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a) ĐKXĐ: \(a>0;a\ne1\)
b) ta có:
\(P=\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{\sqrt{a}+1}{a}\)
\(=\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\frac{\sqrt{a}+1}{a}\)
\(=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right):\frac{\sqrt{a}+1}{a}=\frac{a-1}{\sqrt{a}}.\frac{a}{\sqrt{a}+1}\)
\(=\sqrt{a}\left(\sqrt{a}-1\right)\)
c) ta có:
\(P=\sqrt{a}\left(\sqrt{a}-1\right)=a-\sqrt{a}=a-\sqrt{a}+\frac{1}{4}-\frac{1}{4}\)
\(=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Dấu "=" xảy ra khi : \(a=\frac{1}{4}\)
Vậy min P =-1/4 khi a=1/4
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Ngẫm nghĩ xem quên mật khẩu hay bị hack đây!!
a, ĐK: \(a\ge0;\) a khác 4
b,\(A=\frac{a+3\sqrt{a}+2+2a-4\sqrt{a}-5\sqrt{a}-2}{\left(\sqrt{a}-2\right)\cdot\left(\sqrt{a}+2\right)}=\frac{3a-6\sqrt{a}}{\left(\sqrt{a}+2\right)\cdot\left(\sqrt{a}-2\right)}=\frac{3\sqrt{a}}{\sqrt{a}+2}\)
c, để A= 2 KHI \(\frac{3\sqrt{a}}{\sqrt{a}+2}=2\)
<=>\(3\sqrt{a}=2\sqrt{a}+4\)
<=>\(\sqrt{a}=4\)
<=>a=16
tick nha
\(S=\frac{\left[\frac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right]^3+2a\sqrt{a}+b\sqrt{b}}{3a^2+3b\sqrt{ab}}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)
\(S=\frac{\left(\sqrt{a}-\sqrt{b}\right)^3+2\left(\sqrt{a}\right)^2\sqrt{a}+\left(\sqrt{b}\right)^2\sqrt{b}}{3a^2+3b\sqrt{ab}}+\frac{\sqrt{b}-\sqrt{a}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(S=\frac{\left(\sqrt{a}\right)^3-3\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)-\left(\sqrt{b}\right)^3+2\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{3a^2+3b\sqrt{ab}}-\frac{1}{\sqrt{a}+\sqrt{b}}\)
\(S=\frac{3\left(\sqrt{a}\right)^3-3a\sqrt{b}+3\sqrt{a}b}{3a^2+3b\sqrt{ab}}-\frac{1}{\sqrt{a}+\sqrt{b}}\)
\(S=\frac{\sqrt{a}\left(a-\sqrt{ab}+b\right)}{\sqrt{a}\left[\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3\right]}-\frac{1}{\sqrt{a}+\sqrt{b}}\)
\(S=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{\sqrt{a}+\sqrt{b}}\)
\(S=\frac{1}{\sqrt{a}+\sqrt{b}}-\frac{1}{\sqrt{a}+\sqrt{b}}=0\)
C= \(\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\) - \(\frac{2}{\sqrt{ab}}\); \(\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2\)
= \(\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)- \(\frac{2}{\sqrt{ab}}\).: \(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{ab}\)
= \(\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)-\(\frac{2\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
= \(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
=1
#mã mã#
a)\(\hept{\begin{cases}a\ge0\\\sqrt{a}-2>0\Leftrightarrow\\\sqrt{a}+2>0\end{cases}a>4}\)
b)\(\frac{\sqrt{a}\left(\sqrt{a}+2\right)+\sqrt{a}\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}.\frac{a-4}{2\sqrt{a}}\) \(=\frac{2a}{a-4}.\frac{a-4}{2\sqrt{a}}=\sqrt{a}\)
c)\(\sqrt{a}>3\Leftrightarrow a>9\)