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Câu C : Lần đầu làm dạng này :))
Xét hiệu A - 2 , ta có :
\(A-2=\frac{2\sqrt{a}+2-4a-2}{2a+1}=\frac{2\sqrt{a}-4a}{2a+1}=\frac{2\sqrt{a}\left(1-2\sqrt{a}\right)}{2a+1}\)
Ta thấy :
+) Do \(a\ge0\)\(\Rightarrow2\sqrt{a}\left(1-2\sqrt{a}\right)\le0\)
+) a khác 1 ; \(a\ge0\)=> 2a + 1 > 0
\(\Rightarrow\frac{2\sqrt{a}\left(1-2\sqrt{a}\right)}{2a+1}\le0\)
\(\Leftrightarrow A< 2\)
P/s : sai bỏ qua :))
\(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{1-\sqrt{a}}{\sqrt{a}-1}\right)\div\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}}{\sqrt{a}+1}+\frac{\sqrt{a}}{1-a}\right)\)
ĐKXĐ : \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)
\(A=\left(\frac{\sqrt{a}+1+1-\sqrt{a}}{\sqrt{a}-1}\right)\div\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}}{a-1}\right)\)
\(A=\frac{2}{\sqrt{a}-1}\div\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(A=\frac{2}{\sqrt{a}-1}\div\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(A=\frac{2}{\sqrt{a}-1}\div\left(\frac{a+2\sqrt{a}+1+a-\sqrt{a}-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(A=\frac{2}{\sqrt{a}-1}\div\frac{2a+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(A=\frac{2}{\sqrt{a}-1}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{2a+1}\)
\(A=\frac{2\left(\sqrt{a}+1\right)}{2a+1}\)
b) \(a=1-\frac{\sqrt{3}}{2}=\frac{2}{2}-\frac{\sqrt{3}}{2}=\frac{2-\sqrt{3}}{2}\)( tmđk )
Rồi từ đây thế vô :)
c) Nhờ cao nhân làm tiếp chứ em mới lớp 8 thôi ạ :(
ĐKXĐ : \(a>0,a\ne1\)
a) \(\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}=\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{\sqrt{a}-1}{\sqrt{a}}\)
b) \(B=1-\frac{1}{\sqrt{a}}< 1\)
a)ĐK: \(a>0;a\ne1\)
\(B=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}-1}{\sqrt{a}}\)
b) \(B=\frac{\sqrt{a}-1}{\sqrt{a}}=1-\frac{1}{\sqrt{a}}< 1\)
a) Ta có: \(B=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{1-\sqrt{a}}{\sqrt{a}+1}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}}{\sqrt{a}+1}+\frac{\sqrt{a}}{1-a}\right)\)
\(=\left(\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\left(\frac{a+2\sqrt{a}+1-\left(a-2\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right):\left(\frac{a+2\sqrt{a}+1+a-\sqrt{a}-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}:\frac{2a+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{4\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}{2a+1}\)
\(=\frac{4\sqrt{a}}{2a+1}\)
b) ĐKXĐ: \(0\le a\ne1\)
Ta có: \(a=9-4\sqrt{5}\)
\(=5-2\cdot\sqrt{5}\cdot2+4\)
\(=\left(\sqrt{5}-2\right)^2\)(nhận)
Thay \(a=\left(\sqrt{5}-2\right)^2\) vào biểu thức \(B=\frac{4\sqrt{a}}{2a+1}\),ta được:
\(B=\frac{4\cdot\sqrt{\left(\sqrt{5}-2\right)^2}}{2\cdot\left(\sqrt{5}-2\right)^2+1}\)
\(=\frac{4\cdot\left|\sqrt{5}-2\right|}{2\cdot\left(9-4\sqrt{5}\right)+1}\)
\(=\frac{4\cdot\left(\sqrt{5}-2\right)}{18-8\sqrt{5}+1}\)(Vì \(\sqrt{5}>2\))
\(=\frac{4\sqrt{5}-8}{19-8\sqrt{5}}\)
Vậy: Khi \(a=9-4\sqrt{5}\) thì \(B=\frac{4\sqrt{5}-8}{19-8\sqrt{5}}\)