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1.\(DK:x\le\frac{1}{3}\)
2.\(DK:x\ge-1\)
3.\(DK:-1\le x< 1\)
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4\sqrt{x}-3}{2\sqrt{x}-x}\right):\)\(\left(\frac{\sqrt{x}+2}{\sqrt{x}}-\frac{\sqrt{x}-4}{\sqrt{x}-2}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)\(:\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}:\frac{x-4-x+4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-3}{4}\)
\(b,\)Để \(P>0\Rightarrow\frac{\sqrt{x}-3}{4}>0\)
Mà \(4>0\Rightarrow\sqrt{x}-3>0\Rightarrow\sqrt{x}>3\Rightarrow x>9\)
\(c,\sqrt{P}_{min}=0\Rightarrow\frac{\sqrt{x}-3}{4}=0\)
\(\Leftrightarrow\sqrt{x}-3=0\Rightarrow\sqrt{x}=3\Rightarrow x=9\)
\(dkxd\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}-2\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}}\)
\(A=\left(\frac{\sqrt{x}}{x-4}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}+2}.\)
\(=\left(\frac{\sqrt{x}}{x-4}-\frac{2\left(\sqrt{x}+2\right)}{x-4}+\frac{\sqrt{x}-2}{x-4}\right):\frac{1}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+2}{1}\)
\(=\frac{-6\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=-\frac{6}{\sqrt{x}-2}\)
\(A=\)\(\left(\frac{\sqrt{x}}{x-4}+\frac{2}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\)\(:\frac{1}{\sqrt{x}+2}\)
a,ĐKXĐ:\(\hept{\begin{cases}x\ge0\\2-\sqrt{x}\\x-4\ne0\end{cases}\ne0}\)\(\Rightarrow\)\(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(A=\)\(\left(\frac{\sqrt{x}}{x-4}+\frac{2}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\)\(:\frac{1}{\sqrt{x}+2}\)
\(A=\)\(\left(\frac{\sqrt{x}}{x-4}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right)\)\(.\left(\sqrt{x}+2\right)\)
\(A=\)\(\left(\frac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}-2}{x-4}\right)\)\(.\left(\sqrt{x}+2\right)\)
\(A=\)\(\left(\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{x-4}\right)\)\(.\left(\sqrt{x}+2\right)\)
\(A=\)\(\left(\frac{-6}{x-4}\right)\)\(.\left(\sqrt{x}+2\right)\)
\(A=\)\(\frac{-6}{\sqrt{x}-2}\)
b,\(x=9-4\sqrt{5}\)\(\Rightarrow\)\(A=\)\(\frac{-6}{\sqrt{9-4\sqrt{5}}-2}\)\(=\frac{-6}{\sqrt{5-2.2\sqrt{5}+4}-2}\)
\(A=\)\(\frac{-6}{\sqrt{\left(\sqrt{5}-2\right)^2}-2}\)\(=\frac{-6}{\sqrt{5}-2-2}\)\(=\frac{-6}{\sqrt{5}-4}\)
c,\(A>-1\)\(\Rightarrow\)\(\frac{-6}{\sqrt{x}-2}\)\(>-1\)\(\Rightarrow\)\(\frac{-6}{\sqrt{x}-2}+1>0\)
\(\Leftrightarrow\)\(\frac{-6+\sqrt{x}-2}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\)\(\frac{\sqrt{x}-8}{\sqrt{x}-2}>0\)
ĐK \(x\ne0\)
Chia cả 2 vế cho \(\frac{1}{x}\)ta được
\(\frac{3}{3x-4+\frac{1}{x}}+\frac{13}{3x+2+\frac{1}{x}}=6\)
Đặt \(3x+\frac{1}{x}=y\)
\(\Rightarrow\frac{3}{y-4}+\frac{13}{y+2}=6\)
\(\Leftrightarrow16y-46=6\left(y-4\right)\left(y+2\right)\)
Đến đây tự giải nhé (Phá ngoặc rồi ghép cặp lại)
Lời giải :
a) \(A=3\sqrt{x-1}+7\ge7\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=1\)
b) \(B=\frac{4}{\sqrt{x}+3}\le\frac{4}{3}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
c) \(C=\frac{3\sqrt{x}+8}{\sqrt{x}+3}=\frac{3\left(\sqrt{x}+3\right)-1}{\sqrt{x}+3}=3-\frac{1}{\sqrt{x}+3}\)
Có \(\frac{1}{\sqrt{x}+3}\le\frac{1}{3}\forall x\)
\(\Leftrightarrow-\frac{1}{\sqrt{x}+3}\ge\frac{-1}{3}\)
\(\Leftrightarrow3-\frac{1}{\sqrt{x}+3}\ge3-\frac{1}{3}=\frac{8}{3}\)
\(\Leftrightarrow C\ge\frac{8}{3}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
d) \(D=x-3\sqrt{x}+2\)
\(D=\left(\sqrt{x}\right)^2-2\cdot\sqrt{x}\cdot\frac{3}{2}+\frac{9}{4}-\frac{1}{4}\)
\(D=\left(\sqrt{x}-\frac{3}{2}\right)^2-\frac{1}{4}\ge\frac{-1}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=\frac{3}{2}\Leftrightarrow x=\frac{9}{4}\)
e) \(E=\frac{4}{x-2\sqrt{x}+3}=\frac{4}{\left(\sqrt{x}-1\right)^2+2}\le\frac{4}{2}=2\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
a) Vì \(3\sqrt{x-1}\ge0\forall x\ge1\)
\(\Rightarrow3\sqrt{x-1}+7\ge7\forall x\ge1\)
Dấu "=" xảy ra <=>\(3\sqrt{x-1}=0\Leftrightarrow\sqrt{x-1}=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy Amin =7 tại x=1