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Áp dụng BĐT AM-GM: \(\left(x^2-y^2\right)\left(1-x^2y^2\right)\le\frac{1}{4}\left(x^2-y^2+1-x^2y^2\right)^2=\frac{1}{4}\left(1-y^2\right)^2\left(1+x^2\right)^2\)

\(P\le\frac{1}{4}\frac{\left(1-y^2\right)^2}{\left(1+y^2\right)^2}\)

mà theo BĐT AM-GM:\(\left(1-y\right)\left(1+y\right)\le\frac{1}{4}\left(1-y+1+y\right)^2=1\)

\(\Rightarrow P\le\frac{1}{4}.\frac{1}{\left(1+y^2\right)^2}\le\frac{1}{4}.\frac{1}{1}=\frac{1}{4}\)

Dấu = xảy ra khi x=1;y=0 wait : có gì đó sai sai. số thực

vì \(a,b,c\in\left[0,1\right]\)\(\Rightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge0\)

\(\Leftrightarrow\left(1-a-b+ab\right)\left(1-c\right)\ge0\)

\(\Leftrightarrow1-c-a+ac-b+bc+ab-abc\ge0\)

\(\Leftrightarrow a+b+c-\left(ab+bc+ac\right)\le1-abc\)

mặt khác : \(a.bc\ge0\)

\(\Rightarrow a+b+c-\left(ab+ac+bc\right)\le1-0=1\)

mà \(b,c\in\left[0.1\right]\Rightarrow b^2\le b;c^3\le c\)

vì vậy ta được điều phải chứng minh : 

\(a+b^2+c^3-\left(ab+bc+ac\right)\le1\)

Vì \(b,c\in[0;1]\)

\(\Rightarrow b^2\le b\)

     \(c^3\le c\)

Do đó :  \(a+b^2+c^3-ab-bc-ca\le a+b+c-ab-bc-ca\)          (1)

Và có : \(a+b+c-ab-bc-ca=\left(a-1\right).\left(b-1\right).\left(c-1\right)-abc+1\)             (2)

Theo đề bài ta có : \(a,b,c\in[0;1]\)

\(\Rightarrow\left(a-1\right)\left(b-1\right)\left(c-1\right)\le0\)

và \(-ab\le0\)

Từ (2)

\(\Rightarrow a+b+c-ab-bc-ca\le1\)       (3)

Từ (1) và (3)

\(\Rightarrow a+b^2+c^3-ab-bc-ca\le1\)( đpcm)

a. ĐKXĐ: \(x\ne0;1\)

Ta có: \(A=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)

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

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

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

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

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

b. \(A=4\Leftrightarrow4=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)

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

\(\Leftrightarrow x+1+2\sqrt{x}-4\sqrt{x}=0\)

\(\Leftrightarrow x+1-2\sqrt{x}=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2=0\)

\(\Leftrightarrow\sqrt{x}=1\)

\(\Leftrightarrow x=1\)

Vậy....