Question

[Soul Hopless]

Bài 1: Cho A =\(\dfrac{x}{x+1}-\dfrac{3-3x}{x^2-x+1}+\dfrac{x+4}{x^3+1}\)

a) Rút gọn A

b) CMR A luôn dương với mọi x # -1

Bài 2: Cho M =\(\left(\dfrac{x}{x^2-4}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

a)Rút gọn M

b) Tìm x nguyên để M có giá trị nguyên

c)Tìm giá trị của M tại x =\(\dfrac{1}{2};x=2\)

Giúp mình vs mình cần gấp

Answer 1

[Phùng Khánh Linh]

\(1a.A=\dfrac{x}{x+1}-\dfrac{3-3x}{x^2-x+1}+\dfrac{x+4}{x^3+1}=\dfrac{x\left(x^2-x+1\right)-3\left(1-x^2\right)+x+4}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^3+2x^2+2x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^3+x^2+x^2+x+x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2+x+1}{x^2-x+1}\left(x\ne-1\right)\)

\(b.A=\dfrac{x^2+x+1}{x^2-x+1}=\dfrac{x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+1-\dfrac{1}{4}}{x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+1-\dfrac{1}{4}}=\dfrac{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}>0\left(x\ne-1\right)\)

\(2a.M=\left(\dfrac{x}{x^2-4}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)=\left[\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right]:\dfrac{x^2-4+10-x^2}{x+2}=\dfrac{6}{\left(2-x\right)\left(x+2\right)}.\dfrac{x+2}{6}=\dfrac{1}{2-x}\left(x\ne\pm2\right)\)

\(b.Để:M\in Z\Leftrightarrow\dfrac{1}{2-x}\in Z\Leftrightarrow2-x\in\left\{\pm1\right\}\)

\(\oplus2-x=1\Leftrightarrow x=1\left(TM\right)\)

\(\oplus2-x=-1\Leftrightarrow x=3\left(TM\right)\)

\(c.\circledast x=\dfrac{1}{2}\left(TM\right)\) , ta có :

\(M=\dfrac{1}{2-\dfrac{1}{2}}=\dfrac{2}{3}\)

\(\circledast x=2\left(KTM\right)\) , giá trị của M không xác định tại x = 2