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3 tháng 5 2018
\(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)\)
\(A=\)\(\left[\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right]\left[\frac{a+1}{\sqrt{a}}\right]\)
\(A=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{a-1}.\) \(\frac{a+1}{\sqrt{a}}\)
\(A=\frac{4a\sqrt{a}}{a-1}.\frac{a+1}{\sqrt{a}}\)
\(A=\frac{4a\left(a+1\right)}{a-1}\)
ta có \(a=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(a=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(a=\left(4+\sqrt{15}\right).2\left(4-\sqrt{15}\right)\)
\(a=2\left(16-15\right)\)
\(a=2\)
khi đó \(A=\frac{4.2.\left(2+1\right)}{2-1}=8.3=24\)
vậy.....
6 tháng 9 2020
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 :))
6 tháng 9 2020
\(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{1-\sqrt{a}}{\sqrt{a}-1}\right)\div\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}}{\sqrt{a}+1}+\frac{\sqrt{a}}{1-a}\right)\)
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 :(
5 tháng 7 2019
\(P=\)\(\left(\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{1}{a-1}\right):\frac{a}{2+2\sqrt{a}}.\)
\(=\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{a-1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{a-1}+\frac{1}{a-1}\right):\frac{a}{2\left(\sqrt{a}+1\right)}\)
\(=\frac{a-\sqrt{a}-a-\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\div\frac{a}{2\left(\sqrt{a}+1\right)}\)
\(=\frac{1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{2\left(\sqrt{a}+1\right)}{a}=\frac{2}{a\left(\sqrt{a}-1\right)}\)
\(b,P=\frac{2}{a\left(\sqrt{a}-1\right)}=\frac{2}{25\left(5-1\right)}=\frac{2}{25.4}=\frac{1}{50}\)
Vậy \(P=\frac{1}{50}\)tại \(a=25\)
HT
15 tháng 9 2017
a) A = \(\left(\frac{1}{1-\sqrt{x}}+\frac{1}{1+\sqrt{x}}\right):\left(\frac{1}{1-\sqrt{x}}-\frac{1}{1+\sqrt{x}}\right)+\frac{1}{1-\sqrt{x}}\)
A = \(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{1-x}\right):\left(\frac{1+\sqrt{x}-1+\sqrt{x}}{1-x}\right)+\frac{1}{1-\sqrt{x}}\)
A = \(\left(\frac{2}{1-x}\right):\left(\frac{2\sqrt{x}}{1-x}\right)+\frac{1}{1-\sqrt{x}}\)
A = \(\frac{2}{1-x}.\frac{1-x}{2\sqrt{x}}+\frac{1}{1-\sqrt{x}}\)
A = \(\frac{2}{2\sqrt{x}}+\frac{1}{1-\sqrt{x}}\)
A = \(\frac{1}{\sqrt{x}}+\frac{1}{1-\sqrt{x}}\)
A = \(\frac{1-\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}\)
A = \(\frac{1-\sqrt{x}+\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}\)
A = \(\frac{1}{\sqrt{x}-x}\)
b) Ta tính \(\sqrt{x}=\sqrt{7+4\sqrt{3}}=2+\sqrt{3}\) .
Sau đó thế vào A, ta có \(A=\frac{1}{\sqrt{x}-x}=\frac{1}{2+\sqrt{3}-7-4\sqrt{3}}=\frac{1}{-5-3\sqrt{3}}=-\frac{1}{5+3\sqrt{3}}\)
\(\sqrt{a}-\frac{1}{\sqrt{a}}=\frac{a-1}{\sqrt{a}}\)