Question

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

tìm x để A= -3/4

b) tìm x để biểu thức A có giá tị nguyên

Answer 1

a) \(A=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\left(ĐK:x\ne-3;x\ne2\right)\)

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

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

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

Để \(A=-\frac{3}{4}\)

\(\Leftrightarrow\frac{x-4}{x-2}=-\frac{3}{4}\)

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

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

\(\Leftrightarrow7x=22\Leftrightarrow x=\frac{22}{7}\left(tm\right)\)

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

b) \(A=\frac{x-4}{x-2}=\frac{\left(x-2\right)-2}{x-2}=1-\frac{2}{x-2}\)

Để \(A\in Z\Rightarrow\frac{2}{x-2}\in Z\Rightarrow x-2\inƯ\left(2\right)\)

Mà: \(Ư\left(2\right)=\left\{1;-1;2;-2\right\}\)

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

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

+) \(x-2=-1\Rightarrow x=1\left(tm\right)\)

+) \(x-2=2\Rightarrow x=4\left(tm\right)\)

+) \(x-2=-2\Rightarrow x=0\left(tm\right)\)

Vậy \(x\in\left\{0;1;3;4\right\}\) thì \(A\in Z\)

Answer 2

A=x+2/x+3-5/(x-2)(x+3)-1/x-2

A=(x+2)(x-2)-5-x-3/(x-2)(x+3)

A=x^2-4-5-x-3/(x-2)(x+3)

A=x^2-x-12/(x-2)(x+3)

A=(x+3)(x-4)/(x-2)(x+3)

A=x-4/x-2

Để A=-3/4 thì x-4/x-2=-3/4

Từ đó suy ra (x-4)4=-3(x-2)

4x-16=-3x+6

7x=22

x=22/7

b,Do A nguyên nên x-4/x-2 nguyên(x#2)

suy ra x-4-x+2 chia hết cho x-2

nên 2 chia hết cho x-2

mà ước 2=-2;-1;1;2

nên x=0;1;3;4