A=\(\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\) biết a=\(\sqrt{4+2\sqrt{3}}\) , b=\(\sqrt{4-2\sqrt{3}}\)
Những câu hỏi liên quan
Pleft(frac{sqrt{a}-sqrt{b}}{sqrt{a}+sqrt{b}}-frac{sqrt{a}+sqrt{b}}{sqrt{a}-sqrt{b}}right)sqrt{frac{1}{a}-frac{1}{b}}left(frac{left(sqrt{a}-sqrt{b}right)^2}{a-b}-frac{left(sqrt{a}+sqrt{b}right)^2}{a-b}right).sqrt{frac{b-a}{ab}}frac{a-2sqrt{ab}+b-a-2sqrt{ab}-b}{a-b}.sqrt{frac{b-a}{ab}}frac{-4sqrt{ab}}{a-b}.sqrt{frac{b-a}{ab}}frac{-4sqrt{ab}}{2017-2018}.sqrt{frac{2018-2017}{ab}}4sqrt{ab}.sqrt{frac{1}{ab}}sqrt{frac{16ab}{ab}}4
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\(P=\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right)\sqrt{\frac{1}{a}-\frac{1}{b}}\)
\(=\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{a-b}-\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{a-b}\right).\sqrt{\frac{b-a}{ab}}\)
\(=\frac{a-2\sqrt{ab}+b-a-2\sqrt{ab}-b}{a-b}.\sqrt{\frac{b-a}{ab}}\)
\(=\frac{-4\sqrt{ab}}{a-b}.\sqrt{\frac{b-a}{ab}}\)\(=\frac{-4\sqrt{ab}}{2017-2018}.\sqrt{\frac{2018-2017}{ab}}\)
\(=4\sqrt{ab}.\sqrt{\frac{1}{ab}}\)\(=\sqrt{\frac{16ab}{ab}}\)\(=4\)
sao tổng lại lớn hơn hiệu
Đề bài: Rút gọn biểu thức:
1. frac{sqrt{a^2+x^2}+sqrt{a^2-x^2}}{sqrt{a^2+x^2}-sqrt{a^2-x^2}}-sqrt{^{ }frac{a^4}{x^4}-1}
2. left(frac{1-asqrt{a}}{1-sqrt{a}}+sqrt{a}right) . left(frac{1+asqrt{a}}{1+sqrt{a}}-sqrt{a}right)
3. left(frac{3}{sqrt{1+x}}sqrt{1-x}right) : left(frac{3}{sqrt{1-x^2}}+1right)
4. left(sqrt{a}+frac{b-sqrt{ab}}{sqrt{a}+sqrt{b}}right) : left(frac{a}{sqrt{ab+b}}+frac{b}{sqrt{ab}-a}-frac{a+b}{sqrt{ab}}right)
5. frac{sqrt{a}+sqrt{b}-1}{a+sqrt{ab}} + frac{sqrt{a}-sqrt{b}}{2...
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Đề bài: Rút gọn biểu thức:
1. \(\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\sqrt{^{ }\frac{a^4}{x^4}-1}\)
2. \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\) . \(\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
3. \(\left(\frac{3}{\sqrt{1+x}}\sqrt{1-x}\right)\) : \(\left(\frac{3}{\sqrt{1-x^2}}+1\right)\)
4. \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right)\) : \(\left(\frac{a}{\sqrt{ab+b}}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)\)
5. \(\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}\) + \(\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\) .\(\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
Các bạn giúp tớ nhé, hứa sẽ tick, tớ cảm ơn!!!!
1.
Đặt \(\sqrt{a^2+x^2}=m,\sqrt{a^2-x^2}=n\Rightarrow x^2=\frac{m^2-n^2}{2}\)
\(\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\sqrt{\frac{a^4}{x^4}-1}=\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\sqrt{\frac{(a^2+x^2)(a^2-x^2)}{x^4}}\)
\(=\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\frac{\sqrt{(a^2+x^2)(a^2-x^2)}}{x^2}\)
\(=\frac{m+n}{m-n}-\frac{mn}{\frac{m^2-n^2}{2}}=\frac{(m+n)^2}{m^2-n^2}-\frac{2mn}{m^2-n^2}=\frac{m^2+n^2}{m^2-n^2}\)
\(=\frac{2a^2}{2x^2}=\frac{a^2}{x^2}\)
2.
