1.Chứng minh:
\(\frac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}=1\)
Những câu hỏi liên quan
Chứng minh đẳng thức sau:
\(\dfrac{\left(5\sqrt{3}+\sqrt{50}\right).\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\) = 1
\(A=\dfrac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\\ =\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}\\ =\dfrac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{\sqrt{3}-\sqrt{2}}\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\\ =3-2\\ =1\)
Vậy \(A=1\)
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Chứng minh đẳng thức dưới:
\(\frac{\left(5\sqrt{3}+\sqrt{50}\right).\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\) = 1
Làm ơn giúp mình với T^T
\(VT=\frac{\left(5\sqrt{3}+5\sqrt{2}\right).\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)
\(=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)^2.\left(5-2\sqrt{6}\right)}{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5\sqrt{3}-5\sqrt{2}\right)}\)\(=\frac{\left(75+50\sqrt{6}+50\right).\left(5-2\sqrt{6}\right)}{75-50}\)
\(=\frac{25\left(5+2\sqrt{6}\right).\left(5-2\sqrt{6}\right)}{25}=5^2-\left(2\sqrt{6}\right)^2\)\(=25-24=1=VP\)
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bn chép lại đề nhé
\(=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)
\(=\frac{\left(75+50\sqrt{6}+50\right)\left(\sqrt{3}-\sqrt{2}\right)}{75-50}\)
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a chết bn ơi bài ở trên chưa đúng đâu, bài giải đây nè
\(=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)
\(=\frac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{5\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=3-2=1\left(đpcm\right)\)
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Xem thêm câu trả lời
chứng minh:
\(\frac{\left(5\sqrt{3}+\sqrt{50}\right).\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
cm > hay < ?
Chứng minh đẳng thức dưới bằng 1. Làm ơn giúp tớ với T^T Tớ sẽ tick ngay nha ^^
\(\frac{\left(5\sqrt{3}+\sqrt{50}\right).\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
\(\frac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
\(=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)
\(=\frac{5\left(\sqrt{3}+\sqrt{2}\right)\left(3-2.\sqrt{3}.\sqrt{2}+2\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\frac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{5\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=1\)
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DÀI QUÁ MK KO GHI ĐƯỢC NÊN VIẾT KQ LUÔN NHA !!!
ĐẲNG THỨC ĐÓ = 1 NHA 
Hatsune Miku !
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Nguyễn Thị Thanh Trúc: Vậy nêu bước giải bằng lời giúp mk được ko T^T
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Chứng minh
\(\frac{1}{3\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+\)+\(\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}\)+....+\(\frac{1}{49\left(\sqrt{24}+\sqrt{25}\right)}\)<\(\frac{2}{5}\)
Bài 1. thực hiện phép tínha) sqrt{frac{5+sqrt{21}}{5-sqrt{21}}}+sqrt{frac{5-sqrt{21}}{5+sqrt{21}}} b) sqrt{frac{4+sqrt{7}}{4-sqrt{7}}}+sqrt{frac{4-sqrt{7}}{4+sqrt{7}}}Bài 2. Tính:a) Mleft(2sqrt{2}right)sqrt{2+4sqrt{3+sqrt{2}+sqrt{11-6sqrt{2}}}}b) frac{1}{sqrt{25}+sqrt{24}}+frac{1}{sqrt{24}+sqrt{23}}+...+frac{1}{sqrt{2}+sqrt{1}}4c)frac{left(5sqrt{3}+sqrt{50}right)left(5-sqrt{24}right)}{sqrt{75}0-5sqrt{2}}
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Bài 1. thực hiện phép tính
a) \(\sqrt{\frac{5+\sqrt{21}}{5-\sqrt{21}}}+\sqrt{\frac{5-\sqrt{21}}{5+\sqrt{21}}}\) b) \(\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}+\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\)
Bài 2. Tính:a) \(M=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{11-6\sqrt{2}}}}\)
b) \(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)
c)
\(\frac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}0-5\sqrt{2}}\)
\(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)
\(\Leftrightarrow\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-\sqrt{1}=4\)
\(\Leftrightarrow\sqrt{25}-\sqrt{1}=4\Leftrightarrow5-1=4\)(đúng)
Vậy \(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)(đpcm)
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\(M=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{11-6\sqrt{2}}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{2-6\sqrt{2}+9}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{\left(3-\sqrt{2}\right)^2}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+3-\sqrt{2}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{6}}\)
\(=\sqrt{16+32\sqrt{6}}\)
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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.\(\left(\sqrt{24}-\sqrt{\frac{2}{3}}-\sqrt{\frac{1}{6}}+\sqrt{\frac{3}{2}}\right).\sqrt{6}\)
THỰC HIỆN PHÉP TÍNH:
22) frac{1}{sqrt{5}+sqrt{2}}+frac{1}{sqrt{5}-sqrt{2}}
23) frac{2+sqrt{3}}{sqrt{2}+sqrt{2+sqrt{3}}}+frac{2-sqrt{3}}{sqrt{2}-sqrt{2-sqrt{3}}}
24) frac{sqrt{18}}{sqrt{2}}-frac{sqrt{12}}{sqrt{3}}
25) sqrt{left(sqrt{5}+1right)^2}+sqrt{left(sqrt{5}-1right)^2}
27) sqrt{3-2sqrt{2}}
28) frac{1}{sqrt{8}+sqrt{7}}+sqrt{175}-2sqrt{2}
30) left(2sqrt{1frac{9}{16}}-sqrt{5frac{1}{16}}right):sqrt{16}
34) frac{left(5sqrt{3}+sqrt{50}right)left(5-sqrt{24}right)}{sqrt{75}-5sqrt{2}}
35) l...
