rút gọn bt sau:
\(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a-1}}:\frac{\sqrt{a+1}}{a-2\sqrt{a+1}}\)
Những câu hỏi liên quan
A=\(\left(\frac{1}{2\sqrt{a}-a}+\frac{1}{2-\sqrt{a}}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}}\)Rút gọn bt
\(=\frac{1+a}{2\sqrt{a}-a}.\frac{2\sqrt{a}-a}{-\left(1+\sqrt{a}\right)}=\frac{-\left(1+a\right)}{1+\sqrt{a}}\)
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BT rút gọn với ĐK: a>0 và a khác 1:
M = \(\left(\frac{2+\sqrt{a}}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right)\)\(\frac{a\sqrt{a}+a-\sqrt{a}-1}{\sqrt{a}}\)
N = \(\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\)\(\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\)
\(M=\left(\frac{2+\sqrt{a}}{\left(\sqrt{a}+1\right)^2}-\frac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\frac{a\left(\sqrt{a}+1\right)-\left(\sqrt{a}+1\right)}{a}\)
\(=\frac{\left(2+\sqrt{a}\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(a-1\right)}{a}\)
\(=\frac{2\sqrt{a}-2+a-\sqrt{a}-a-\sqrt{a}+2\sqrt{a}+2}{\left(\sqrt{a}+1\right)\left(a-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(a-1\right)}{a}\)
\(=\frac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(a-1\right)}{a}\)
\(=\frac{2\sqrt{a}\left(\sqrt{a-1}\right)}{a\left(\sqrt{a}+1\right)}=\frac{2\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\)
\(N=\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}\)
\(=\left(\frac{a+1+2\sqrt{a}-a-1+2\sqrt{a}}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}\)
\(=\left(\frac{4\sqrt{a}}{a-1}+4\sqrt{a}\right)\cdot\frac{a-1}{\sqrt{a}}=4\sqrt{a}\left(\frac{1}{a-1}+1\right)\cdot\frac{a-1}{\sqrt{a}}=4\cdot\left(a-1\right)\left(\frac{1}{a-1}+1\right)\)
\(=4\cdot\left(a-1\right)\)
vừa tham khảo cách làm vừa check lại hộ tớ với nhé :33
\(Với\)\(a>0\);\(a\ne1\)ta có:
\(M=(\frac{2+\sqrt{a}}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}).(\frac{a\sqrt{a}+a-\sqrt{a}-1}{\sqrt{a}})\)
\(=[\frac{\sqrt{a}+2}{(\sqrt{a}+1)^2}-\frac{\sqrt{a}-2}{(\sqrt{a}+1)(\sqrt{a}-1)}].\frac{(a\sqrt{a}-\sqrt{a})+(\sqrt{a}-1)}{\sqrt{a}}\)
\(=[\frac{(\sqrt{a}-2).(\sqrt{a}-1)}{(\sqrt{a}+1)^2.(\sqrt{a}-1)}-\frac{(\sqrt{a}-2).(\sqrt{a}+1)}{(\sqrt{a}+1)^2.(\sqrt{a}-1)}].\frac{\sqrt{a}(a-1)+(a-1)}{\sqrt{a}}\)
\(=[\frac{a+\sqrt{a}-2}{(\sqrt{a}+1)(a-1)}-\frac{a-\sqrt{a}-2}{(\sqrt{a}+1)(a-1)}].\frac{(a-1).(\sqrt{a}+1)}{\sqrt{a}}\)
\(=\frac{a+\sqrt{a}-2-a+\sqrt{a}+2}{(a-1).(\sqrt{a}+1)}.\frac{(a-1)(\sqrt{a}+1)}{\sqrt{a}}\)
\(=\frac{2\sqrt{a}}{(a-1)(\sqrt{a}+1)}.\frac{(a-1)(\sqrt{a}+1)}{\sqrt{a}}\)
\(=2\)
Vậy \(M=2\)
\(Với\)\(a>0;a\ne1:\)\(N=(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}).(\sqrt{a}-\frac{1}{\sqrt{a}})\)
\(=[\frac{(\sqrt{a}+1).(\sqrt{a}+1)}{\left(\sqrt{a}-1\right).(\sqrt{a}+1)}-\frac{(\sqrt{a}-1).(\sqrt{a}-1)}{(\sqrt{a}-1).(\sqrt{a}+1)}+\frac{4\sqrt{a}(a-1)}{(\sqrt{a}-1).(\sqrt{a}+1)}].\frac{a-1}{\sqrt{a}}\)
\(=\frac{(\sqrt{a}+1)^2-(\sqrt{a}-1)^2+(4a\sqrt{a}-4\sqrt{a})}{(\sqrt{a}-1).(\sqrt{a}+1)}.\frac{a-1}{\sqrt{a}}\)
\(=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{a-1}.\frac{a-1}{\sqrt{a}}\)
\(=\frac{4a\sqrt{a}}{a-1}.\frac{a-1}{\sqrt{a}}\)\(=4a\)
Vậy \(N=4a\)
Xem thêm câu trả lời
cho bt: \(A=\frac{1}{2\sqrt{2}-2}-\frac{1}{2\sqrt{2}+2}+\frac{\sqrt{a}}{1-a}\)
a, rút gọn bt
b, tính giá trị A biết a=4/9
c, tìm a để |A| = 1/2
Bạn ơi, mk làm câu a), các câu sau bạn tự làm dc k ???
