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a: \(S=\dfrac{\sqrt{x}-1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-4-x+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{-3}\)
\(=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b: Để S=0 thì \(\sqrt{x}-2=0\)
hay x=4(loại)
a: \(B=\dfrac{x-\sqrt{x}+7+\sqrt{x}+2}{x-4}:\dfrac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-2\sqrt{x}}{x-4}\)
\(=\dfrac{x+9}{x-4}\cdot\dfrac{x-4}{6\sqrt{x}}\)
\(=\dfrac{x+9}{6\sqrt{x}}\)
b: \(B-\dfrac{1}{B}=\dfrac{x+9}{6\sqrt{x}}-\dfrac{6\sqrt{x}}{x+9}\)
\(=\dfrac{x^2+18x+81-36x}{6\sqrt{x}\left(x+9\right)}=\dfrac{x^2-18x+81}{6\sqrt{x}\left(x+9\right)}=\dfrac{\left(x-9\right)^2}{6\sqrt{x}\left(x+9\right)}>0\)
=>B>1/B
a) điều kiện \(x>0;x\ne1\)
\(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{1-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(\Leftrightarrow\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(\Leftrightarrow\left(\dfrac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\left(\sqrt{x}+1\right)^2+\left(1-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(\Leftrightarrow\dfrac{x+2\sqrt{x}+1+x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x+2\sqrt{x}+1+\sqrt{x}-1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow\dfrac{x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow\dfrac{x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{4\sqrt{x}}=\dfrac{x+1}{4\sqrt{x}}\)
a: \(S=\dfrac{x+1}{\sqrt{x}}:\dfrac{x-1+1-\sqrt{x}}{x+\sqrt{x}}\)
\(=\dfrac{x+1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(x+1\right)\cdot\left(\sqrt{x}+1\right)}{\sqrt{x}\left(x-1\right)}\)
b: Khi \(x=\dfrac{2}{2+\sqrt{3}}=4-2\sqrt{3}\) vào S, ta được:
\(S=\dfrac{\left(4-2\sqrt{3}+1\right)\left(\sqrt{3}-1+1\right)}{\left(\sqrt{3}-1\right)\left(4-2\sqrt{3}-1\right)}\)
\(=\dfrac{\left(5-2\sqrt{3}\right)\cdot\sqrt{3}}{\left(\sqrt{3}-1\right)\left(3-2\sqrt{3}\right)}\)
a) điều kiện xác định : \(x\ge0;x\ne\dfrac{9}{4}\)
ta có : \(P=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(\Leftrightarrow P=\left(\dfrac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\) \(\Leftrightarrow P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1+\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{2x+3\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{2\sqrt{x}-3}{2\sqrt{x}+1}\right)=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\)b) thay \(x=\dfrac{3-2\sqrt{2}}{4}=\left(\dfrac{\sqrt{2}-1}{2}\right)^2\) vào \(P\) ta có :
\(P=\dfrac{3\sqrt{\left(\dfrac{\sqrt{2}-1}{2}\right)^2}-5}{2\sqrt{\left(\dfrac{\sqrt{2}-1}{2}\right)^2}+1}=\dfrac{3\left(\dfrac{\sqrt{2}-1}{2}\right)-5}{2\left(\dfrac{\sqrt{2}-1}{2}\right)+1}=\dfrac{6-13\sqrt{2}}{4}\)
c) ta có : \(P-\dfrac{3}{2}=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}-\dfrac{3}{2}=\dfrac{3\sqrt{x}-5-3\sqrt{x}-\dfrac{3}{2}}{2\sqrt{x}+1}\)
\(=\dfrac{\dfrac{-13}{2}}{2\sqrt{x}+1}< 0\) \(\Rightarrow P< \dfrac{3}{2}\)
a) Ta có:
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{2\sqrt{x}\left(\sqrt{x-3}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x-3}\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3x+3}{x-9}\right):\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\left(\frac{2x-6}{x-9}+\frac{x+3\sqrt{x}}{x-9}-\frac{3x+3}{x-9}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2x-6+x+3\sqrt{x}-3x-3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\)
b) \(P< \frac{-1}{2}\Rightarrow\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}< \frac{-1}{2}\)
.....Chưa nghĩ ra....
c) Ta có: \(\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-3=0\Rightarrow x=9\)
Vậy Min P = 0 khi x =9.
k - kb với tớ nhia mn!
Bài 1:
\(M=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(x-1\right)}{\sqrt{x}}\)
=2
Bài 2:
\(P=\dfrac{x+1+\sqrt{x}}{x+1}:\dfrac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{x+1}\cdot\dfrac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne0\\x\ne1\end{matrix}\right.\)
a )
\(S=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
b )
\(S=3\)
\(\Leftrightarrow x-\sqrt{x}+1=3\)
\(\Leftrightarrow x-\sqrt{x}-2=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=4\end{matrix}\right.\)
Vậy \(x=4\)
a: \(B=\dfrac{2x+\sqrt{x}-1}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\)
\(=\left(2x+\sqrt{x}-1\right)\left(\dfrac{-1}{x-1}+\dfrac{\sqrt{x}}{x\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\cdot\left(\dfrac{-x+\sqrt{x}-1+x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=-\dfrac{\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\)
\(A=\dfrac{-\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}{2\sqrt{x}-1}\)
\(=\dfrac{-x+\sqrt{x}-1}{\sqrt{x}}\)
b: Khi \(x=17-12\sqrt{2}=\left(3-2\sqrt{2}\right)^2\) thì
\(A=\dfrac{-17+12\sqrt{2}+3-2\sqrt{2}-1}{3-2\sqrt{2}}=-5\)
c: \(A=\dfrac{-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{-\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{3}{4}}{\sqrt{x}}< 0\)
=>căn A không tồn tại