\(\sqrt{\frac{-2\sqrt{6}+\sqrt{23}}{-x+5}}\) ; \(\sqrt{x^2-3}\)
ĐKXĐ của x để căn thức có nghĩa.
Những câu hỏi liên quan
Tính
\(E=\frac{-23}{\sqrt{6}+\sqrt{3}+\sqrt{2}}+\frac{6}{\sqrt{3}-\sqrt{2}}-\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{5\sqrt{6}}{\sqrt{6}-1}\)
Help me
Tìm điều kiện củasqrt{A}
1) sqrt{sqrt{5}-sqrt{3}x}
2) sqrt{frac{-2sqrt{6}+sqrt{23}}{-x+5}}
3) sqrt{2011-m}
4)sqrt{frac{2sqrt{15}-sqrt{59}}{x-7}}
5)sqrt{49x^2-24x+4}
6)sqrt{frac{12x+5}{sqrt{3}}}
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Help me
Tìm điều kiện của\(\sqrt{A}\)
1) \(\sqrt{\sqrt{5}-\sqrt{3}x}\)
2) \(\sqrt{\frac{-2\sqrt{6}+\sqrt{23}}{-x+5}}\)
3) \(\sqrt{2011-m}\)
4)\(\sqrt{\frac{2\sqrt{15}-\sqrt{59}}{x-7}}\)
5)\(\sqrt{49x^2-24x+4}\)
6)\(\sqrt{\frac{12x+5}{\sqrt{3}}}\)
Help meTìm điều kiện củasqrt{A}1) sqrt{sqrt{5}-sqrt{3}x}2) sqrt{frac{-2sqrt{6}+sqrt{23}}{-x+5}}3) sqrt{2011-m}4)sqrt{frac{2sqrt{15}-sqrt{59}}{x-7}}5)sqrt{49x^2-24x+4}6)sqrt{frac{12x+5}{sqrt{3}}}
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Help me
Tìm điều kiện của\(\sqrt{A}\)
1) \(\sqrt{\sqrt{5}-\sqrt{3}x}\)
2) \(\sqrt{\frac{-2\sqrt{6}+\sqrt{23}}{-x+5}}\)
3) \(\sqrt{2011-m}\)
4)\(\sqrt{\frac{2\sqrt{15}-\sqrt{59}}{x-7}}\)
5)\(\sqrt{49x^2-24x+4}\)
6)\(\sqrt{\frac{12x+5}{\sqrt{3}}}\)
Bài 1. Tính:
a) sqrt{frac{5+sqrt{21}}{5-sqrt{21}}}+sqrt{frac{5-sqrt{21}}{5+sqrt{21}}} b) left(2-sqrt{2}right)sqrt{2+4sqrt{3+sqrt{2}+sqrt{11-6sqrt{2}}}}
c) frac{left(5sqrt{3}+sqrt{50}right)left(5-sqrt{24}right)}{sqrt{75}-5sqrt{2}} d) frac{1}{sqrt{25}+sqrt{24}}+frac{1}{sqrt{24}+sqrt{23}}+...+frac{1}{sqrt{2}+sqrt{1}}
Bài 2. Tìm x
a) sqrt{x-2sqrt{x-1}}2 b) sqrt{x^2-x}+sqrt{x^2+x-2}0
giúp tớ với, chiều nay tớ p ik hok rồi, thầy ktra...
