Rút gọn
\(A=4\sqrt{32}+2\sqrt{50}-8\sqrt{2}-2\sqrt{98}\)
\(B=\frac{1}{\sqrt{6}+\sqrt{10}}-\frac{1}{\sqrt{6}-\sqrt{10}}\)
Những câu hỏi liên quan
Câu 1: Rút gọn:
a) \(2\sqrt{18}-4\sqrt{50}-3\sqrt{32}\)
b) \(\sqrt{14-6\sqrt{5}}+\sqrt{6+2\sqrt{5}}\)
c) \(\dfrac{\sqrt{10}+10}{1+\sqrt{10}}-\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}\)
Câu 2: Giải phương trình:
\(\sqrt{9x^2-30x+25}=5\)
1.
a, \(2\sqrt{18}-4\sqrt{50}-3\sqrt{32}=6\sqrt{2}-20\sqrt{2}-12\sqrt{2}=-2\sqrt{2}\)
b, \(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(\sqrt{5}+3\right)^2}\)
\(=\left|\sqrt{5}-3\right|+\left|\sqrt{5}+3\right|\)
\(=-\sqrt{5}+3+\sqrt{5}+3=6\)
c, \(\dfrac{\sqrt{10}+10}{1+\sqrt{10}}-\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}=\dfrac{\sqrt{10}\left(1+\sqrt{10}\right)}{1+\sqrt{10}}-\dfrac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}\)
\(=\sqrt{10}-\sqrt{10}=0\)
2.
ĐK: \(x\in R\)
\(\sqrt{9x^2-30x+25}=5\)
\(\Leftrightarrow\sqrt{\left(3x-5\right)^2}=5\)
\(\Leftrightarrow\left|3x-5\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=5\\3x-5=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=0\end{matrix}\right.\)
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Rút gọn các biểu thức sau:
a) \(0,2\sqrt{\left(-10\right)^2.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\) b) \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
c) \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right):\sqrt{6}\) d) \(\frac{\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
bài 1:a) Rút gọn biểu thức : sqrt{frac{2sqrt{10}+sqrt{30}-2sqrt{2}-sqrt{6}}{2sqrt{10}-2sqrt{2}}}:frac{2}{sqrt{3}-1}b) giải phương trình sau: sqrt{frac{1}{4}x^2+x+1}-sqrt{6-2sqrt{5}}0c) tính A left(sqrt{4+sqrt{7}}-sqrt{4-sqrt{7}}right)^3d) rút gọn biểu thức B sqrt{3-2sqrt{2}}+sqrt{3+2sqrt{2}}
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bài 1:
a) Rút gọn biểu thức : \(\sqrt{\frac{2\sqrt{10}+\sqrt{30}-2\sqrt{2}-\sqrt{6}}{2\sqrt{10}-2\sqrt{2}}}:\frac{2}{\sqrt{3}-1}\)
b) giải phương trình sau: \(\sqrt{\frac{1}{4}x^2+x+1}-\sqrt{6-2\sqrt{5}}=0\)
c) tính A= \(\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^3\)
d) rút gọn biểu thức B= \(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}\)
Rút gọn biểu thức 1) frac{sqrt{5+2sqrt{6}}+sqrt{8+2sqrt{15}}}{sqrt{7+2sqrt{10}}}2) left(2+frac{3+sqrt{3}}{sqrt{3}+1}right)left(2+frac{3-sqrt{3}}{sqrt{3}-1}right):left(sqrt{5}-2right)3) left(frac{15}{sqrt{6}+1}+frac{4}{sqrt{6}-2}-frac{12}{3-sqrt{6}}right).left(sqrt{6}+11right)4) frac{1}{1+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+frac{1}{sqrt{3}+sqrt{4}}+...+frac{1}{sqrt{99}+sqrt{100}}5) frac{1}{1-sqrt{2}}-frac{1}{sqrt{2}-sqrt{3}}+frac{1}{sqrt{3}-sqrt{4}}-...-frac{1}{sqrt{98}-sqrt{99}}+frac{1}{sqrt{99}-s...