\(=\left[\frac{(1-\sqrt{a})(1+\sqrt{a}+a)}{1-\sqrt{a}}+\sqrt{a}\right].\left[\frac{(1+\sqrt{a})(1-\sqrt{a}+a)}{1+\sqrt{a}}-\sqrt{a}\right]\)
\(=(1+\sqrt{a}+a+\sqrt{a})(1-\sqrt{a}+a-\sqrt{a})\)
\(=(a+2\sqrt{a}+1)(a-2\sqrt{a}+1)=(\sqrt{a}+1)^2(\sqrt{a}-1)^2\)
\(=(a-1)^2\)
3.
\(=\frac{3(1-x)}{\sqrt{1+x}.\sqrt{1-x}}:\frac{3+\sqrt{1-x^2}}{\sqrt{1-x^2}}=\frac{3(1-x)}{\sqrt{1-x^2}}.\frac{\sqrt{1-x^2}}{3+\sqrt{1-x^2}}=\frac{3(1-x)}{3+\sqrt{1-x^2}}\)
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4. Bạn xem lại đề xem đã đúng chưa?
5.
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\frac{\sqrt{b}(a+\sqrt{ab})+\sqrt{b}(a-\sqrt{ab})}{(a-\sqrt{ab})(a+\sqrt{ab})}\)
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\frac{2a\sqrt{b}}{a^2-ab}\)
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}}.\frac{1}{a-b}\)
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}\)
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{1}{a+\sqrt{ab}}=\frac{\sqrt{a}+\sqrt{b}}{a+\sqrt{ab}}=\frac{1}{\sqrt{a}}\)
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Câu 1 : Rút gọn biểu thức
a, \(\frac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\frac{2}{3}\sqrt{12}.\)b, \(\frac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\frac{3}{3+3\sqrt{6}}.\)
c\(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}.\)Với a>0;b>0
\(\frac{3}{a}.\sqrt{\frac{a^3}{b}}-\frac{1}{2}\sqrt{\frac{4}{ab}}-2\sqrt{\frac{b}{a}}:\sqrt{\frac{1}{ab}}\)
Thu gọn biểu thức
a, A = \(\frac{2\sqrt{3-\sqrt{3+\sqrt{3+\sqrt{48}}}}}{\sqrt{6}-2}\)
b, B = \(\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
RÚT GỌN CÁC BIỂU THỨC SAU
\(A=\frac{-2}{3}\sqrt{\frac{\left(a-b\right)^3.b^5}{c}}.\frac{9}{4}\sqrt{\frac{c^3}{2\left(a-b\right)}}.\sqrt{98b}\)
\(B=\left(\sqrt{ab}+2\sqrt{\frac{b}{a}}-\sqrt{\frac{a}{b}+\sqrt{\frac{1}{ab}}}\right).\sqrt{ab}\)
1.So sánh
a) sqrt{2002}+sqrt{2004} và 2sqrt{2003}
b)sqrt{4+sqrt{7}}-sqrt{4-sqrt{7}} và sqrt{2}
2. Rút gọn
a) frac{a^2-sqrt{a}}{a+sqrt{a}+1}-frac{a^2+sqrt{a}}{a-sqrt{a}+1} với 0 ≤ a ≥ 1
b) frac{asqrt{b}-bsqrt{a}}{sqrt{a}-sqrt{b}}-sqrt{ab}
c) frac{sqrt{a}+sqrt{b}-1}{a+sqrt{ab}}+frac{sqrt{a}-sqrt{b}}{2sqrt{ab}}left(frac{sqrt{b}}{a-sqrt{ab}}+frac{sqrt{b}}{a+sqrt{ab}}right)
d) frac{a+b+2sqrt{ab}}{sqrt{a}+sqrt{b}}-frac{a-b}{sqrt{a}-sqrt{b}}
e)frac{sqrt{a}-1}{asqrt{a}-a+sqrt{a}}:frac{1}{a^2+sqr...