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Bài 1. Tính:
a) \(\sqrt{\frac{5+\sqrt{21}}{5-\sqrt{21}}}+\sqrt{\frac{5-\sqrt{21}}{5+\sqrt{21}}}\) b) \(\left(2-\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{11-6\sqrt{2}}}}\)
c) \(\frac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\) d) \(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}\)
Bài 2. Tìm x
a) \(\sqrt{x-2\sqrt{x-1}}=2\) b) \(\sqrt{x^2-x}+\sqrt{x^2+x-2}=0\)
giúp tớ với, chiều nay tớ p ik hok rồi, thầy ktra bài tập nx
Bài 1:
a/ \(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}\)
\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{4}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{4}}=\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}\)
\(=\frac{10}{2}=5\)
b/ \(=\left(2-\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{\left(3-\sqrt{2}\right)^2}}}\)
\(=\left(2-\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+3-\sqrt{2}}}\)
\(=\left(2-\sqrt{3}\right)\sqrt{2+4\sqrt{6}}\)
Bạn coi lại đề, tới đây ko rút gọn được nữa nên chắc bạn ghi đề nhầm ở chỗ nào đó
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c/ \(=\frac{5\left(\sqrt{3}+\sqrt{2}\right)\left(5-\sqrt{24}\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-\sqrt{24}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\left(5+2\sqrt{6}\right)\left(5-\sqrt{24}\right)=\left(5+\sqrt{24}\right)\left(5-\sqrt{24}\right)=1\)
d/ Nhân cả tử và mẫu của từng phân số với liên hợp của mẫu, mẫu số sẽ thành 1 hết:
\(=\frac{\sqrt{25}-\sqrt{24}}{\left(\sqrt{25}+\sqrt{24}\right)\left(\sqrt{25}-\sqrt{24}\right)}+\frac{\sqrt{24}-\sqrt{23}}{\left(\sqrt{24}+\sqrt{23}\right)\left(\sqrt{24}-\sqrt{23}\right)}+...+\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)
\(=\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-1\)
\(=\sqrt{25}-1=5-1=4\)
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Câu 2:
a/ ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{x-1-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x-1}-1\right|=2\)
TH1: \(\sqrt{x-1}-1\ge0\Rightarrow x\ge2\) pt trở thành:
\(\sqrt{x-1}-1=2\Leftrightarrow\sqrt{x-1}=3\)
\(\Rightarrow x-1=9\Rightarrow x=10\) (nhận)
TH2: \(\sqrt{x-1}-1< 0\Rightarrow1\le x< 2\) pt trở thành:
\(1-\sqrt{x-1}=2\Rightarrow\sqrt{x-1}=-1< 0\) (vô nghiệm)
b/
\(\sqrt{x^2-x}+\sqrt{x^2+x-2}=0\)
Do \(\left\{{}\begin{matrix}\sqrt{x^2-x}\ge0\\\sqrt{x^2+x-2}\ge0\end{matrix}\right.\) nên đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\sqrt{x^2-x}=0\\\sqrt{x^2+x-2}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2-x=0\\x^2+x-2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow x=1\)
