Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1 :
\(6xy\cdot\sqrt{\frac{9x^2}{16y^2}}=6xy\cdot\frac{3x}{4y}=\frac{18x^2y}{4y}=\frac{9}{2}x^2\)
\(\sqrt{\frac{4+20a+25a^2}{b^4}}=\sqrt{\frac{\left(2+5a\right)^2}{\left(b^2\right)^2}}=\frac{2+5a}{b^2}\)
\(\left(m-n\right).\sqrt{\frac{m-n}{\left(m-n\right)^2}}=\sqrt{\left(m-n\right)^2}\cdot\sqrt{\frac{1}{m-n}}=\sqrt{\frac{\left(m-n\right)^2}{m-n}}=\sqrt{m-n}\)
Bài 2 :
1. \(\left(2\sqrt{3}-\sqrt{12}\right):5\sqrt{3}=\left(2\sqrt{3}-2\sqrt{3}\right):5\sqrt{3}=0:5\sqrt{3}=0\)
2. \(\sqrt{\frac{317^2-302^2}{1013^2-1012^2}}=\frac{\sqrt{\left(317+302\right)\left(317-302\right)}}{\sqrt{\left(1013+1012\right)\left(1013-1012\right)}}=\frac{\sqrt{619}\cdot\sqrt{15}}{\sqrt{2025}}=\sqrt{\frac{619}{135}}\)(check lại)
3. \(\sqrt{27\left(1-\sqrt{3}\right)^2}:3\sqrt{75}\)
\(=\sqrt{27}\left(1-\sqrt{3}\right):15\sqrt{3}\)
\(=3\sqrt{3}\left(1-\sqrt{3}\right):15\sqrt{3}\)
\(=\frac{1-\sqrt{3}}{5}\)
4.\(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\frac{5}{\sqrt{5}}+\frac{\sqrt{20}}{2}-\frac{\frac{5}{4}\cdot2}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\sqrt{5}+\frac{2\sqrt{5}}{2}-\frac{\frac{5}{2}}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\sqrt{5}+\sqrt{5}+\frac{\sqrt{5}}{2}+\sqrt{5}\right):2\sqrt{5}\)
\(=\frac{7}{2}\sqrt{5}:2\sqrt{5}\)
\(=\frac{7}{4}\)
a) \(\left(\sqrt{8}+\sqrt{72}-\sqrt{2}\right).\sqrt{2}\)
\(=\left(2\sqrt{2}+6\sqrt{2}-\sqrt{2}\right).\sqrt{2}\)
\(=7\sqrt{2}.\sqrt{2}=7.2=14\)
b) \(\left(\sqrt{5}+\sqrt{2}+1\right)\left(\sqrt{5}-1\right)\)
\(=5-\sqrt{5}+\sqrt{10}-\sqrt{2}+\sqrt{5}-1\)
\(=4+\sqrt{10}-\sqrt{2}\)
c) \(\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2\)
\(=\left(\sqrt{4+\sqrt{7}}\right)^2-2\sqrt{4+\sqrt{7}}\sqrt{4-\sqrt{7}}+\left(\sqrt{4-\sqrt{7}}\right)^2\)
\(=\left(4+\sqrt{7}\right)-6+\left(4-\sqrt{7}\right)\)
\(=4+\sqrt{7}-6+4-\sqrt{7}=2\)
d) \(\left(\sqrt{2}+1+\sqrt{3}\right).\left(\sqrt{2}+1-\sqrt{3}\right)\)
\(=\left(\sqrt{2}+1\right)^2-3=2+2\sqrt{2}+1-3=2\sqrt{2}\)
e) \(\left(\sqrt{\frac{9}{2}}+\sqrt{\frac{1}{2}}-\sqrt{2}\right).\sqrt{2}\)
\(=3+1-2=2\)(nhân vào)
f) \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
\(=\left(\sqrt{75}+\sqrt{45}\right):\sqrt{15}=\sqrt{5}+\sqrt{3}\)(chia đa tức cho đơn thức)
có sai xót mong m.n bỏ qa cho ♥
Với mọi n nguyên dương ta có:
\(\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=1\Rightarrow\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}\)
Với k nguyên dương thì
\(\frac{1}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k+1}+\sqrt{k}}\Rightarrow\frac{2}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k-1}+\sqrt{k}}+\frac{1}{\sqrt{k+1}+\sqrt{k}}=\sqrt{k}-\sqrt{k-1}+\sqrt{k+1}-\sqrt{k}\)
\(=\sqrt{k+1}-\sqrt{k-1}\)(*)
Đặt A = vế trái. Áp dụng (*) ta có:
\(\frac{2}{\sqrt{1}+\sqrt{2}}>\sqrt{3}-\sqrt{1}\)
\(\frac{2}{\sqrt{3}+\sqrt{4}}>\sqrt{5}-\sqrt{3}\)
...
