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\(A=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(A=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(A=\sqrt{a}\left(\sqrt{a}+1\right)-\left(2\sqrt{a}+1\right)+1\)
\(A=a+\sqrt{a}-2\sqrt{a}-1+1\)
\(A=a-\sqrt{a}\)
a) Ta có: \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1+1\)
\(=a-\sqrt{a}\)
c) Để A=2 thì \(a-\sqrt{a}-2=0\)
\(\Leftrightarrow\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)=0\)
\(\Leftrightarrow a=4\)
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 )
\(\Leftrightarrow a=4\)
a) ĐK: \(a>0\)
\(A=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\frac{\sqrt{a}.\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=\sqrt{a}.\left(\sqrt{a}+1\right)-\left(2\sqrt{a}+1\right)+1\)
\(=a+\sqrt{a}-2\sqrt{a}=a-\sqrt{a}\)
Giúp m với