Question

mình cần câu d ạ:(((

Answer 1

d. Khúc này mình rút gọn B ở câu c nhé.

\(B=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\) (ĐK đã cho)

\(=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-1\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\)

Ta có : \(M=A.B\)

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

\(=\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)

Xét \(x+\sqrt{x}+1\)

\(=x+2.\sqrt{x}\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

\(=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Mà \(\left(\sqrt{x}+\dfrac{1}{2}\right)^2\ge0\) với mọi x không âm.

\(\Rightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

Lại có : \(M=\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\) có 2 > 0.

Vậy : \(M=A.B>0\) (đpcm).

Answer 2

lx