x-2}\) đạt giá trị nguyên

\(\Leftrightarrow x+1⋮x-2\)

\(\Leftrightarrow x-2+3⋮x-2\)

\(\Leftrightarrow3⋮x-2\) \(\Leftrightarrow\left(x-2\right)\in\left\{-3;-1;1;3\right\}\)

\(\Leftrightarrow x\in\left\{-1;1;3;5\right\}\)

d) \(B=-\frac{3}{4}\Leftrightarrow\frac{x+1}{x-2}=\frac{-3}{4}\)

\(\Leftrightarrow4\left(x+1\right)=-3\left(x-2\right)\)

\(\Leftrightarrow4x+4=-3x+6\)

\(\Leftrightarrow7x=2\Leftrightarrow x=\frac{2}{7}\)