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Ta có: \(C=\frac{a\sqrt{a}-1}{a-\sqrt{a}}+\frac{\sqrt{a}-1}{\sqrt{a}}\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
\(=\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{\sqrt{a}-1}{\sqrt{a}}\cdot\left(\frac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\frac{a+\sqrt{a}+1}{\sqrt{a}}+\frac{\sqrt{a}-1}{\sqrt{a}}\cdot\frac{2a+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\frac{2\left(a+1\right)}{\sqrt{a}\cdot\left(\sqrt{a}+1\right)}+\frac{a+\sqrt{a}+1}{\sqrt{a}}\)
\(=\frac{2\left(a+1\right)}{\sqrt{a}\cdot\left(\sqrt{a}+1\right)}+\frac{\left(\sqrt{a}+1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\)
\(=\frac{2a+2+a\sqrt{a}+2a+2\sqrt{a}+1}{\sqrt{a}\cdot\left(\sqrt{a}+1\right)}\)
\(=\frac{a\sqrt{a}+4a+2\sqrt{a}+3}{\sqrt{a}\cdot\left(\sqrt{a}+1\right)}\)
\(\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{\sqrt{a}+a}\right)+\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\cdot\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\dfrac{\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{a-1}\)
\(=2+\dfrac{1}{\sqrt{a}+1}\cdot\dfrac{2a+2}{\sqrt{a}}\)
\(=\dfrac{2a+2\sqrt{a}+2a+2}{\sqrt{a}\left(\sqrt{a}+1\right)}=\dfrac{4a+2\sqrt{a}+2}{\sqrt{a}\left(\sqrt{a}+1\right)}\)
Bạn cần viết đề bằng công thức toán (biểu tượng $\sum$ bên trái khung soạn thảo) để được hỗ trợ tốt hơn.
Ta có:a, |2x-1|= |2x+3|
<=> 2x - 1 = -(2x + 3)
=> 2x + 2x = 3 + 1
=> 4x = 4
=> x = 1
\(\dfrac{\sqrt{a}-1}{\sqrt{a}-a}+\dfrac{\sqrt{a}-1}{a^2+\sqrt{a}}\)
\(=\dfrac{-1}{\sqrt{a}}+\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a\sqrt{a}+1\right)}\)
\(=\dfrac{-a\sqrt{a}-1+\sqrt{a}-1}{\sqrt{a}\left(a\sqrt{a}+1\right)}=\dfrac{-a\sqrt{a}+\sqrt{a}-2}{\sqrt{a}\left(a\sqrt{a}+1\right)}\)
có bộ gõ kí hiệu Toán mà :))
ĐK : a >= 0 ; a khác 36
\(K=\left[\frac{a+14\sqrt{a}+100}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}+\frac{\left(\sqrt{a}+6\right)\left(\sqrt{a}-6\right)}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}-\frac{\left(\sqrt{a}-7\right)\left(\sqrt{a}+7\right)}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}\right]\div\left(\frac{\sqrt{a}-6}{\sqrt{a}-6}-\frac{\sqrt{a}-7}{\sqrt{a}-6}\right)\)
\(=\frac{a+14\sqrt{a}+100+a-36-a+49}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}\div\frac{1}{\sqrt{a}-6}\)
\(=\frac{a+14\sqrt{a}+113}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}\cdot\left(\sqrt{a}-6\right)=\frac{a+14\sqrt{a}+113}{\sqrt{a}+7}\)
Để K = 2 thì \(\frac{a+14\sqrt{a}+113}{\sqrt{a}+7}=2\Rightarrow a+14\sqrt{a}+113=2\sqrt{a}+14\Leftrightarrow a+12\sqrt{a}+99=0\)
Với a >= 0 thì \(a+12\sqrt{a}+99\ge99>0\)=> Không có giá trị x thỏa mãn K = 2
Ta có : \(K=\frac{a+14\sqrt{a}+113}{\sqrt{a}+7}=\frac{\left(a+14\sqrt{a}+49\right)+64}{\sqrt{a}+7}=\frac{\left(\sqrt{a}+7\right)^2+64}{\sqrt{a}+7}\)
\(=\left(\sqrt{a}+7\right)+\frac{64}{\sqrt{a}+7}\ge2\sqrt{\left(\sqrt{a}+7\right)\cdot\frac{64}{\sqrt{a}+7}}=16\)( bđt AM-GM )
Dấu "=" xảy ra <=> \(\sqrt{a}+7=\frac{64}{\sqrt{a}+7}\Rightarrow a=1\left(tm\right)\). Vậy MinK = 16