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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.\(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\frac{5}{\sqrt{5}}+\frac{\sqrt{20}}{2}-\frac{\frac{5}{4}\cdot2}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\sqrt{5}+\frac{2\sqrt{5}}{2}-\frac{\frac{5}{2}}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\sqrt{5}+\sqrt{5}+\frac{\sqrt{5}}{2}+\sqrt{5}\right):2\sqrt{5}\)
\(=\frac{7}{2}\sqrt{5}:2\sqrt{5}\)
\(=\frac{7}{4}\)
cho S=1-3+32-33+...+398-399
a. Chứng minh: S chia hêt cho 20
b. Rút gọn S, từ đó suy ra 3100 chia 4 dư 1
chịu
Ta có: \(VT=\sqrt{\left(2n+1\right)^2}+\sqrt{4n^2}=\sqrt{\left(2n+1\right)^2}+\sqrt{\left(2n\right)^2}\)
\(=\left|2n+1\right|+\left|2n\right|\)
Vì \(n\inℕ\)\(\Rightarrow2n+1>0\); \(2n\ge0\)
\(\Rightarrow\left|2n+1\right|=2n+1\)và \(\left|2n\right|=2n\)
\(\Rightarrow VT=2n+1+2n=4n+1\)
Ta có: \(VP=\left(2n+1\right)^2-4n^2=\left(2n+1\right)^2-\left(2n\right)^2\)
\(=\left(2n+1-2n\right)\left(2n+1+2n\right)=4n+1\)
\(\Rightarrow VT=VP\)\(\Rightarrowđpcm\)
Ta đi chứng minh công thức tổng quát: \(f\left(n\right)=\frac{2n+1+\sqrt{n\left(n+1\right)}}{\sqrt{n}+\sqrt{n+1}}=\left(n+1\right)\sqrt{n+1}-n\sqrt{n}\)
Thật vậy: \(\left[\left(n+1\right)\sqrt{n+1}-n\sqrt{n}\right]\left(\sqrt{n}+\sqrt{n+1}\right)=\left(n+1\right)\sqrt{n\left(n+1\right)}-n^2+\left(n+1\right)^2-n\sqrt{n\left(n+1\right)}=2n+1+\sqrt{n\left(n+1\right)}\)Áp dụng, ta được: \(f\left(1\right)+f\left(2\right)+...+f\left(2020\right)=\left(2\sqrt{2}-1\sqrt{1}\right)+\left(3\sqrt{3}-2\sqrt{2}\right)+\left(4\sqrt{4}-3\sqrt{3}\right)+...+\left(2021\sqrt{2021}-2020\sqrt{2020}\right)=2021\sqrt{2021}-1\)