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\(P=\left(\frac{2\left(\sqrt{x}+2\right)+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\frac{x+2\sqrt{x}}{2\sqrt{x}}\) điều kiện x >0
\(P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}.\frac{x+2\sqrt{x}}{2\sqrt{x}}\)
\(P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}=1+\frac{4+x}{2\sqrt{x}}.\)
b) P = 3
\(\Leftrightarrow1+\frac{4+x}{2\sqrt{x}}=3\Leftrightarrow\frac{4+x}{2\sqrt{x}}=2\)
\(\Leftrightarrow4+x=4\sqrt{x}\Leftrightarrow4+x-4\sqrt{x}=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-2=0\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)
Ngô Văn Tuyên cảm ơn bạn nha. Nhưng cho mình hỏi tí sao bạn lại tách ra thành \(1+\frac{4-x}{2\sqrt{x}}\)
giải thích hộ mình với nhé. Cảm ơn nhiều !!
Ta có: \(1+\left(\dfrac{2a+\sqrt{a}-1}{1-a}-\dfrac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right)\cdot\dfrac{a-\sqrt{a}}{2\sqrt{a}-1}\)
\(=1+\left(\dfrac{-2\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\cdot\dfrac{a-\sqrt{a}}{2\sqrt{a}-1}\)
\(=1+\left(\dfrac{-\left(2\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)+\sqrt{a}\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\dfrac{\left(2\sqrt{a}-1\right)\left(-a-\sqrt{a}-1+a+\sqrt{a}\right)}{a+\sqrt{a}+1}\cdot\dfrac{\sqrt{a}}{2\sqrt{a}-1}\)
\(=1+\dfrac{-\sqrt{a}}{a+\sqrt{a}+1}\)
\(=\dfrac{a+\sqrt{a}+1-\sqrt{a}}{a+\sqrt{a}+1}\)
\(=\dfrac{a+1}{a+\sqrt{a}+1}\)
ĐKXĐ : \(x\ge0\)
\(A=\frac{2}{3}.\frac{2+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)^2+\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2}{\left[1+\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2\right]\left[1+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)^2\right]}.\frac{2010}{x+1}\)
\(A=\frac{2}{3}.\frac{2+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}+\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2-2\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)}{\left[1+\frac{\left(2\sqrt{x}+1\right)^2}{3}\right]\left[1+\frac{\left(2\sqrt{x}-1\right)^2}{3}\right]}.\frac{2010}{x+1}\)
\(A=\frac{2}{3}.\frac{2+\left(\frac{4\sqrt{x}}{\sqrt{3}}\right)^2-\frac{2\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}{3}}{\left(\frac{4x+4\sqrt{x}+4}{3}\right)\left(\frac{4x-4\sqrt{x}+4}{3}\right)}.\frac{2010}{x+1}\)
\(A=\frac{2}{3}.\frac{2+\frac{16x}{3}-\frac{2\left(4x-1\right)}{3}}{\frac{16\left(x+1+\sqrt{x}\right)\left(x+1-\sqrt{x}\right)}{9}}.\frac{2010}{x+1}\)
\(A=\frac{2}{3}.\frac{\frac{6+16x-8x+2}{3}}{\frac{16\left(x+1\right)^2-16x}{9}}.\frac{2010}{x+1}\)
\(A=\frac{x+1}{x^2+x+1}.\frac{2010}{x+1}=\frac{2010}{x^2+x+1}\le2010\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=0\)
...
câu 2
\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)
câu 1
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
a, ĐK \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(Q=\left(1+\frac{\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\frac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\)\(=\frac{x+\sqrt{x}+1}{x+1}.\frac{\left(\sqrt{x}-1\right)\left(x+1\right)}{\left(\sqrt{x}-1\right)^2}=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
b. \(Q>1\Rightarrow Q-1>0\Rightarrow\frac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}>0\)
\(\Rightarrow\frac{x+2}{\sqrt{x}-1}>0\)
TH1 \(\hept{\begin{cases}x+2>0\\\sqrt{x}-1>0\end{cases}\Rightarrow\hept{\begin{cases}x>-2\\x>1\end{cases}\Rightarrow}x>1}\)
TH2 \(\hept{\begin{cases}x+2< 0\\\sqrt{x}-1< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\0\le x< 1\end{cases}\left(l\right)}}\)
Vậy \(x>1\)thì \(Q>1\)
\(\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^1}}\right):\frac{b}{a-\sqrt{a^2-b^2}}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\left(\frac{\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}+\frac{a}{\sqrt{a^2-b^2}}\right).\frac{a-\sqrt{a^2-b^2}}{b}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}.\frac{a-\sqrt{a^2-b^2}}{b}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-\left(a^2-b^2\right)}{b.\sqrt{a^2-b^2}}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-a^2+b^2}{b\sqrt{a^2-b^2}}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{b^2}{b\sqrt{a^2-b^2}}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{b}{\sqrt{a^2-b^2}}\)
\(=\frac{a-b}{\sqrt{a^2-b^2}}\)
\(=\frac{a-b}{\sqrt{a-b}.\sqrt{a+b}}\)
\(=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)
\(=\frac{\sqrt{a^2-b^2}}{a+b}\)