So sánh \(A=\sqrt{20+1}+\sqrt{40+2}+\sqrt{60+3}\) và \(B=\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{20}+\sqrt{40}+\sqrt{60}\right)\)
Những câu hỏi liên quan
So sánh:a)Asqrt[]{21}+sqrt{42}+sqrt{63}Bsqrt{1}+sqrt{2}+sqrt{3}+sqrt{20}+sqrt{40}+sqrt{60}b)Aleft(1-frac{1}{sqrt{4}}right)left(1-frac{1}{sqrt{16}}right)left(1-frac{1}{sqrt{100}}right)Bsqrt{0,1}c) Afrac{1}{sqrt{1}}+frac{1}{sqrt{2}}+...+frac{1}{sqrt{100}}B10
Đọc tiếp
So sánh:
a)\(A=\sqrt[]{21}+\sqrt{42}+\sqrt{63}\)
\(B=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{20}+\sqrt{40}+\sqrt{60}\)
b)\(A=\left(1-\frac{1}{\sqrt{4}}\right)\left(1-\frac{1}{\sqrt{16}}\right)\left(1-\frac{1}{\sqrt{100}}\right)\)
\(B=\sqrt{0,1}\)
c) \(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}\)
\(B=10\)
So sánh A và B :
a)
\(A=\sqrt{20+1}+\sqrt{40+2}+\sqrt{60+3}\)
\(B=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{20}+\sqrt{40}+\sqrt{60}\)
b)
\(A=\frac{1}{\sqrt{121}}+\frac{1}{\sqrt{12321}}+\frac{1}{\sqrt{1234321}}+...+\frac{1}{\sqrt{12345678987654321}}\)
\(B=0,111111111\)
Rút gọn biểu thức
a) A= \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
b) B= \(\sqrt{10+\sqrt{60}-\sqrt{24}-\sqrt{40}}\)
c) C= \(\frac{\sqrt{x-\sqrt{4\left(x-1\right)}}+\sqrt{x+\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}\)
Lời giải:
a)
\(\frac{2A}{\sqrt{2}}=\frac{4+2\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{4-2\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}=\frac{3+1+2\sqrt{3}}{2+\sqrt{3+1+2\sqrt{3}}}+\frac{3+1-2\sqrt{3}}{2-\sqrt{3+1-2\sqrt{3}}}\)
\(=\frac{(\sqrt{3}+1)^2}{2+\sqrt{(\sqrt{3}+1)^2}}+\frac{(\sqrt{3}-1)^2}{2-\sqrt{(\sqrt{3}-1)^2}}=\frac{(\sqrt{3}+1)^2}{2+\sqrt{3}+1}+\frac{(\sqrt{3}-1)^2}{2-(\sqrt{3}-1)}\)
\(=\frac{(\sqrt{3}+1)^2}{\sqrt{3}(\sqrt{3}+1)}+\frac{(\sqrt{3}-1)^2}{\sqrt{3}(\sqrt{3}-1)}=\frac{\sqrt{3}+1}{\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}}=2\)
$\Rightarrow A=\sqrt{2}$
b)
\(B=\sqrt{10+2\sqrt{15}-2\sqrt{6}-2\sqrt{10}}=\sqrt{(8+2\sqrt{15})+2-2\sqrt{2}(\sqrt{3}+\sqrt{5})}\)
\(=\sqrt{(\sqrt{3}+\sqrt{5})^2+2-2\sqrt{2}(\sqrt{3}+\sqrt{5})}\)
\(=\sqrt{(\sqrt{3}+\sqrt{5}-\sqrt{2})^2}=\sqrt{3}+\sqrt{5}-\sqrt{2}\)
c)
\(C=\frac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4x+4}}=\frac{\sqrt{(x-1)-2\sqrt{x-1}+1}+\sqrt{(x-1)+2\sqrt{x-1}+1}}{\sqrt{(x-2)^2}}\)
\(=\frac{\sqrt{(\sqrt{x-1}-1)^2}+\sqrt{(\sqrt{x-1}+1)^2}}{|x-2|}=\frac{|\sqrt{x-1}-1|+|\sqrt{x-1}+1|}{|x-2|}\)
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\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=2006\left(2x-1\right)+\sqrt{2}+\sqrt{3}+\sqrt{5}\)
Bài 1:Thực hiện phép tính
1. \(\sqrt{27}-3\sqrt{48}-2\sqrt{75}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
2.\(\left(\sqrt{8}-5\sqrt{2}+\sqrt{20}\right).\sqrt{5}+\left(40\sqrt{\frac{1}{10}}-10\right)\)
3.\(\left(\sqrt{24}-\sqrt{\frac{2}{3}}-\sqrt{\frac{1}{6}}+\sqrt{\frac{3}{2}}\right).\sqrt{6}\)
a)3sqrt{40sqrt{12}}+4sqrt{sqrt{75}}-5sqrt{5sqrt{48}}b)sqrt{8sqrt{3}}+3sqrt{20sqrt{3}}-2sqrt{45sqrt{3}}c)left(sqrt{x}-1right).left(x+sqrt{x}+1right)left(xge0;yge0right)d)left(sqrt{x}+1right)left(x+1-sqrt{x}right)left(xge0;yge0right)e)left(sqrt{x}+yright)left(x+y^2-ysqrt{2}right)left(xge0;yge0right)
Đọc tiếp
a)\(3\sqrt{40\sqrt{12}}+4\sqrt{\sqrt{75}}-5\)\(\sqrt{5\sqrt{48}}\)
b)\(\sqrt{8\sqrt{3}}+3\sqrt{20\sqrt{3}}-2\sqrt{45\sqrt{3}}\)
c)\(\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)\left(x\ge0;y\ge0\right)\)
