\(\hept{\begin{cases}2\sqrt{2xy-y}+2x+y=10\left(1\right)\\\sqrt{3y+4}-\sqrt{2y+1}+2\sqrt{2x-1}=3\left(2\right)\end{cases}}\)

\(ĐK:x\ge\frac{1}{2};y\ge0\)

\(\left(1\right)\Leftrightarrow\left(\sqrt{2x-1}+\sqrt{y}\right)^2=9\Leftrightarrow\sqrt{2x-1}+\sqrt{y}=3\)

\(\Leftrightarrow\sqrt{2x-1}=3-\sqrt{y}\)(*)

Thay \(\sqrt{2x-1}=3-\sqrt{y}\)vào (2), ta được: \(\sqrt{3y+4}-\sqrt{2y+1}-2\left(\sqrt{y}-2\right)-1=0\)

\(\Leftrightarrow\left(\sqrt{3y+4}-4\right)-\left(\sqrt{2y+1}-3\right)-2\left(\sqrt{y}-2\right)=0\)

\(\Leftrightarrow\frac{3\left(y-4\right)}{\sqrt{3y+4}+4}-\frac{2\left(y-4\right)}{\sqrt{2y+1}+3}-\frac{2\left(y-4\right)}{\sqrt{y}+2}=0\)

\(\Leftrightarrow\left(y-4\right)\left(\frac{3}{\sqrt{3y+4}+4}-\frac{2}{\sqrt{2y+1}+3}-\frac{2}{\sqrt{y}+2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}y=4\Rightarrow x=1\\\frac{3}{\sqrt{3y+4}+4}=\frac{2}{\sqrt{2y+1}+3}+\frac{2}{\sqrt{y}+2}\left(3\right)\end{cases}}\)

Với \(y\ge0\)thì \(\frac{3}{\sqrt{3y+4}+4}\le\frac{1}{2}\)

Từ (*) suy ra \(y\le9\Rightarrow\frac{2}{\sqrt{2y+1}+3}+\frac{2}{\sqrt{y}+2}>\frac{1}{2}\)

Suy ra (3) vô nghiệm

Vậy hệ có cặp nghiệm duy nhất \(\left(x,y\right)=\left(1,4\right)\)

a/ Dặt \(\sqrt{x+1}=a\ge0\)

\(\Rightarrow4\sqrt{x+1}=x^2+5x+4\)

\(\Leftrightarrow4\sqrt{x+1}=\left(x+1\right)^2+3\left(x+1\right)\)

\(\Leftrightarrow4a=a^4+3a^2\)

\(\Leftrightarrow a\left(a-1\right)\left(a^2+a+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=0\\a=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+1}=0\\\sqrt{x+1}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)

b/ Đặt \(\hept{\begin{cases}\sqrt{4x+1}=a\ge0\\\sqrt{3x-2}=b\ge0\end{cases}}\)

\(\Rightarrow a^2-b^2=x+3\)

Từ đây ta có:

\(a-b=\frac{a^2-b^2}{5}\)

\(\Leftrightarrow\left(a-b\right)\left(5-a-b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=b\left(1\right)\\a+b=5\left(2\right)\end{cases}}\)

Thế vô làm tiếp

\(P=\frac{\sqrt{x}+1}{2\sqrt{x}+5}\left(ĐK:x\ge0\right)\)

\(2P=\frac{2\sqrt{x}+4}{2\sqrt{x}+5}\)\(=\frac{2\sqrt{x}+5-1}{2\sqrt{x}+5}\)

\(=1-\frac{1}{2\sqrt{x}+5}\)

\(P\inℤ\Leftrightarrow2P\inℤ\Leftrightarrow\frac{1}{2\sqrt{x}+5}\inℤ\)

\(\Leftrightarrow\left(2\sqrt{x}+5\right)\inƯ\left(1\right)=\left\{\pm1\right\}\)

Mà \(2\sqrt{x}+5\ge5\)(Do \(x\ge0\)) nên P không thể đạt giá trị nguyên

Mk mới 8 chịu thôi, bạn vô đây học nè www.slideshare.net/cunbeo/bd-hsgchuyen-de24nguyenlydirichletvoicacbaitoandaisohinhhoc9667

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