Question

Bài 1: Cho P= \(\left(\frac{x}{x-1}+\frac{1}{x^2-x}\right)\) : \(\left(\frac{1}{x+1}+\frac{2}{x^2+1}\right)\)

a, Rút gọn P

b, Tìm x để P>-1

c, Tìm IPI = 2

Bài 2: Cho a>0, b>0 và a² + b² = 10

 Tìm minQ = \(\frac{1}{a^2}+\frac{1}{b^2}\)

Các bạn giúp mình nha mình cần gấp. Cảm ơn!

Answer 1

Xình lỗi bài 1 đề \(\frac{2}{x^2-1}\) nha !

Answer 2

2) bổ đề : \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)  (x,y > 0)

\(< =>\frac{\left(x+y\right)^2-4xy}{xy\left(x+y\right)}\ge0< =>\frac{\left(x-y\right)^2}{xy\left(x+y\right)}\ge0\)

Dấu "=" xảy ra <=> x=y

Có \(Q=\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{4}{a^2+b^2}=\frac{4}{10}=\frac{2}{5}\)

Dấu "=" xảy ra <=> \(a^2=b^2\)

Ta có hệ \(\hept{\begin{cases}a^2=b^2\\a^2+b^2=10\end{cases}}< =>a=b=\sqrt{5}\left(do.a>b>0\right)\)

Vậy minQ=2/5 khi \(a=b=\sqrt{5}\)

Answer 3

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

\(=\left[\frac{x}{x-1}+\frac{1}{x\left(x-1\right)}\right]:\left[\frac{1}{x+1}+\frac{2}{\left(x-1\right)\left(x+1\right)}\right]\)

\(=\left[\frac{x^2+1}{x\left(x-1\right)}\right]:\left[\frac{x-1+2}{\left(x-1\right)\left(x+1\right)}\right]=\frac{\left(x^2+1\right).\left(x-1\right)\left(x+1\right)}{x\left(x-1\right)\left(x+1\right)}=\frac{x^2+1}{x}\)