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THỰC HIỆN PHÉP TÍNH:
22) \(\frac{1}{\sqrt{5}+\sqrt{2}}+\frac{1}{\sqrt{5}-\sqrt{2}}\)
23) \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
24) \(\frac{\sqrt{18}}{\sqrt{2}}-\frac{\sqrt{12}}{\sqrt{3}}\)
25) \(\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
27) \(\sqrt{3-2\sqrt{2}}\)
28) \(\frac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-2\sqrt{2}\)
30) \(\left(2\sqrt{1\frac{9}{16}}-\sqrt{5\frac{1}{16}}\right):\sqrt{16}\)
34) \(\frac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
35) \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\frac{1}{4}\sqrt{8}\right).3\sqrt{6}\)
36) \(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
39) \(\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}+\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}}\)
45) \(\frac{\sqrt{6-2\sqrt{5}}}{2-\sqrt{20}}\)
22) \(\frac{1}{\sqrt{5}+\sqrt{2}}+\frac{1}{\sqrt{5}-\sqrt{2}}\)
\(=\frac{\left(\sqrt{5}-\sqrt{2}\right)+\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}\)
\(=\frac{2\sqrt{5}}{\sqrt{5^2}-\sqrt{2^2}}\)
\(=\frac{2\sqrt{5}}{5-2}=\frac{2\sqrt{5}}{3}\)
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Tinh
a,sqrt{75}-sqrt{5frac{1}{3}}+frac{9}{2}sqrt{2frac{2}{3}}+2sqrt{27}
b,sqrt{48}+sqrt{5frac{1}{3}}+2sqrt{75}-5sqrt{1frac{1}{3}}
c,left(sqrt{15}+2sqrt{3}right)^2+12sqrt{5}
d,left(sqrt{6}+2right)left(sqrt{3}-sqrt{2}right)
e,left(sqrt{3}+1right)^2-2sqrt{3}+4
f,frac{1}{7+4sqrt{3}}+frac{1}{7-4sqrt{3}}
g,left(frac{1}{sqrt{5}-sqrt{2}}-frac{1}{sqrt{5}+sqrt{2}}+1right)frac{1}{left(sqrt{2}+1right)^2}
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Tinh
\(a,\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}\)
\(b,\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}\)
\(c,\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}\)
\(d,\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)\)
\(e,\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4\)
\(f,\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
\(g,\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right)\frac{1}{\left(\sqrt{2}+1\right)^2}\)
a)\(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}=5\sqrt{3}-\frac{\sqrt{15}}{3}+3\sqrt{3}+6\sqrt{3}=14\sqrt{3}-\frac{\sqrt{15}}{3}\)
b) \(\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}=4\sqrt{3}+\frac{\sqrt{15}}{3}+10\sqrt{3}-\frac{5\sqrt{3}}{3}=\frac{12\sqrt{3}+30\sqrt{3}-5\sqrt{3}}{3}+\frac{\sqrt{15}}{3}=\frac{37\sqrt{3}+\sqrt{15}}{3}\)
c) \(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}=\left[\left(\sqrt{15}\right)^2+4\sqrt{45}+\left(2\sqrt{3}\right)^2\right]+12\sqrt{5}=15+12\sqrt{5}+12+12\sqrt{5}=27+24\sqrt{5}\)
d) \(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{18}-\sqrt{12}+\sqrt{6}-2\sqrt{2}=3\sqrt{2}-2\sqrt{3}+\sqrt{6}-2\sqrt{2}=\sqrt{2}-2\sqrt{3}+\sqrt{6}\)
e) \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=\left(\sqrt{3}\right)^2+2\sqrt{3}+1-2\sqrt{3}+4=3+2\sqrt{3}+1-2\sqrt{3}+4=8\)
f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)
g) \(\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right)\frac{1}{\left(\sqrt{2}+1\right)^2}=\left(\frac{\sqrt{5}+2-\sqrt{5}+2+5-2}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\right)\frac{1}{3+2\sqrt{2}}=\frac{7}{3}.\frac{1}{3+2\sqrt{2}}=\frac{7}{9+6\sqrt{2}}\)
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