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rút gọn các bt sau:
(\(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\)) (\(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\))
\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}\)
\(\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
rút gọn bt sau : \(\sqrt{1-a}\sqrt{a\left(a-1\right)}+a\sqrt{\frac{a-1}{a}}\)
ĐK a < 0 hoặc a = 1
\(=\sqrt{1-a}+\sqrt{a\left(a-1\right)}-\sqrt{\frac{a^2\left(a-1\right)}{a}}=\sqrt{1-a}+\sqrt{a\left(a-1\right)}-\sqrt{a\left(a-1\right)}=\sqrt{1-a}\)
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Đk \(1-a\ge0;a\left(1-a\right)\ge0;\frac{a}{a-1}\ge0\)
=> \(a\le1;a\ge1ora\le0;a\ge1ora
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Cho bt
R=\(\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{1-\sqrt{a}}{\sqrt{a}+1}\right)\div\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}}{\sqrt{a}+1}+\frac{\sqrt{a}}{1-\sqrt{a}}\right)\)
a) rút gọn b) tính giá trị của R khi a=\(1-\frac{\sqrt{3}}{2}\) c) o sánh R với 2
B=\(\left(\frac{1}{1-\sqrt{a}}-\frac{1}{1+\sqrt{a}}\right)\left(\frac{1}{\sqrt{a}}+1\right)\) . RÚT GỌN BT
\(B=\left(\frac{1}{1-\sqrt{a}}-\frac{1}{1+\sqrt{a}}\right)\left(\frac{1}{\sqrt{a}}+1\right)\)
\(=\left(\frac{1+\sqrt{a}}{1-a}-\frac{1-\sqrt{a}}{1-a}\right)\left(\frac{\sqrt{a}}{a}+\frac{a}{a}\right)\)
\(=\frac{1+\sqrt{a}-1+\sqrt{a}}{1-a}.\frac{\sqrt{a}+a}{a}\)
\(=\frac{2\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\frac{\sqrt{a}.\left(1+\sqrt{a}\right)}{a}\)
\(=\frac{2}{1-\sqrt{a}}\)
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Rút gọn BT P=\(\left(\frac{1}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\frac{2}{\sqrt{1-a^2}}+1\right)\)
rút gọn biểu thúc sau A= \(\left(\frac{2\sqrt{a}}{a\sqrt{a}+a+\sqrt{a}+1}+\frac{1}{\sqrt{a}+1}\right)\div\left(1+\frac{\sqrt{a}}{a+\text{1}}\right)\)
\(ĐKXĐ:a\ge0\)
\(A=\left(\frac{2\sqrt{a}}{a\sqrt{a}+a+\sqrt{a}+1}+\frac{1}{\sqrt{a}+1}\right):\left(1+\frac{\sqrt{a}}{a+1}\right)\)
\(\Leftrightarrow A=\left(\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}+\frac{1}{\sqrt{a}+1}\right):\frac{a+\sqrt{a}+1}{a+1}\)
\(\Leftrightarrow A=\frac{2\sqrt{a}+a+1}{\left(a+1\right)\left(\sqrt{a}+1\right)}\cdot\frac{a+1}{a+\sqrt{a}+1}\)
\(\Leftrightarrow A=\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(a+\sqrt{a}+1\right)}\)
\(\Leftrightarrow A=\frac{\sqrt{a}+1}{a+\sqrt{a}+1}\)