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1. sqrt{x-2sqrt{x-1}}+sqrt{x+3-4sqrt{x-1}}left(2 x 5right)2. frac{6}{1-sqrt{3}}-frac{3sqrt{3}-1}{sqrt{3}+1}+sqrt{3}3. sqrt{29-12sqrt{5}+sqrt{24-8sqrt{3}}}4. sqrt{frac{4}{9-4sqrt{5}}}-sqrt{frac{4}{9+4sqrt{5}}}5. 5sqrt{frac{1}{5}}+frac{1}{2}sqrt{x}-frac{5}{4}sqrt{frac{4}{5}+sqrt{5}}6. frac{6-sqrt{6}}{sqrt{6}-1}-9sqrt{frac{2}{3}}-frac{4}{2-sqrt{6}}7. left(frac{sqrt{x}-2}{x-1}-frac{sqrt{x}+2}{x+2sqrt{x}+1}right)frac{left(sqrt{x}-1right)^2}{2}left(xge0,xne1right)
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1. \(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+3-4\sqrt{x-1}}\left(2< x< 5\right)\)
2. \(\frac{6}{1-\sqrt{3}}-\frac{3\sqrt{3}-1}{\sqrt{3}+1}+\sqrt{3}\)
3. \(\sqrt{29-12\sqrt{5}+\sqrt{24-8\sqrt{3}}}\)
4. \(\sqrt{\frac{4}{9-4\sqrt{5}}}-\sqrt{\frac{4}{9+4\sqrt{5}}}\)
5. \(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{x}-\frac{5}{4}\sqrt{\frac{4}{5}+\sqrt{5}}\)
6. \(\frac{6-\sqrt{6}}{\sqrt{6}-1}-9\sqrt{\frac{2}{3}}-\frac{4}{2-\sqrt{6}}\)
7. \(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\frac{\left(\sqrt{x}-1\right)^2}{2}\left(x\ge0,x\ne1\right)\)
Trả lời nhanh giúp mình với mình cần gấp lắm
Thực hiện phép tính:
a) \(\left(\frac{1}{7-4\sqrt{3}}+\frac{3}{7+4\sqrt{3}}\right)\left(7+2\sqrt{3}\right)\)
b)\(\left(\frac{3\sqrt{5}-\sqrt{15}}{\sqrt{27}-3}+\frac{2\sqrt{5}}{\sqrt{3}}\right).4\sqrt{15}\)
c)\(\sqrt{5-2\sqrt{6-25-\sqrt{96}}}\)
d)\(\sqrt{23-2\sqrt{112}}+\sqrt{23+2\sqrt{112}}\)
a/ \(=\left(7+4\sqrt{3}+3\left(7-4\sqrt{3}\right)\right)\left(7+2\sqrt{3}\right)\)
\(=\left(28-8\sqrt{3}\right)\left(7+2\sqrt{3}\right)\)
\(=4\left(7-2\sqrt{3}\right)\left(7+2\sqrt{3}\right)\)
\(=4\left(49-12\right)=...\)
b/ \(=\left(\frac{\sqrt{15}\left(\sqrt{3}-1\right)}{3\left(\sqrt{3}-1\right)}+\frac{2\sqrt{15}}{3}\right).4\sqrt{15}\)
\(=\left(\frac{\sqrt{15}}{3}+\frac{2\sqrt{15}}{3}\right).4\sqrt{15}\)
\(=\sqrt{15}.4\sqrt{15}=4.15=...\)
c/ Bạn coi lại đề
d/ \(\sqrt{23-2\sqrt{112}}+\sqrt{23+2\sqrt{112}}\)
\(=\sqrt{\left(4-\sqrt{7}\right)^2}+\sqrt{\left(4+\sqrt{7}\right)^2}\)
\(=4-\sqrt{7}+4+\sqrt{7}=8\)
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left(5+4sqrt{2}right)left(3+2sqrt{1+sqrt{2}}right)left(3-2sqrt{1+sqrt{2}}right) sqrt{frac{9}{4}-sqrt{2}} Sosanh2sqrt{27}vasqrt{147}
2sqrt{15}vasqrt{59}
2sqrt{2}-1va2
frac{sqrt{3}}{2}va1
-frac{sqrt{10}}{2}va-2sqrt{5}
sqrt{6}-1va3
2sqrt{5}-5sqrt{2}va1
frac{sqrt{8}}{3}vafrac{3}{4}
-2sqrt{6}va-sqrt{23}
2sqrt{6}-2va3
sqrt{111}-7va4
Xếp theo thứ tự tăng dần: 21,2sqrt{7},15sqrt{3},-sqrt{123} ; 28sqrt{2},sqrt{14},2sqrt{147},36sqrt{4}
giảm dần: 6sqrt{frac{1}{4}},4sqrt{frac{1}{2}...
