Question

[Le DuyDucc_]

Giúp mình với

Answer 1

[ミ★Zero ❄ ( Hoàng Nhật )]

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

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

Xét \(\frac{1}{1-x}+\frac{\sqrt{x}}{1+x\sqrt{x}}=\frac{\left(x\sqrt{x}+1\right)+\sqrt{x}-x\sqrt{x}}{\left(1-x\right)\left(1+x\sqrt{x}\right)}=\frac{1+\sqrt{x}}{\left(1-x\right)\left(1+x\sqrt{x}\right)}\)

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

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

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

b, Đặt \(\sqrt{x}=a,\left(a\ge0\right)\)\(\Rightarrow P=\frac{1+a^3}{a^2-a}\), để chứng minh P > 1

thì ta chứng minh \(1+a^3>a^2-a\)

\(\Leftrightarrow a^3-a^2+a+1>0\Leftrightarrow\left(a-1\right)^3+2\left(a^2-a+1\right)>0\)

mà \(a^2-a+1=\left(a-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall a\)

\(\Rightarrow2\left(a^2-a+1\right)\ge\frac{3}{2},a\ge0\)nên \(\left(a-1\right)^3\ge1\Rightarrow a^3-a^2+a+1\ge\frac{1}{2}\)hay \(P>1\)