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Rút gọn biểu thức
1) \(\frac{\sqrt{5+2\sqrt{6}}+\sqrt{8+2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
2) \(\left(2+\frac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\frac{3-\sqrt{3}}{\sqrt{3}-1}\right):\left(\sqrt{5}-2\right)\)
3) \(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
4) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
5) \(\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-...-\frac{1}{\sqrt{98}-\sqrt{99}}+\frac{1}{\sqrt{99}-\sqrt{100}}\)
6) \(\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
7)\(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{3}{2}}+2\right)\left(\frac{\sqrt{2}+\sqrt{3}}{4\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right)\left(24+8\sqrt{6}\right)\left(\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}}+\frac{\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)\)
Câu 1,2,3 Ez quá rồi :3
Câu 4:
Tổng quát:
\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v
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Câu 5 ko khác câu 4 lắm :v
Câu 5:
Tổng quát:
\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v
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Sao làm hổng ai bảo đú.n/g vậy :(((
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Rút gọn căn thức :
A = \(\frac{\sqrt{10}+2\sqrt{6}+\sqrt{10}.\sqrt{4+\sqrt{15}}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\)\(\frac{\sqrt{10}+2\sqrt{6}+\sqrt{10}.\sqrt{4+\sqrt{15}}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\)
rút gọn các biểu thức a/ Asqrt{7+2sqrt{6}}+sqrt{7-2sqrt{6}}-2sqrt{6}b/ B sqrt{7+2sqrt{10}}-sqrt{7+2sqrt{10}}-sqrt{8}c/ Csqrt{14+4sqrt{6}}-sqrt{14-4sqrt{6}}d/ Dsqrt{3+sqrt{5}}-sqrt{3-sqrt{5}}e/ Efrac{sqrt{2-sqrt{3}}}{sqrt{2}}f/ Ffrac{sqrt{x-sqrt{2x-1}}}{sqrt{2}}vớifrac{1}{2}le x 1
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rút gọn các biểu thức
a/ A=\(\sqrt{7+2\sqrt{6}}+\sqrt{7-2\sqrt{6}}-2\sqrt{6}\)
b/ B= \(\sqrt{7+2\sqrt{10}}-\sqrt{7+2\sqrt{10}}-\sqrt{8}\)
c/ C=\(\sqrt{14+4\sqrt{6}}-\sqrt{14-4\sqrt{6}}\)
d/ D=\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\)
e/ E=\(\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2}}\)
f/ F=\(\frac{\sqrt{x-\sqrt{2x-1}}}{\sqrt{2}}với\frac{1}{2}\le x< 1\)
Tinh
a, \(3\sqrt{50}-2\sqrt{98}-5\sqrt{18}-\sqrt{63}-2\sqrt{28}\)
b, \(\sqrt{42-10\sqrt{17}}+\sqrt{3-8\sqrt{17}}\)
c, \(\frac{4}{\sqrt{3}+1}+\frac{6}{\sqrt{3}-3}-\frac{5}{\sqrt{3}-2}\)
Rút gọn
a)\(\frac{\sqrt{12}+\sqrt{18}-\sqrt{6}}{2+\sqrt{6}-\sqrt{2}}\)
b)\(\frac{\left(\sqrt{10}-1\right)^2-3}{\sqrt{10}+\sqrt{3}-1}\)
c)\(\sqrt{\sqrt{3}+\sqrt{2}}+\sqrt{\sqrt{3}-\sqrt{2}}\)
Rút gọn : \(\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{9}}\)
với n >0, ta có :
\(\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=n+1-n=1\Rightarrow\frac{1}{\sqrt{n+1}-\sqrt{n}}=\sqrt{n+1}+\sqrt{n}\)
Gọi biểu thức đã cho là A
\(A=\frac{1}{-\left(\sqrt{2}-\sqrt{1}\right)}-\frac{1}{-\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{1}{-\left(\sqrt{8}-\sqrt{7}\right)}-\frac{1}{-\left(\sqrt{9}-\sqrt{8}\right)}\)
\(A=-\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}-...-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{9}-\sqrt{8}}\)
\(A=-\left(\sqrt{2}+\sqrt{1}\right)+\left(\sqrt{3}+\sqrt{2}\right)-...-\left(\sqrt{8}+\sqrt{7}\right)+\left(\sqrt{9}+\sqrt{8}\right)\)
\(A=-\sqrt{1}+\sqrt{9}=2\)
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\(\frac{1}{\sqrt{n}-\sqrt{n+1}}=\frac{\sqrt{n}+\sqrt{n+1}}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n}-\sqrt{n+1}\right)}=-\sqrt{n}-\sqrt{n+1}\)
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