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a) điều kiện xác định : \(a>0\)
ta có : \(D=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(\Leftrightarrow D=\dfrac{\left(a+\sqrt{a}\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)\(\Leftrightarrow D=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
b) ta có : \(D=2\Leftrightarrow x-\sqrt{x}=2\Leftrightarrow x-\sqrt{x}-2=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\\sqrt{x}-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=4\end{matrix}\right.\)
vậy \(x=4\)
c) ta có : \(a>1\Leftrightarrow a-1>0\Leftrightarrow\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)>0\)
\(\Leftrightarrow\sqrt{a}-1>0\Leftrightarrow\sqrt{a}\left(\sqrt{a}-1\right)>0\Leftrightarrow a-\sqrt{a}>0\)
\(\Rightarrow\left|D\right|=\left|a-\sqrt{a}\right|=a-\sqrt{a}=D\) vậy \(D=\left|D\right|\)
d) ta có : \(D=a-\sqrt{a}\Leftrightarrow a-\sqrt{a}-D=0\)
phương trình này luôn có nghiệm \(\Rightarrow\Delta\ge0\)
\(\Leftrightarrow1^2-4\left(-D\right)=4D+1\ge0\Leftrightarrow D\ge\dfrac{-1}{4}\)
\(\Rightarrow D_{min}=\dfrac{-1}{4}\) khi \(\sqrt{a}=\dfrac{-b}{2a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\)
vậy \(D_{min}=\dfrac{-1}{4}\) khi \(a=\dfrac{1}{4}\)
a) \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\left(\sqrt{a}+\sqrt{b}\right).\left(\sqrt{a}-\sqrt{b}\right)=a-b\)
b) đề sai rồi nha
c) \(\dfrac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}=\dfrac{a\sqrt{a}-4\sqrt{a}+2a-8}{a-4}\)
\(=\dfrac{\sqrt{a}\left(a-4\right)+2\left(a-4\right)}{a-4}=\dfrac{\left(\sqrt{a}+2\right)\left(a-4\right)}{a-4}=\sqrt{a}+2\)
điều kiện a> 0
\(D=\frac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1..\)
\(=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\left(a-\sqrt{a}+1\right)}-\left(2\sqrt{a}+1\right)+1\)
\(\sqrt{a}\left(\sqrt{a}+1\right)-2\sqrt{a}-1+1=a-\sqrt{a}.\)
b, D = 2 => \(a-\sqrt{a}=2\Leftrightarrow a-\sqrt{a}-2=0\)
\(\Leftrightarrow\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)=0\Leftrightarrow\sqrt{a}-1=0\)( vì a > 0 nên \(\sqrt{a}+1>0\))
\(\Leftrightarrow a=1\)
c, a > 1 => \(\sqrt{a}>1\Rightarrow\sqrt{a}-1>0\)
\(\Rightarrow D=a-\sqrt{a}=\sqrt{a}\left(\sqrt{a}-1\right)>0\)
Vậy D = | D | > 0
d, \(D=a-\sqrt{a}=a-\sqrt{a}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)với mọi a > 0
vậy Dmin = - 1/4 khi a = 1/4
xin lỗi phàn b anh làm sai. Sửa lại như sau :
b, D = 2 => \(a-\sqrt{a}=2\Rightarrow a-\sqrt{a}-2=0\Leftrightarrow\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)=0.\)
\(\Leftrightarrow\sqrt{a}-2=0\)( vì a > 0, nên căn a + 1 > 0 )
\(\Leftrightarrow a=4\)
1a)
\(D=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\left(ĐK:a\ge0\right)\)
\(=\frac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\left(2\sqrt{a}+1\right)+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
2:
a: \(E=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(=\dfrac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b: a^2+3a=0
=>a(a+3)=0
=>a=0(nhận) hoặc a=-3(loại)
Khi a=0 thì \(E=\dfrac{-4}{-2}=2\)
ĐKXĐ: \(a>0\)
\(D=\frac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\left(2\sqrt{a}+1\right)+1\)
\(=a+\sqrt{a}-2\sqrt{a}=a-\sqrt{a}\)
\(D=2\Rightarrow a-\sqrt{a}=2\)
\(\Rightarrow a-\sqrt{a}-2=0\Rightarrow\left[{}\begin{matrix}\sqrt{a}=-1\left(l\right)\\\sqrt{a}=2\end{matrix}\right.\) \(\Rightarrow a=4\)
\(D=a-\sqrt{a}=\sqrt{a}\left(\sqrt{a}-1\right)\)
Với \(a>1\Rightarrow\sqrt{a}-1>0\Rightarrow D>0\Rightarrow D=\left|D\right|\)
\(D=a-\sqrt{a}=a-\sqrt{a}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
\(\Rightarrow D_{min}=-\frac{1}{4}\) khi \(\sqrt{a}=\frac{1}{2}\Rightarrow a=\frac{1}{4}\)
a: \(=\dfrac{\sqrt{x}+1-2x+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-2x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-2}{\sqrt{x}+1}\)
b: Để D=-1/2 thì \(\sqrt{x}+1=4\)
=>x=9
a) điều kiện : \(a>0\)
ta có : \(P=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(\Leftrightarrow P=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{2a+2\sqrt{a}}{\sqrt{a}}+1\)
\(\Leftrightarrow P=a+\sqrt{a}-2\sqrt{a}-2=a-\sqrt{a}-2\)
b) ta có : \(a>1\Rightarrow a>\sqrt{a}>1\Rightarrow a-\sqrt{a}>0\Rightarrow a-\sqrt{a}-2>-2\)
\(\Rightarrow\left|P\right|\ge P\) dấu "=" xảy ra khi \(a-\sqrt{a}>2\)
c) ta có : \(P=2\Leftrightarrow a-\sqrt{a}-2=2\Leftrightarrow a-\sqrt{a}-4=0\)
ta có : \(\Delta=1^2-4\left(-4\right)=17>0\) \(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(a=\dfrac{1+\sqrt{17}}{2};a=\dfrac{1-\sqrt{17}}{2}\)
vậy.................................................................................................................
d) ta có : \(P=a-\sqrt{a}-2\Leftrightarrow a-\sqrt{a}-2-P=0\)
ta có phương trình này luôn có nghiệm \(\Rightarrow\Delta\ge0\Leftrightarrow1^2-4\left(-2-P\right)\ge0\)
\(\Leftrightarrow4P+9\ge0\Leftrightarrow P\ge\dfrac{-9}{4}\)
\(\Rightarrow\) giá trị nhỏ nhất của \(P\) là \(\dfrac{-9}{4}\) ; dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{-b}{2a}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)
a: ĐKXĐ: a>0
\(A=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
b: Để D=2 thì a-căn a-2=0
=>a=4
c: Khi a>1 thì D>0
=>D=|D|