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\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\\ \\ \\ \sqrt{\frac{9}{4}-\sqrt{2}}\\ \\ \\ Sosanh2\sqrt{27}va\sqrt{147}\\ \\ \\ 2\sqrt{15}va\sqrt{59}\\ \\ \\ 2\sqrt{2}-1va2\\ \\ \\ \frac{\sqrt{3}}{2}va1\\ \\ \\ -\frac{\sqrt{10}}{2}va-2\sqrt{5}\\ \\ \\ \sqrt{6}-1va3\\ \\ \\ 2\sqrt{5}-5\sqrt{2}va1\\ \\ \\ \frac{\sqrt{8}}{3}va\frac{3}{4}\\ \\ \\ -2\sqrt{6}va-\sqrt{23}\\ \\ \\ 2\sqrt{6}-2va3\\ \\ \\ \sqrt{111}-7va4\)
Xếp theo thứ tự tăng dần: \(21,2\sqrt{7},15\sqrt{3},-\sqrt{123}\) ; \(28\sqrt{2},\sqrt{14},2\sqrt{147},36\sqrt{4}\)
giảm dần: \(6\sqrt{\frac{1}{4}},4\sqrt{\frac{1}{2}},-\sqrt{132},2\sqrt{3},\sqrt{\frac{15}{5}}\); \(-27,4\sqrt{3},16\sqrt{5},21\sqrt{2}\)
a,\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4\left(1+\sqrt{2}\right)\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(5-4\sqrt{2}\right)=25-\left(4\sqrt{2}\right)^2\)
=-7
b, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{9-4\sqrt{2}}}{2}=\frac{\sqrt{9-2\sqrt{8}}}{2}=\frac{\sqrt{\left(\sqrt{8}-1\right)^2}}{2}=\frac{\left|\sqrt{8}-1\right|}{2}=\frac{\sqrt{8}-1}{2}\)
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So sánh:
1) \(2\sqrt{27}\) và \(\sqrt{147}\)
+ \(2\sqrt{27}\) = \(6\sqrt{3}\)
+ \(\sqrt{147}\) = \(7\sqrt{3}\)
⇒ \(6\sqrt{3}\) < \(7\sqrt{3}\)
Vậy: \(2\sqrt{27}\)< \(\sqrt{147}\)
2) \(2\sqrt{15}\) và \(\sqrt{59}\)
+ \(2\sqrt{15}\) = \(\sqrt{60}\)
⇒ \(\sqrt{60}\) > \(\sqrt{59}\)
Vậy: \(2\sqrt{15}\) > \(\sqrt{59}\)
3) \(2\sqrt{2}-1\) và 2
\(giống\left(-1\right)\left\{{}\begin{matrix}3-1\\2\sqrt{2}-1\end{matrix}\right.\)
So sánh: 3 và \(2\sqrt{2}\)
+ 3 = \(\sqrt{9}\)
+ \(2\sqrt{2}=\sqrt{8}\)
⇒ \(\sqrt{8}\) < \(\sqrt{9}\)
⇒ \(\sqrt{8}\) -1 < \(\sqrt{9}\) -1
⇒ \(2\sqrt{2}\) - 1 < 3 - 1
Vậy: \(2\sqrt{2}-1< 2\)
4) \(\frac{\sqrt{3}}{2}\) và 1
+ 1 = \(\frac{2}{2}\)
⇒ \(\frac{\sqrt{3}}{2}\) < \(\frac{2}{2}\)
Vậy: \(\frac{\sqrt{3}}{2}\) < 1
5) \(\frac{-\sqrt{10}}{2}\) và \(-2\sqrt{5}\)
+ \(-2\sqrt{5}\) = \(\frac{-4\sqrt{5}}{2}\) = \(\frac{-\sqrt{80}}{2}\)
⇒ \(\frac{-\sqrt{10}}{2}\) > \(\frac{-\sqrt{80}}{2}\)
Vậy: \(\frac{-\sqrt{10}}{2}\) > \(-2\sqrt{5}\)
