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a/ Ta có
P = \(\frac{1+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) - \(\frac{2+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) - \(\frac{1+\sqrt{x}}{x+\sqrt{x}+1}\)
= \(\frac{-\sqrt{x}}{1+\sqrt{x}+x}\)
a)ĐKXĐ : x > 0
P = \(\left(\frac{x-1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}\left(1+\sqrt{x}\right)}\right)\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{1}{\sqrt{x}}.\left(\sqrt{x}-1+\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{\sqrt{x}-1}{\sqrt{x}}.\left(1-\frac{1}{\sqrt{x}+1}\right)\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{\left(\sqrt{x}-1\right).\sqrt{x}}{\sqrt{x}}\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
Vậy P = \(\frac{\sqrt{x}+1}{\sqrt{x}}\)
b) x = \(\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\frac{2.\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{3}-1\)
=> P = \(\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{\sqrt{3}-1+1}{\sqrt{3}-1}=\frac{\sqrt{3}}{\sqrt{3}-1}\)
= \(\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3+1}\right)}=\frac{3+\sqrt{3}}{3-1}=\frac{3+\sqrt{3}}{2}\)
c)\(P\sqrt{x}=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Leftrightarrow\frac{\left(\sqrt{x}+1\right)\sqrt{x}}{\sqrt{x}}=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Leftrightarrow\sqrt{x}+1=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Leftrightarrow\sqrt{x-4}=5\sqrt{x-4}\)
Đặt \(\hept{\begin{cases}a=\sqrt{x}\\b=\sqrt{x-4}\end{cases}\Rightarrow a^2+b^2=x-\left(x-4\right)=4}\)
\(\Rightarrow\hept{\begin{cases}a^2-b^2=4\\b=5a-4\end{cases}\Rightarrow\hept{\begin{cases}a^2-\left(5a-4\right)^2=4\left(^∗\right)\\b=5a-4\end{cases}}}\)
Từ (*) <=> a2 -(25a2 -40a + 16 ) =4
<=> -24a2 + 40a - 20 = 0
=> \(\Delta'=-80< 0\)
=> PT vô nghiệm
=> ko tồn tại x thỏa mãn
\(M=\left(\frac{x+2}{x\sqrt{x-1}}+\frac{\sqrt{x}}{x+\sqrt{x+1}}-\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}-1}{7}\)
\(=\left[\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right]:\frac{\sqrt{x}-1}{7}\)
\(=\left[\frac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]:\frac{\sqrt{x}-1}{7}\)
\(=\left[\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]:\frac{\sqrt{x}-1}{7}\)
\(=\frac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{7}{\sqrt{x}-1}\)
\(=\frac{\left(\sqrt{x}-1\right)^2.7}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}\)
\(=\frac{7}{x+\sqrt{x}+1}\)
a) ĐK: \(x\ge0;x\ne1\)
Ta có: \(x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(x+\sqrt{x}-2=\left(x-1\right)+\left(\sqrt{x}-1\right)=\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\)
=> \(P=\frac{3\left(\sqrt{x}+1\right)+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{2+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{4\sqrt{x}}{\sqrt{x}+1}\)
b) \(P=\sqrt{x}-1\)
<=> \(\frac{4\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-1\)
<=> \(x-4\sqrt{x}-1=0\)
<=> \(\orbr{\begin{cases}\sqrt{x}=2+\sqrt{5}\\\sqrt{x}=2-\sqrt{5}< 0\left(loại\right)\end{cases}}\)
<=> \(x=9+4\sqrt{5}\)thỏa mãn
Cái thứ 2 là cộng hay trừ vậy bạn ?
cộng
giúp mình nha