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a) \(\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\left(2\sqrt{a}+1\right)+1\)
\(=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-2\sqrt{a}-1+1\)
\(=\frac{a^2-\sqrt{a}}{a-\sqrt{a}+1}-2\sqrt{a}\)
b) \(\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-2\sqrt{a}=2\)
\(\Leftrightarrow a^2+\sqrt{a}.\left(a-\sqrt{a}+1\right)-2\sqrt{a}.\left(a-\sqrt{a}+1\right)=2\left(a-\sqrt{a}+1\right)\)
\(\Leftrightarrow a^2-2\sqrt{a}.a+2a-\sqrt{a}-2a=2a-2\sqrt{a}+2\)
\(\Leftrightarrow a^2-2\sqrt{a}.a+2a-\sqrt{a}-2a=-2\sqrt{a}+2\)
\(\Leftrightarrow-2\sqrt{a}.a+2a-\sqrt{a}-2a=-2\sqrt{a}+2-a^2\)
\(\Leftrightarrow-2\sqrt{a}.a-\sqrt{a}=-2\sqrt{a}+2-a^2\)
\(\Leftrightarrow-2a\sqrt{a}+\sqrt{a}=2-a^2\)
\(\Leftrightarrow\sqrt{a}.\left(2a+1\right)=2-a^2\)
\(\Leftrightarrow\left[\sqrt{a}.\left(2a+1\right)\right]^2=\left(2-a^2\right)^2\)
\(\Leftrightarrow4a^3-4a^2+a=4-4a^2+a^4\)
\(\Leftrightarrow\orbr{\begin{cases}a=4\left(\text{thỏa mãn}\right)\\a=1\left(\text{loại}\right)\end{cases}}\)
=> a = 4
Cách ngắn hơn :
\(đkxđ\Leftrightarrow x\ge0\)
\(A=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\frac{\sqrt{a}\left(\sqrt{a}^3+1\right)}{a-\sqrt{a}+1}-\left(2\sqrt{a}+1\right)+1\)
\(=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}\)\(-2\sqrt{a}-1+1\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)-2\sqrt{a}\)
\(=a+\sqrt{a}-2\sqrt{a}=a-\sqrt{a}\)
\(b,A=2\Rightarrow a-\sqrt{a}=2\)
\(\Rightarrow a-\sqrt{a}-2=0\)
\(\Rightarrow a+\sqrt{a}-2\sqrt{a}-2=0\)
\(\Rightarrow\sqrt{a}\left(\sqrt{a}+1\right)-2\left(\sqrt{a}+1\right)=0\)
\(\Rightarrow\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{a}=2\\\sqrt{a}=-1\end{cases}\Rightarrow\orbr{\begin{cases}a=4\\a\in\varnothing\end{cases}}}\)
\(\Rightarrow a=4\)
\(c,A=a-\sqrt{a}=\sqrt{a}^2-2.\sqrt{a}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\)
\(=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\)
\(\Rightarrow A_{min}=-\frac{1}{4}\Leftrightarrow\left(\sqrt{a}-\frac{1}{2}\right)^2=0\)
\(\Rightarrow\sqrt{a}=\frac{1}{2}\Rightarrow a=\frac{1}{4}\)
Vậy với \(a=\frac{1}{4}\)thì A có giá trị nhỏ nhất là \(-\frac{1}{4}\)
a) ĐK: \(a>0\)
\(A=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\frac{\sqrt{a}.\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=\sqrt{a}.\left(\sqrt{a}+1\right)-\left(2\sqrt{a}+1\right)+1\)
\(=a+\sqrt{a}-2\sqrt{a}=a-\sqrt{a}\)
ĐK \(a\ge0\)
\(A=\frac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\left(2\sqrt{a}+1\right)+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
b. \(A=2\Rightarrow a-\sqrt{a}-2=0\Rightarrow\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{a}=2\\\sqrt{a}=-1\left(l\right)\end{cases}\Rightarrow a=4}\)
Vậy a=4 thì A=2
c. \(A=a-\sqrt{a}=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall a\Rightarrow A\ge-\frac{1}{4}\)
Vậy \(MinA=-\frac{1}{4}\Rightarrow\sqrt{a}=\frac{1}{2}\Rightarrow a=\frac{1}{4}\)
a) \(P=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\frac{\sqrt{a}.\left[\left(\sqrt{a}\right)^3+1\right]}{a-\sqrt{a}+1}-\frac{\sqrt{a}.\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=\frac{\sqrt{a}.\left(\sqrt{a}+1\right).\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-2\sqrt{a}-1+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1=a-\sqrt{a}\)
b)Ta có a>0 do đó: \(P=a-\sqrt{a}\ge0\)
Dấu "=" xảy ra khi a=1
c) Ta thấy \(P\ge0\)
=>P2\(\ge\)P
=>P\(\ge\)\(\sqrt{P}\)
ĐKXĐ: \(a>0\)
a/ \(P=\frac{\sqrt{a}\left(\sqrt{a}^3+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\left(2\sqrt{a}+1\right)+1\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)-2\sqrt{a}-1+1\)
\(=a-\sqrt{a}\)
b/ Ta có: \(\hept{\begin{cases}a>0\\\sqrt{a}\ge0\end{cases}\Rightarrow a-\sqrt{a}\ge0}\)
MinP = 0 khi \(\sqrt{a}=0\Rightarrow a=0\)
c/ \(P\ge\sqrt{P}\)
Bài 1 :
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
\(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}-2}\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
b) Để \(A< -1\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< -1\)
\(\Leftrightarrow\sqrt{x}-2< -\sqrt{x}-1\)
\(\Leftrightarrow2\sqrt{x}< 1\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{2}\)
\(\Leftrightarrow x< \frac{1}{4}\)
Vậy để \(A< -1\Leftrightarrow x< \frac{1}{4}\)
1,
\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)
\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{a^2-a}.\frac{a-2}{a+2}\)
\(=\frac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{a\left(a-1\right)}.\frac{a-2}{a+2}\)
\(=\frac{2a\left(a-1\right)\left(a-2\right)}{a\left(a-1\right)\left(a+2\right)}=\frac{2\left(a-2\right)}{a+2}\)
Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)
2,
a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)
b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)
\(=\frac{2\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}-\frac{\sqrt{x}-2}{x-4}\)
\(=\frac{2\sqrt{x}+2+2}{x-4}=\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{2}{\sqrt{x}-2}\)
c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
a) \(A=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\left(ĐK:a\ge0\right)\)
\(=\frac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\left(2\sqrt{a}+1\right)+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
b) Để A=2 \(\Leftrightarrow a-\sqrt{a}=2\)
\(\Leftrightarrow a-\sqrt{a}-2=0\)
\(\Leftrightarrow\left(\sqrt{a}+1\right)\left(\sqrt{a}-2\right)=0\)
\(\Leftrightarrow\sqrt{a}-2=0\left(Vì\sqrt{a}+1\ne0\right)\)
\(\Leftrightarrow a=4\) (TM)
Vậy a=4 thì A=2
c) \(A=a-\sqrt{a}=a-\sqrt{a}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\)
Vì: \(\left(\sqrt{a}-\frac{1}{2}\right)^2\ge0\)
=> \(\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Vậy GTNN của A là \(-\frac{1}{4}\) khi \(a=\frac{1}{4}\)
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