Question

[nguy hiem qua]

1/x(x+1) +1/(x+1)(x+2) +1/(x+2)(x+3) = 3/10.    Cho mình câu trả lời dễ hiểu nhé (=.=)

Answer 1

[Thắng Nguyễn]

\(\Leftrightarrow\frac{1}{x^2+5x+6}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+x}=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{x^2+5x+6}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+x}-\frac{3}{10}=0\)

\(\Leftrightarrow-\frac{3\left(x^2+3x-10\right)}{10x\left(x+3\right)}=0\)

\(\Leftrightarrow3\left(x^2+3x-10\right)=0\)

\(\Leftrightarrow x^2+3x-10=0\)

\(\Leftrightarrow x^2-2x+5x-10=0\)

\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow x-2=0\)hoặc\(x+5=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)

Answer 2

[Hoàng Phúc]

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+3}=\frac{3}{10}\Leftrightarrow\frac{\left(x+3\right)-x}{x\left(x+3\right)}=\frac{3}{10}\Leftrightarrow\frac{3}{x\left(x+3\right)}=\frac{3}{10}\)

<=>x(x+3)=10 <=> x²+3x=10 <=> x²+3x-10=0

<=>-(x²-3x+10)=0

<=>x²-3x+10=0

<=>x²-2.x.\(\frac{3}{2}\)+ \(\left(\frac{3}{2}\right)^2+\frac{31}{4}\)=0

<=> \(\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\)=0

Vì \(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\ge\frac{31}{4}>0\) (với mọi x)

=>PT vô nghiệm