1. comfortable 

3. unhealthy

4. illness

 \(a)P=\left(\dfrac{x^2+2}{x^3-1}+\dfrac{x+1}{x^2+x+1}+\dfrac{1}{1-x}\right).\left(\dfrac{x^2}{x+1}+1\right).\left(x\ne1;x\ne-1\right).\\ P=\dfrac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}.\dfrac{x^2+x+1}{x+1}.\\ P=\dfrac{x^2-x}{x-1}.\dfrac{1}{x+1}.\\ P=\dfrac{x\left(x-1\right)}{x-1}.\dfrac{1}{x+1}.\\ P=x.\dfrac{1}{x+1}.\\ P=\dfrac{x}{x+1}.\)

\(P=\dfrac{1}{4}.\Rightarrow\dfrac{x}{x+1}=\dfrac{1}{4}.\\ \Leftrightarrow4x-x-1=0.\\ \Leftrightarrow3x-1=0.\\ \Leftrightarrow x=\dfrac{1}{3}\left(TM\right).\)

\(ĐK:x\ne0\)

\(\dfrac{x-1}{3}+\dfrac{x+3}{x}=2\)

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

\(\Leftrightarrow x\left(x-1\right)+3\left(x+3\right)=6x\)

\(\Leftrightarrow x^2-x+3x+9-6x=0\)

\(\Leftrightarrow x^2-4x+9=0\)

Ta có: \(x^2-4x+9=x^2-4x+4+5=\left(x-2\right)^2+5\ge5>0\)

Vậy pt vô nghiệm

THAM KHẢO

Gọi x là v.tốc dự định của xe(x>0, km/h)

Nửa quãng đường xe đi là: 120:2=60(km)

=> Vận tốc đi nửa quãng đường là: 120x(h)120x(h)

Vì nửa qquangx đường sau xe đi với thời gian là: 60x+60x+10=120x−0.560x+60x+10=120x−0.5

Gải được x=40(tmđk)

Vậy v.tốc dự định là 40km/h

\(P=\left(\dfrac{x^2+1}{x^2-9}-\dfrac{x}{x+3}+\dfrac{5}{3-x}\right):\left(\dfrac{2x+10}{x+3}-1\right)\)

\(=\left(\dfrac{x^2+1}{\left(x-3\right)\left(x+3\right)}-\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{2x+10}{x+3}-\dfrac{x+3}{x+3}\right)\)

\(=\left(\dfrac{x^2+1-x^2+3x-5x-15}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{2x+10-x-3}{x+3}\right)\)

\(=\left(\dfrac{-2x-14}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{x+7}{x+3}\right)\)

\(=\dfrac{-2\left(x+7\right)}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+7}\)

\(=\dfrac{-2}{x-3}\)

đk : x khác -3 ; 3 ; -7 

\(P=\left(\dfrac{x^2+1+x\left(x-3\right)+5x+15}{x^2-9}\right):\left(\dfrac{2x+10-x-3}{x+3}\right)\)

\(=\dfrac{2x^2+1+2x+15}{x^2-9}:\dfrac{x+7}{x+3}=\dfrac{2x^2+2x+16}{\left(x-3\right)\left(x+7\right)}\)

\(\widehat{BAx}=\dfrac{1}{2}sđ\stackrel\frown{AB}\)

\(\widehat{BAC}=\dfrac{1}{2}sđ\stackrel\frown{BC}\)

\(\widehat{ABC}=90^0\) (góc nội tiếp chắn nửa đường tròn)

http://baigiang.violet.vn/present/show/entry_id/6146495

Tham khảo đi. cho cj nhé