Question

2) Cho biểu thức D = (1+\(\frac{x}{x^2+1}\)) : (\(\frac{1}{x-1}-\frac{2x}{x^3+x-x^2-1}\))

a. Rút gọn biểu thức D

b. Tìm giá trị của x sao cho D < 1

c. Tìm giá trị nguyên của x để D có giá trị nguyên

Answer 1

ĐK : x \(\ne\) 1
a) D = \(\left(1+\frac{x}{x^2+1}\right):\left(\frac{1}{x-1}-\frac{2x}{x^3+x-x^2-1}\right)=\left(\frac{x^2+1}{x^2+1}+\frac{x}{x^2+1}\right):\left(\frac{x^2+1}{\left(X^2+1\right)\left(x-1\right)}-\frac{2x}{x^2\left(x-1\right)+\left(x-1\right)}\right)\)

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

b)

D <1

=> \(x^2+x+1< x-1\Rightarrow x^2+x+1-x+1< 0\Rightarrow x^2+2< 0\) ( vô lí )

Vậy D > 1, không có x thỏa mãn

c) D thuộc Z

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

Vì x thuộc Z nên D thuộc Z khi

\(x-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

* x -1 = 1 => x= 2 (tm)

* x-1 = -1 => x = 0 (tm)

* x-1 =3 => x = 4 (tm)

* x-1 = -3 => x = -2 ( tm )

Answer 2

\(ĐKXD:x\ne1\)

\(a,D=\left(1+\frac{x}{x^2+1}\right):\left(\frac{1}{x-1}-\frac{2x}{x^3+x-x^2-1}\right)=\frac{x^2+x+1}{x^2+1}:\left(\frac{1}{\left(x-1\right)}-\frac{2x}{\left(x-1\right)\left(x^2+1\right)}\right)=\frac{x^2+x+1}{x^2+1}:\left(\frac{x^2+1}{\left(x-1\right)\left(x^2+1\right)}-\frac{2x}{\left(x-1\right)\left(x^2+1\right)}\right)=\frac{x^2+x+1}{x^2+1}:\left(\frac{x^2-2x+1}{\left(x-1\right)\left(x^2+1\right)}\right)=\frac{x^2+x+1}{x^2+1}:\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+1\right)}=\frac{x^2+x+1}{x^2+1}:\frac{x-1}{x^2+1}=\frac{\left(x^2+x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x-1\right)}=\frac{x^2+x+1}{x-1}\)

\(D< 1\Leftrightarrow x^2+x+1< x-1\Leftrightarrow\left(x-1\right)-\left(x^2+x+1\right)>0\Leftrightarrow x-1-x^2-x-1>0\Leftrightarrow-\left(x^2+2\right)>0\left(\text{ vô lí}\right).\text{ Nên không tìm được x thỏa mãn}\)

\(ĐểDnguyênthì:x^2+x+1⋮x-1\Leftrightarrow x\left(x-1\right)+2x+1⋮x-1\Leftrightarrow\left(x+2\right)\left(x-1\right)+3⋮x-1\Leftrightarrow3⋮x-1\left(\text{ vì: (x+2)(x-1) chia hết cho x-1}\right)\Leftrightarrow x-1\in\left\{-1;1;-3;3\right\}\Leftrightarrow x\in\left\{0;2;-2;4\right\}.Vậy:x\in\left\{0;2;-2;4\right\}thìDnguyên\)