\(\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-\sqrt{79}\)
Cộng tất cả lại
\(2A=\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{4}}+....+\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-1=8\Rightarrow A>4\left(đpcm\right)\)
3.
Theo bất đẳng thức cô si ta có:
\(\sqrt{b-1}=\sqrt{1.\left(b-1\right)}\le\frac{1+b-1}{2}=\frac{b}{2}\Rightarrow a.\sqrt{b-1}\le\frac{a.b}{2}\)
Tương tự \(\Rightarrow b.\sqrt{a-1}\le\frac{a.b}{2}\Rightarrow a.\sqrt{b-1}+b.\sqrt{a-1}\le a.b\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=2\)
\(a,\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
Ta có
:\(VT=\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)
\(=|2-\sqrt{5}|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}\)
\(=-2=VP\left(đpcm\right)\)
\(b,\frac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}\)
Ta có:
\(VT=\frac{\sqrt{2}+1}{\sqrt{2}-1}\)
\(=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(=\frac{2+\sqrt{2}+\sqrt{2}+1}{\sqrt{2}^2-1^2}\)
\(=\frac{3+2\sqrt{2}}{2-1}\)
\(=3+2\sqrt{2}=VP\left(đpcm\right)\)
c,Bạn xem lại đề
\(d,\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}=8\)
Ta có:
\(VT=\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)
\(=\sqrt{\frac{2^2}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{2^2}{\left(2+\sqrt{5}\right)^2}}\)
\(=\frac{2}{|2-\sqrt{5}|}-\frac{2}{|2+\sqrt{5}|}\)
\(=\frac{2\left(2+\sqrt{5}\right)}{\left(\sqrt{5}-2\right)\left(2+\sqrt{5}\right)}-\frac{2\left(\sqrt{5}-2\right)}{\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)}\)
\(=\frac{4+2\sqrt{5}-2\sqrt{5}+4}{\sqrt{5}^2-2^2}\)
\(=\frac{8}{5-4}\)
\(=8=VP\left(đpcm\right)\)
ax cau nhu vay ma cung hoi
choc choi ak
\(a=\left[\left(\frac{4+\sqrt{5}}{3+\sqrt{5}}\right)^2+\left(\frac{4-\sqrt{5}}{3-\sqrt{5}}\right)^2\right]^2-2\left(\frac{4+\sqrt{5}}{3+\sqrt{5}}\right)^2\left(\frac{4-\sqrt{5}}{3-\sqrt{5}}\right)^2\)
\(a=\left[\frac{21+8\sqrt{5}}{14+6\sqrt{5}}+\frac{21-8\sqrt{5}}{14-6\sqrt{5}}\right]^2-2\left(\frac{21-8\sqrt{5}}{14-6\sqrt{5}}\right)\left(\frac{21+8\sqrt{5}}{14+6\sqrt{5}}\right)\)
\(a=\left[\frac{\left(21+8\sqrt{5}\right)\left(14-6\sqrt{5}\right)+\left(21-8\sqrt{5}\right)\left(14+6\sqrt{5}\right)}{\left(14+6\sqrt{5}\right)\left(14-6\sqrt{5}\right)}\right]^2-\frac{121}{8}\)
\(a=\left(\frac{54-14\sqrt{5}+54+14\sqrt{5}}{16}\right)^2-\frac{121}{8}=\left(\frac{108}{16}\right)^2-\frac{121}{8}=\frac{487}{16}\)