\(\frac{3\sqrt{2}}{2}x\frac{2\sqrt{ }2}{2}\)
Những câu hỏi liên quan
\(x=\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+2}}}}}}}}}}}}}{\frac{2}{\sqrt{3+\sqrt{\frac{2}{\sqrt{3+\sqrt{\frac{2}{\sqrt{3+\sqrt{\frac{2}{\sqrt{3+\sqrt{\frac{2}{\sqrt{3+1}}}}}}}}}}}}}}}\)
Tính \(A=\left(\sqrt{x}^{1000}+x^{500}\right)^{2000}\)
Dễ thấy x có tử = 2; mẫu = 1. Vậy x = 2.
\(A=\left(2^{500}+2^{500}\right)^{2000}=2^{501.2000}\)
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eo ơi Mr Lazy nhìn sao ra tớ nhìn ko hỉu nỗi
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chứng minh rằng
a, \(\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}=1\)
b, \(\frac{1}{x+\sqrt{x}}+\frac{2\sqrt{x}}{x-1}-\frac{1}{x-\sqrt{x}}=\frac{2}{\sqrt[]{x}}\)
a, \(\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-\sqrt{3}+1}\)
\(=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{6+\sqrt{3}-3+6-\sqrt{3}-3}{9-3}=\frac{6}{6}=1\)
b, \(\frac{1}{x+\sqrt{x}}+\frac{2\sqrt{x}}{x-1}-\frac{1}{x-\sqrt{x}}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x-1}\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-1+2x-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{2\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=\frac{2}{\sqrt{x}}\)
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\(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3x+9}{x-9}\)
\(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}-\frac{3\sqrt{x}}{x+\sqrt{x}-2}\)
\(\frac{2}{\sqrt{x}-1}+\frac{2}{\sqrt{x}+1}-\frac{5-\sqrt{x}}{x-1}\)
\(\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
trục căn thức ở mẫu :
a,frac{3}{sqrt{5}};frac{2sqrt{3}}{sqrt{2}};frac{a}{sqrt{b}};frac{x+1}{sqrt{x^2-1}}
b,frac{1}{sqrt{3}+sqrt{2}};frac{2}{2-sqrt{3}};frac{sqrt{2}+1}{sqrt{2}-1};frac{3sqrt{2}}{sqrt{3}+1}
c,frac{1}{1+sqrt{2}+sqrt{3}}
d,frac{1}{sqrt{2sqrt{3}-sqrt{2}}.sqrt{2}.sqrt{sqrt{2}+sqrt{3}}}
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trục căn thức ở mẫu :
a,\(\frac{3}{\sqrt{5}};\frac{2\sqrt{3}}{\sqrt{2}};\frac{a}{\sqrt{b}};\frac{x+1}{\sqrt{x^2-1}}\)
b,\(\frac{1}{\sqrt{3}+\sqrt{2}};\frac{2}{2-\sqrt{3}};\frac{\sqrt{2}+1}{\sqrt{2}-1};\frac{3\sqrt{2}}{\sqrt{3}+1}\)
c,\(\frac{1}{1+\sqrt{2}+\sqrt{3}}\)
d,\(\frac{1}{\sqrt{2\sqrt{3}-\sqrt{2}}.\sqrt{2}.\sqrt{\sqrt{2}+\sqrt{3}}}\)
a) \(\frac{3}{\sqrt{5}}=\frac{3\sqrt{5}}{\sqrt{5}.\sqrt{5}}=\frac{3\sqrt{5}}{5}\)
\(\frac{2\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{2\sqrt{6}}{2}=\sqrt{6}\)
\(\frac{a}{\sqrt{b}}=\frac{a\sqrt{b}}{\sqrt{b}.\sqrt{b}}=\frac{a\sqrt{b}}{b}\)
\(\frac{x+1}{\sqrt{x^2-1}}=\frac{\left(x+1\right)\left(\sqrt{x^2-1}\right)}{\left(\sqrt{x^2-1}\right)\left(\sqrt{x^2-1}\right)}\) = \(\frac{\left(\sqrt{x^2-1}\right)\left(x+1\right)}{x^2-1}\)
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câu c chắc là như này
\(\frac{1}{1+\sqrt{2}+\sqrt{3}}=1+\frac{1}{\sqrt{2}+\sqrt{3}}\) = \(1+\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}\)
= \(1+\frac{\sqrt{2}-\sqrt{3}}{2-3}=1+\frac{\sqrt{2}-\sqrt{3}}{-1}\) = \(1-\sqrt{2}+\sqrt{3}\)
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rút gọn biểu thức
a) A 2sqrt{frac{1}{2}}+sqrt{18}
b) B frac{5+3sqrt{5}}{sqrt{5}}+frac{3+sqrt{3}}{sqrt{3}+1}-left(sqrt{5+3}right)
c) C frac{1}{x+sqrt{x}}+frac{2sqrt{x}}{x-1}-frac{1}{x-sqrt{x}}left(x0,xne1right)
d) D left(frac{sqrt{x}+2}{x+2sqrt{x}+1}-frac{sqrt{x-2}}{x-1}right)left(x+sqrt{x}right)left(x0,xne1right)
e) E frac{2+sqrt{3}}{sqrt{2}+sqrt{2+sqrt{3}}}+frac{2-sqrt{3}}{sqrt{2}-sqrt{2-sqrt{3}}}
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rút gọn biểu thức
a) A= \(2\sqrt{\frac{1}{2}}+\sqrt{18}\)