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1.So sánh
a) \(\sqrt{2002}+\sqrt{2004}\) và \(2\sqrt{2003}\)
b)\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\) và \(\sqrt{2}\)
2. Rút gọn
a) \(\frac{a^2-\sqrt{a}}{a+\sqrt{a}+1}-\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}\) với 0 ≤ a ≥ 1
b) \(\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\)
c) \(\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
d) \(\frac{a+b+2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a-b}{\sqrt{a}-\sqrt{b}}\)
e)\(\frac{\sqrt{a}-1}{a\sqrt{a}-a+\sqrt{a}}:\frac{1}{a^2+\sqrt{a}}\)
3. Giải phương trình
a)\(\frac{\sqrt{27x}}{\sqrt{3}}=6\)
b)\(\sqrt{x+1}=3-\sqrt{x}\)
c) \(\sqrt{2x+1}=2+\sqrt{x-3}\)
d) \(\sqrt{x-5}-\frac{x-14}{3+\sqrt{x-5}}=3\)
Bài 1:
b) Ta có: \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(=\frac{\sqrt{2\left(4+\sqrt{7}\right)}}{\sqrt{2}}-\frac{\sqrt{2\left(4-\sqrt{7}\right)}}{\sqrt{2}}\)
\(=\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}-\frac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\frac{\sqrt{7+2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}-\frac{\sqrt{7-2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{7}+1\right|}{\sqrt{2}}-\frac{\left|\sqrt{7}-1\right|}{\sqrt{2}}\)
\(=\frac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Bài 2:
a) Ta có: \(\frac{a^2-\sqrt{a}}{a+\sqrt{a}+1}-\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}\)
\(=\frac{\sqrt{a}\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{a+\sqrt{a}+1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}\)
\(=\sqrt{a}\left(\sqrt{a}-1\right)-\sqrt{a}\left(\sqrt{a}+1\right)\)
\(=a-\sqrt{a}-a-\sqrt{a}\)
\(=-2\sqrt{a}\)
b) Ta có: \(\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\)
\(=\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\)
\(=\sqrt{ab}-\sqrt{ab}=0\)
d) Ta có: \(\frac{a+b+2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a-b}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\sqrt{a}+\sqrt{b}-\left(\sqrt{a}+\sqrt{b}\right)\)
=0
Bài 3:
a) ĐKXĐ: x≥0
Ta có: \(\frac{\sqrt{27x}}{\sqrt{3}}=6\)
\(\Leftrightarrow\frac{\sqrt{27}\cdot\sqrt{x}}{\sqrt{3}}=6\)
\(\Leftrightarrow3\cdot\sqrt{x}=6\)
\(\Leftrightarrow\sqrt{x}=\frac{6}{3}=2\)
hay \(x=4\)(thỏa mãn)
Vậy: S={4}
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x+1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-1\end{matrix}\right.\Leftrightarrow x\ge0\)
Ta có: \(\sqrt{x+1}=3-\sqrt{x}\)
\(\Leftrightarrow\left(\sqrt{x+1}\right)^2=\left(3-\sqrt{x}\right)^2\)
\(\Leftrightarrow x+1=9-6\sqrt{x}+x\)
\(\Leftrightarrow x+1-9+6\sqrt{x}-x=0\)
\(\Leftrightarrow-8+6\sqrt{x}=0\)
\(\Leftrightarrow6\sqrt{x}=8\)
\(\Leftrightarrow\sqrt{x}=\frac{8}{6}=\frac{4}{3}\)
hay \(x=\frac{16}{9}\)(thỏa mãn)
Vậy: \(S=\left\{\frac{16}{9}\right\}\)
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cho B=\(\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}-b}+\frac{a-b}{\sqrt{ab}}\)
a) Rút gọn B
b) Tính giá trị của B khi a=\(\sqrt{4+2\sqrt{3}}\), b=\(\sqrt{4-2\sqrt{3}}\)
Cho B=(\(\frac{\sqrt{a}+1}{\sqrt{a}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\)1):\(\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}-\frac{\sqrt{ab}-\sqrt{a}}{\sqrt{ab}-1}+1\right)\)
a.rút gọn B
b.tính B biết a=\(2-\sqrt{3}\),b=\(\frac{\sqrt{3}-1}{1-\sqrt{3}}\)
Tui cx đang có câu như thế mà k bt làm đây
Hu hu
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