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a: ĐKXĐ: x>=0; x<>1
b: \(B=\dfrac{2\sqrt{x}+2+x-\sqrt{x}}{x-1}\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}+2}=\dfrac{\sqrt{x}}{x-1}\)
Khi x=3+2căn2 thì \(B=\dfrac{\sqrt{2}+1}{2+2\sqrt{2}}=\dfrac{1}{2}\)
a: \(A=\dfrac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}:\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}:\dfrac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}-5}{\left(2\sqrt{x}-3\right)}\cdot\dfrac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
b: Thay \(x=\dfrac{\left(\sqrt{2}-1\right)^2}{4}\) vào A, ta được:
\(A=\left(3\cdot\dfrac{\sqrt{2}-1}{2}-5\right):\left(2\cdot\dfrac{\sqrt{2}-1}{2}+1\right)\)
\(=\dfrac{3\sqrt{2}-3-10}{2}:\dfrac{2\sqrt{2}-2+2}{2}\)
\(=\dfrac{3\sqrt{2}-13}{2\sqrt{2}}=\dfrac{6-13\sqrt{2}}{4}\)
a: \(A=\dfrac{x+4\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{x-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{x+4\sqrt{x}-2-x+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{4\sqrt{x}-1+x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{x+4\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\)
b: \(B=\dfrac{x\sqrt{x}+26\sqrt{x}-19}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{2x+6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)
\(M=x+\sqrt{x+\dfrac{1}{4}+2.\sqrt{x+\dfrac{1}{4}}.\dfrac{1}{2}+\dfrac{1}{4}}\)
\(M=x+\sqrt{\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2}=x+\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\text{}\text{}\) (do căn + 1 số luôn dương)
\(M=x+\dfrac{1}{4}+2.\sqrt{x+\dfrac{1}{4}}.\dfrac{1}{2}+\dfrac{1}{4}\)
\(M=\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2\)
ta thấy \(\sqrt{x+\dfrac{1}{4}}\ge0\)
\(\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2\ge\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
Dấu = xảy ra khi \(x+\dfrac{1}{4}=0\Leftrightarrow x=-\dfrac{1}{4}\)
vậy \(M_{min}=\dfrac{1}{4}\) khi \(x=-\dfrac{1}{4}\)
b) ta có \(M=\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2\)
ĐK \(x\ge\dfrac{-1}{4}\)
\(\Leftrightarrow\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2=9\)
\(\Leftrightarrow\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2-9=0\)
\(\Leftrightarrow\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}-3\right)\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+\dfrac{1}{4}}=\dfrac{5}{2}\\\sqrt{x+\dfrac{1}{4}}=\dfrac{-7}{2}\left(vl\right)\end{matrix}\right.\)
\(\Leftrightarrow x+\dfrac{1}{4}=\dfrac{25}{4}\Rightarrow x=6\)
\(P=\frac{B}{A}=\frac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)}=\frac{3\sqrt{x}}{\sqrt{x}-2}\)
Để \(\left|P\right|>P\Rightarrow P< 0\)
\(\Rightarrow\frac{3\sqrt{x}}{\sqrt{x}-2}< 0\Rightarrow\sqrt{x}-2< 0\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)
Mà \(x\) nguyên \(\Rightarrow x=\left\{2;3\right\}\)
a, Biến đổi ta được E = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b, Ta có E = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\) = \(1+\dfrac{4}{\sqrt{x}-3}\) .
. Nếu x không là số chính phương thì \(\sqrt{x}\) là số vô tỉ . Suy ra E là số vô tỉ ( loại )
. Nếu x là số chính phươn thì \(\sqrt{x}\) là số nguyên nên để E có giá trị nguyên thì \(4⋮\left(\sqrt{x}-3\right)\) .
Mà \(\sqrt{x}-3\ge-3\) nên \(\left(\sqrt{x}-3\right)\in\left\{-2;-1;1;2;4\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{1;2;4;5;7\right\}\Rightarrow x\in\left\{1;4;16;25;49\right\}\)
Kết hợp với ĐKXĐ ta được x = 1 ; 16 ; 25 ; 49