b) B= \(\frac{5+3\sqrt{5}}{\sqrt{5}}+\frac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5+3}\right)\)
c) C= \(\frac{1}{x+\sqrt{x}}+\frac{2\sqrt{x}}{x-1}-\frac{1}{x-\sqrt{x}}\left(x>0,x\ne1\right)\)
d) D = \(\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x-2}}{x-1}\right)\left(x+\sqrt{x}\right)\left(x>0,x\ne1\right)\)
e) E = \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
Dleft ( frac{1}{3sqrt{x}-6} +frac{1}{x-2sqrt{x}}right )left ( frac{1}{6} +frac{1}{2sqrt{x}}right ) Dleft ( frac{1}{3left ( sqrt{x}-2 right )} +frac{1}{sqrt{x}left ( sqrt{x}-2 right )}right ).frac{sqrt{x}+3}{6sqrt{x}} Dfrac{sqrt{x}+3}{3sqrt{x}left ( sqrt{x}-2 right )}.frac{sqrt{x}+3}{6sqrt{x}} Dfrac{left ( sqrt{x}+3 right )^{2}}{18xleft ( sqrt{x}-2 right )} Dfrac{x+6sqrt{x}+9}{18xsqrt{x}-36x}
A/ Đúng
B/ Sai
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\[D=\left ( \frac{1}{3\sqrt{x}-6} +\frac{1}{x-2\sqrt{x}}\right )\left ( \frac{1}{6} +\frac{1}{2\sqrt{x}}\right )\\ D=\left ( \frac{1}{3\left ( \sqrt{x}-2 \right )} +\frac{1}{\sqrt{x}\left ( \sqrt{x}-2 \right )}\right ).\frac{\sqrt{x}+3}{6\sqrt{x}}\\ D=\frac{\sqrt{x}+3}{3\sqrt{x}\left ( \sqrt{x}-2 \right )}.\frac{\sqrt{x}+3}{6\sqrt{x}}\\ D=\frac{\left ( \sqrt{x}+3 \right )^{2}}{18x\left ( \sqrt{x}-2 \right )}\\ D=\frac{x+6\sqrt{x}+9}{18x\sqrt{x}-36x}\]
A/ Đúng
B/ Sai
\(A=\left(\frac{\sqrt{x}-2}{2\sqrt{x}-2}+\frac{3}{2\sqrt{x}+2}-\frac{\sqrt{x}+3}{2\sqrt{x}+2}\right):\left(1-\frac{\sqrt{x}-3}{x-1}\right)\)
\(A=\left(\frac{\sqrt{x}-2}{2\left(\sqrt{x}-1\right)}+\frac{3}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+3}{2\left(\sqrt{x}+1\right)}\right):\left(\frac{x-1}{x-1}-\frac{\sqrt{x}-3}{x-1}\right)\)
\(A=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2-\sqrt{x}}{x-1}\right)\)
\(A=\left(\frac{x-\sqrt{x}-2+3\sqrt{x}-1-x-2\sqrt{x}-3}{2x-2}\right):\frac{x+2-\sqrt{x}}{x-1}\)
\(A=\frac{-3}{x-1}.\frac{x-1}{x+2-\sqrt{x}}\)
\(A=\frac{-3}{x+2-\sqrt{x}}\)
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Pleft(frac{sqrt{x}+2}{sqrt{x}+1}-frac{x-sqrt{x}-3}{x-sqrt{x}-2}right):left(frac{x-sqrt{x}}{x-sqrt{x}-2}+frac{2}{sqrt{x}-2}right)frac{left(sqrt{x}+2right)left(sqrt{x}-2right)-x+sqrt{x}+3}{left(sqrt{x}+1right)left(sqrt{x}-2right)}:frac{x-sqrt{x}+2left(sqrt{x}+1right)}{left(sqrt{x}+1right)left(sqrt{x}-2right)}frac{x-4-x+3+sqrt{x}}{left(sqrt{x}+1right)left(sqrt{x}-2right)}.frac{left(sqrt{x}+1right)left(sqrt{x}-2right)}{x-sqrt{x}+2sqrt{x}+2}frac{sqrt{x}-1}{x+sqrt{x}+2}
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\(P=\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{x-\sqrt{x}-3}{x-\sqrt{x}-2}\right):\left(\frac{x-\sqrt{x}}{x-\sqrt{x}-2}+\frac{2}{\sqrt{x}-2}\right)\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-x+\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{x-\sqrt{x}+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4-x+3+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{x-\sqrt{x}+2\sqrt{x}+2}\)
\(=\frac{\sqrt{x}-1}{x+\sqrt{x}+2}\)
#)Hỏi j đi bn, bn ph hỏi cái j chứ làm lun rùi còn để cộng đồng ngắm ak ???
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Bó cả tay lẫn chân !!! Bất lực như gặp cực hình !
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Chắc là bạn ấy hỏi bạn ấy làm có đúng ko ha gì đó ?
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Xem thêm câu trả lời
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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Cho A= \(\left(\frac{\sqrt{x}-2}{2\sqrt{x}-2}+\frac{3}{2\sqrt{x}+2}-\frac{\sqrt{x}-3}{2\sqrt{x}+2}\right):\left(1-\frac{\sqrt{x}-3}{x-1}\right)\)