chứng minh:
\(\frac{\left(5\sqrt{3}+\sqrt{50}\right).\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
1.Chứng minh:
\(\frac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}=1\)
\(\frac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(2-2\sqrt{2}.\sqrt{3}+3\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}=\frac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{5\left(\sqrt{3}-\sqrt{2}\right)}=\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2=3-2=1\)
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
\(VT=\frac{\left(5\sqrt{3}+5\sqrt{2}\right).\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)
\(=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)^2.\left(5-2\sqrt{6}\right)}{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5\sqrt{3}-5\sqrt{2}\right)}\)\(=\frac{\left(75+50\sqrt{6}+50\right).\left(5-2\sqrt{6}\right)}{75-50}\)
\(=\frac{25\left(5+2\sqrt{6}\right).\left(5-2\sqrt{6}\right)}{25}=5^2-\left(2\sqrt{6}\right)^2\)\(=25-24=1=VP\)
bn chép lại đề nhé
\(=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)
\(=\frac{\left(75+50\sqrt{6}+50\right)\left(\sqrt{3}-\sqrt{2}\right)}{75-50}\)
a chết bn ơi bài ở trên chưa đúng đâu, bài giải đây nè
\(=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)
\(=\frac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{5\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=3-2=1\left(đpcm\right)\)
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 \(A=1\)
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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ĐẲNG THỨC ĐÓ = 1 NHA Hatsune Miku !
Nguyễn Thị Thanh Trúc: Vậy nêu bước giải bằng lời giúp mk được ko T^T
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 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)
\(M=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{11-6\sqrt{2}}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{2-6\sqrt{2}+9}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{\left(3-\sqrt{2}\right)^2}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+3-\sqrt{2}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{6}}\)
\(=\sqrt{16+32\sqrt{6}}\)
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) \(\frac{1}{\sqrt{5}+\sqrt{2}}+\frac{1}{\sqrt{5}-\sqrt{2}}\)
\(=\frac{\left(\sqrt{5}-\sqrt{2}\right)+\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}\)
\(=\frac{2\sqrt{5}}{\sqrt{5^2}-\sqrt{2^2}}\)
\(=\frac{2\sqrt{5}}{5-2}=\frac{2\sqrt{5}}{3}\)
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.\(\left(\sqrt{24}-\sqrt{\frac{2}{3}}-\sqrt{\frac{1}{6}}+\sqrt{\frac{3}{2}}\right).\sqrt{6}\)
11) \(\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}\)
12) \(\frac{6}{3\sqrt{2}+2\sqrt{3}}\)
13) \(\left(\sqrt{75}-3\sqrt{2}-\sqrt{12}\right)\left(\sqrt{3}+\sqrt{2}\right)\)
14)\(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
15)\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
16)\(\frac{\sqrt{2}}{2\sqrt{3}+4\sqrt{2}}\)
17) \(\frac{1}{4-3\sqrt{2}}-\frac{1}{4+3\sqrt{2}}\)
18)\(\frac{6}{\sqrt{2}-\sqrt{3}+3}\)
19)\(\frac{\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}}{\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}}\)
20)\(\sqrt{24}+6\sqrt{\frac{2}{3}}+\frac{10}{\sqrt{6}-1}\)
21)\(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{58}}\)
22)\(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\frac{1}{5}}\)
23)\(\left(3\sqrt{8}-2\sqrt{12}+\sqrt{20}\right):\left(3\sqrt{18}-2\sqrt{27}+\sqrt{45}\right)\)
24)\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
25)\(\left(\sqrt{7}-\sqrt{5}\right)^2+2\sqrt{35}\)
26)\(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}+\frac{3\sqrt{45}+\sqrt{243}}{\sqrt{5}+\sqrt{3}}\)
27)\(\frac{1}{\sqrt{7-\sqrt{24}}+1}-\frac{1}{\sqrt{7+\sqrt{24}}-1}\)
28)\(\frac{1}{2+\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{2}{3+\sqrt{3}}\)
29)\(\frac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
30)\(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
31)\(\left(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right).\frac{1}{\sqrt{3}+5}\)
32)\(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}-\sqrt{10}\)