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GIÚP EM ĐI ẠTÍNH:frac{3-sqrt{6+sqrt{3+sqrt{6+sqrt{3}}}}}{3-sqrt{3+sqrt{6+sqrt{3}}}}+frac{2+sqrt{6+sqrt{3+sqrt{6+sqrt{3}}}}}{3+sqrt{6+sqrt{3+sqrt{6+sqrt{3}}}}}frac{1+frac{sqrt{3}}{2}}{1+sqrt{1+frac{sqrt{3}}{2}}}+frac{1-frac{sqrt{3}}{2}}{1-sqrt{1-frac{sqrt{3}}{2}}}frac{1}{sqrt{frac{5}{13}}+sqrt{frac{5}{7}}+1}+frac{1}{sqrt{frac{7}{5}}+sqrt{frac{7}{13}}+1}+frac{1}{sqrt{1frac{6}{7}}+1+sqrt{2frac{3}{5}}}RÚT GỌNsqrt{left(x-1right)^2}-x với x lớn hơn 1GIẢI PHƯƠNG TRÌNHsqrt{x^2-3x+2}+sqrt{x^2-4x+3}0
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GIÚP EM ĐI Ạ
TÍNH:
\(\frac{3-\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}{3-\sqrt{3+\sqrt{6+\sqrt{3}}}}+\frac{2+\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}{3+\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}\)
\(\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}\)
\(\frac{1}{\sqrt{\frac{5}{13}}+\sqrt{\frac{5}{7}}+1}+\frac{1}{\sqrt{\frac{7}{5}}+\sqrt{\frac{7}{13}}+1}+\frac{1}{\sqrt{1\frac{6}{7}}+1+\sqrt{2\frac{3}{5}}}\)
RÚT GỌN
\(\sqrt{\left(x-1\right)^2}-x\) với x lớn hơn 1
GIẢI PHƯƠNG TRÌNH
\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)
Bài rút gọn
\(\sqrt{\left(x-1\right)^2}-x=\left|x-1\right|-x\)
\(=\left(x-1\right)-x=x-1-x=-1\left(x>1\right)\)
Bài gpt:
\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)
Đk:\(-1\le x\le3\)
\(pt\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}+\sqrt{x-3}\right)=0\)
Dễ thấy:\(\sqrt{x-2}+\sqrt{x-3}=0\) vô nghiệm
Nên \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)
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Tínha.left(sqrt{10}-sqrt{2}right)left(3+sqrt{5}right)sqrt{27-9sqrt{5}}b.sqrt{frac{3sqrt{3}-4}{2sqrt{3}+1}}-sqrt{frac{4+sqrt{3}}{5-2sqrt{3}}}c.frac{3-4sqrt{3}}{sqrt{6}-sqrt{2}-sqrt{5}}d.left(sqrt{11}-sqrt{3}right)left(sqrt{13-sqrt{6}+2sqrt{30-sqrt{45}}}+sqrt{11}-sqrt{10-sqrt{6}}right)e.frac{sqrt{4+sqrt{5}}+sqrt{4-sqrt{5}}}{sqrt{4}+sqrt{11}}-frac{sqrt{20-4sqrt{23}}}{sqrt{5+sqrt{2}}-sqrt{5-sqrt{2}}}
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\(Tính\)
\(a.\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\sqrt{27-9\sqrt{5}}\)
\(b.\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{4+\sqrt{3}}{5-2\sqrt{3}}}\)
\(c.\frac{3-4\sqrt{3}}{\sqrt{6}-\sqrt{2}-\sqrt{5}}\)
\(d.\left(\sqrt{11}-\sqrt{3}\right)\left(\sqrt{13-\sqrt{6}+2\sqrt{30-\sqrt{45}}}+\sqrt{11}-\sqrt{10-\sqrt{6}}\right)\)
\(e.\frac{\sqrt{4+\sqrt{5}}+\sqrt{4-\sqrt{5}}}{\sqrt{4}+\sqrt{11}}-\frac{\sqrt{20-4\sqrt{23}}}{\sqrt{5+\sqrt{2}}-\sqrt{5-\sqrt{2}}}\)