Cho biểu thức \(A=\left(1-\frac{2-2\sqrt{a}}{a-1}\right);\left(\frac{1}{1+\sqrt{a}}-\frac{a}{1+a\sqrt{a}}\right)\)
a, Rút gọn A
b, Tìm giá trị của a để A đật giá trị lớn nhất
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30 tháng 7 2019
\(A=\left(1-\frac{2\sqrt{a}-2}{a-1}\right):\left(\frac{1}{1+\sqrt{a}}-\frac{a}{1+a\sqrt{a}}\right)\)
\(=\left(\frac{a-1-\left(2\sqrt{a}-2\right)}{a-1}\right):\)\(\left(\frac{1}{\sqrt{a}+1}-\frac{a}{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}\right)\)
\(=\left(\frac{a-1-2\sqrt{a}-2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}+1\right)}\right):\)\(\left(\frac{a-\sqrt{a}+1-a}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{\left(a-2\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}:\frac{-\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=-\frac{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=-\left(\sqrt{a}-1\right)=1-\sqrt{a}\)
YY
26 tháng 5 2018
\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\frac{\left(a-1\right)^2}{4a}\left(\frac{\left(\sqrt{a}-1-\sqrt{a}-1\right)\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right)\)
\(=\frac{\left(a-1\right)\left(-2\right)2\sqrt{a}}{4a}=-\frac{\left(a-1\right)}{\sqrt{a}}\)
Y
6 tháng 7 2019
a. ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(A=\left(\frac{\left(\sqrt{a}\right)^2-1}{2\sqrt{a}}\right)^2\cdot\left(\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\\ =\left(\frac{a-1}{2\sqrt{a}}\right)^2\cdot\left(\frac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\\ =\frac{\left(a-1\right)^2}{4a}\cdot\frac{-4\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\\ =-\frac{a-1}{\sqrt{a}}=\frac{1-a}{\sqrt{a}}\)
b. Để A < 0 thì 1 - a <0 ( vì mẫu \(\sqrt{a}\ge0\forall a\) ) <=> -a < -1 <=> a > 1
1 tháng 8 2019
\(đkxđ\Leftrightarrow\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)
\(A=\)\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\)\(\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\left(\frac{\sqrt{a}.\sqrt{a}}{2\sqrt{a}}-\frac{1}{2\sqrt{a}}\right)^2\)\(\left(\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{\left(a-1\right)^2}{\left(2\sqrt{a}\right)^2}\left(\frac{a-2\sqrt{a}+1-a-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\frac{\left(a-1\right)^2.-4\sqrt{a}}{4a\left(a-1\right)}=\frac{a-1}{\sqrt{a}}\)
\(b,A< 0\Rightarrow\frac{a-1}{\sqrt{a}}< 0\)
Mà \(\sqrt{a}\ge0\Rightarrow a-1\le0\Rightarrow a\le1\)
\(A=2\Rightarrow\frac{a-1}{\sqrt{a}}=2\)
\(\Rightarrow a-1=2\sqrt{a}\Rightarrow a-2\sqrt{a}-1=0\)
\(\Rightarrow a-2\sqrt{a}+1-2=0\)
\(\Rightarrow\left(\sqrt{a}-1\right)^2-\sqrt{2}^2=0\)
\(\Rightarrow\left(\sqrt{a}-1-\sqrt{2}\right)\left(\sqrt{a}-1+\sqrt{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{a}=1+\sqrt{2}\\\sqrt{a}=1-\sqrt{2}\end{cases}\Rightarrow\orbr{\begin{cases}a=\left(1+\sqrt{2}\right)^2=3+2\sqrt{2}\\a=\left(1-\sqrt{2}\right)^2=3-2\sqrt{2}\end{cases}}}\)
BA
1 tháng 8 2019
\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\frac{\left(a-1\right)^2}{4a}.\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\left(a-1\right)^2}{4a}.\frac{\left(\sqrt{a}-1+\sqrt{a}+1\right)\left(\sqrt{a}-1-\sqrt{a}-1\right)}{a-1}\)
\(=\frac{a-1}{4a}.\frac{2\sqrt{a}.\left(-2\right)}{1}\)
\(=\frac{a-1}{4a}.\frac{-4\sqrt{a}.}{1}\)
\(=\frac{1-a}{\sqrt{a}}\)
đặt \(x=\sqrt{a}\Rightarrow x^2=a;x^3=a\sqrt{a}.\)