d)\(\left(\sqrt{x}+1\right)\left(x+1-\sqrt{x}\right)\left(x\ge0;y\ge0\right)\)
e)\(\left(\sqrt{x}+y\right)\left(x+y^2-y\sqrt{2}\right)\left(x\ge0;y\ge0\right)\)
22222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222
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* rút gọn biểu thức:
a, sqrt{frac{3}{20}}+sqrt{frac{1}{60}}-2sqrt{frac{1}{15}}
b, left(frac{1}{sqrt{5}-sqrt{3}}+frac{1}{sqrt{5}+sqrt{3}}right).sqrt{5}
c, left(5sqrt{3}+3sqrt{5}right):sqrt{15}
d, left(2+sqrt{5}right)^2-left(2+sqrt{5}right)^2
e, frac{1}{3}sqrt{48}+3sqrt{75}-sqrt{27}-10sqrt{1frac{1}{3}}
Đọc tiếp
* rút gọn biểu thức:
a, \(\sqrt{\frac{3}{20}}+\sqrt{\frac{1}{60}}-2\sqrt{\frac{1}{15}}\)
b, \(\left(\frac{1}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{3}}\right).\sqrt{5}\)
c, \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
d, \(\left(2+\sqrt{5}\right)^2-\left(2+\sqrt{5}\right)^2\)
e, \(\frac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\frac{1}{3}}\)
a, = \(\frac{\sqrt{15}}{10}\) + \(\frac{\sqrt{15}}{30}\) - \(\frac{2\sqrt{15}}{15}\)
= \(\sqrt{15}\left(\frac{1}{10}+\frac{1}{30}-\frac{2}{15}\right)\)
= \(\sqrt{15}.0\)
= 0
b, = \(\left(\frac{\sqrt{5}+\sqrt{3}}{5-3}+\frac{\sqrt{5}-\sqrt{3}}{5-3}\right).\sqrt{5}\)
= \(\frac{\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}}{2}.\sqrt{5}\)
= \(\frac{2\sqrt{5}}{2}.\sqrt{5}\)
= \(\sqrt{5}.\sqrt{5}\)
= 5
c, = \(\frac{5\sqrt{3}}{\sqrt{15}}+\frac{3\sqrt{5}}{\sqrt{15}}\)
= \(\sqrt{5}+\sqrt{3}\)
d, Mình sửa lại đề bài cho bạn : \(\left(2+\sqrt{5}\right)^2-\left(2-\sqrt{5}\right)^2\)
= \(\left(2+\sqrt{5}-2+\sqrt{5}\right)\left(2+\sqrt{5}+2-\sqrt{5}\right)\)
= \(2\sqrt{5}.4\)
= \(8\sqrt{5}\)
e, = \(\frac{4\sqrt{3}}{3}+15\sqrt{3}-3\sqrt{3}-\frac{20\sqrt{3}}{3}\)
= \(\sqrt{3}.\left(\frac{4}{3}+15-3-\frac{20}{3}\right)\)
= \(\sqrt{3}.\frac{20}{3}\)
= \(\frac{20\sqrt{3}}{3}\)
a, 320+160−2115
b, (15−3+15+3).5
c, (53+35):15
d, (2+5)2−(2+5)2
e, 1348+375−27−10113
So sánh:
A = (\(\sqrt{1}\) + \(\sqrt{2}\) + \(\sqrt{3}\) ) + (\(\sqrt{20}\) + \(\sqrt{40}\) + \(\sqrt{60}\) )
B = \(\sqrt{20+1}\) + \(\sqrt{40+2}\) + \(\sqrt{60+3}\)
A= ( \(\sqrt{1}\)+\(\sqrt{2}\)+\(\sqrt{3}\) ) + (\(\sqrt{20}\) + \(\sqrt{40}\) + \(\sqrt{60}\))
= (1+1,4+1,7)+(4,4+6,3+7,7)
= 4,1+18,4
=22,5
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Tính:Asqrt{20}-10sqrt{dfrac{1}{5}}+sqrt{left(sqrt{5}-1right)^2}B2sqrt{32}+5sqrt{8}-4sqrt{32}Csqrt{left(3-sqrt{2}^2right)}-sqrt{left(1-sqrt{2}right)^2}Dsqrt{left(5-1right)^2}+sqrt{left(sqrt{5}-3right)^2}Eleft(3+dfrac{5-sqrt{5}}{sqrt{5}-1}right)left(3-dfrac{5+sqrt{5}}{sqrt{5}-1}right)Fsqrt{6+2sqrt{5}}-sqrt{9-4sqrt{5}}Gdfrac{3sqrt{2}-2sqrt{3}}{sqrt{3}-sqrt{2}}+dfrac{sqrt{15}-sqrt{5}}{sqrt{3}-1}Hdfrac{10}{sqrt{3}-1}-dfrac{55}{2sqrt{3}+1} help
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Tính:
\(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(C=\sqrt{\left(3-\sqrt{2}^2\right)}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(D=\sqrt{\left(5-1\right)^2}+\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(E=\left(3+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3-\dfrac{5+\sqrt{5}}{\sqrt{5}-1}\right)\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(G=\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\)
\(H=\dfrac{10}{\sqrt{3}-1}-\dfrac{55}{2\sqrt{3}+1}\)
help
a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)
\(=\sqrt{5}-1\)
b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
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