\(1-\frac{a^2b}{2+a^2b}\ge1-\frac{a^2b}{3.\sqrt[3]{a^2b}}\)\(\rightarrow1-3\sqrt[3]{a^4b^2}=3.\sqrt[3]{ab.ab.a^2}\rightarrow.....\)

BĐT cần chứng minh tương đương với \(\frac{a^2b}{2+a^2b}+\frac{b^2c}{2+b^2c}+\frac{c^2a}{2+c^2a}\le1\)

Áp dụng BĐT Cauchy ta có: \(2+a^2b=1+1+a^2b\ge3\sqrt[3]{a^2b}\)

Do đó ta được \(\frac{a^2b}{1+a^2b}\le\frac{a^2b}{3\sqrt[3]{a^2b}}=\frac{a\sqrt[3]{ab^2}}{3}\)

Hoàn toàn tương tự ta được \(\frac{a^2b}{2+a^2b}+\frac{b^2c}{2+b^2c}+\frac{c^2a}{2+c^2a}\le\frac{a\sqrt[3]{ab^2}+b\sqrt[3]{bc^2}+c\sqrt[3]{ca}}{3}\)

Cũng theo BĐT Cauchy ta được \(\sqrt[3]{ab^2}\le\frac{a+b+b}{3}=\frac{a+2b}{3}\)

\(\Rightarrow a\sqrt[3]{ab^2}\le\frac{a\left(a+2b\right)}{3}=\frac{a^2+2ab}{3}\)

Tương tự cũng được \(a\sqrt[3]{ab^2}+b\sqrt[3]{bc^2}+c\sqrt[3]{ca}\le\frac{\left(a+b+c\right)^2}{3}=3\)

Từ đó ta được\(\frac{a^2b}{2+a^2b}+\frac{b^2c}{2+b^2c}+\frac{c^2a}{2+c^2a}\le1\)

Vậy BĐT được chứng minh. Dấu "=" xảy ra <=> a=b=c=1

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

Đặt A=\(\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-...+\frac{197}{4851}-\frac{199}{4950}\)

\(\frac{A}{2}=\frac{13}{42}-\frac{15}{56}+\frac{17}{72}-...+\frac{197}{9702}-\frac{199}{4950}\)

\(=\frac{6+7}{6.7}-\frac{7+8}{7.8}+\frac{8+9}{8.9}-...+\frac{98+99}{98.99}-\frac{99+100}{99.100}\)

\(=\frac{1}{7}+\frac{1}{6}-\frac{1}{8}-\frac{1}{7}+\frac{1}{9}+\frac{1}{8}-...+\frac{1}{99}+\frac{1}{98}-\frac{1}{100}+\frac{1}{99}\)

\(=\frac{1}{6}-\frac{1}{100}=\frac{47}{300}\)

\(\Rightarrow A=\frac{47}{300}.2=\frac{47}{150}\)

\(\Rightarrow Q=\frac{85}{25}+\frac{9}{10}-\frac{11}{5}+\frac{47}{150}=\frac{181}{75}\)

Vậy Q=\(\frac{181}{75}\).

\(2y^2+x-2y+5=xy\)

\(\Leftrightarrow8y^2-4xy+4x-8y+20=0\)

\(\Leftrightarrow\left(4y^2-4xy+x^2\right)-\left(x^2-4x+4\right)+\left(4y^2-8y+4\right)=-20\)

\(\Leftrightarrow\left(2y-x\right)^2-\left(x-2\right)^2+\left(2y-2\right)^2=-20\)

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Làm tiếp bài bạn ɱ√ρ︵ƤUɮĞツ『ღƤℓαէїŋʉɱ ₣їɾεツ』⁀ᶜᵘᵗᵉ

\(\left(2y-x\right)^2-\left(x-2\right)^2+\left(2y-2\right)^2=-20\)

\(\Leftrightarrow\left(2y-2x-2\right)\left(2y-2\right)+\left(2y-2\right)^2=-20\)

\(\Leftrightarrow\left(2y-2\right)\left(2y-2x-2+2y-2\right)=-20\)

\(\Leftrightarrow2\left(y-1\right)\left(4y-2x-4\right)=-20\)

\(\Leftrightarrow\left(y-1\right)\left(2y-x-2\right)=-5\)

Đến đây đơn giản rồi