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Bài 84 (Đề thi chọn Học sinh giỏi THCS Thành phố Lào Cai 2024 - 2025)

08-12-2024 | 1 cách giải | Unknow | Độ khó: 2 | Loại: Tự luận | Lượt xem: 1106

Cách giải 1

Ý 1. a) Rút gọn A ;

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

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

b) Ta có

\(12=\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}} \geqslant \dfrac{2}{\sqrt[4]{\mathrm{xy}}}=2 \cdot \sqrt{A} \Rightarrow 6 \geqslant \sqrt{A} \Rightarrow 36 \geqslant A\)

A đạt GTLN \(=36\) khi \(x=y=\dfrac{1}{36}\)

Ý 2.

\(\dfrac{c}{b}=\dfrac{\sqrt{c^2+1}}{b}-\dfrac{a c}{c+\sqrt{c^2+1}} \Rightarrow 0=\dfrac{\sqrt{c^2+1}-c}{b}-\dfrac{a c\left(-c+\sqrt{c^2+1}\right)}{1} \Rightarrow\left(\sqrt{c^2+1}-c\right)\left(\dfrac{1}{b}-a c\right)=0\)

Do \(\left(\sqrt{c^2+1}-c\right)>|c|-c \geq 0\) nên \(\dfrac{1}{b}-a c=0\) hay \(a b c=1\)

\[P=\dfrac{1}{\sqrt{a b}+a \sqrt{b c}+1}+\dfrac{1}{\sqrt{b c}+\sqrt{b}+1}+\dfrac{1}{\sqrt{c a}+\sqrt{c}+1}\]

\[=\dfrac{1}{\sqrt{a b}+\sqrt{a} \sqrt{a b c}+1}+\dfrac{1}{\sqrt{b c}+\sqrt{b}+1}+\dfrac{1}{\sqrt{c a}+\sqrt{c}+1}\]

\[=\dfrac{1}{\sqrt{a b}+\sqrt{a}+1}+\dfrac{1}{\sqrt{b c}+\sqrt{b}+1}+\dfrac{1}{\sqrt{c a}+\sqrt{c}+1}\]

Đặt \(\sqrt{a}=x ; \sqrt{b}=y ; \sqrt{c}=z\). Ta có \(x y z=1\). Ta có

\[P=\dfrac{1}{x y+x+1}+\dfrac{1}{y z+y+1}+\dfrac{1}{z x+z+1} \]

\[= \dfrac{1}{x y+x+1}+\dfrac{x}{x y z+x y+x}+\dfrac{x y}{x^2 y z+x y z+x y}\]

\[= \dfrac{1}{x y+x+1}+\dfrac{x}{1+x y+x}+\dfrac{x y}{x+1+x y}=1\]

Vậy \(P=1\)