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\(\sqrt{1}-\sqrt{4}+\sqrt{9}-\sqrt{16}+\sqrt{25}-\sqrt{36}+...-\sqrt{400}\)

\(=1-2+3-4+5-6+...-400\)

\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(399-400\right)\)

\(=-1+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\) (200 số hạng -1)

\(=\left(-1\right).200=-200\)

\(\sqrt{\frac{1}{9}+\frac{1}{16}}\)

\(=\frac{1}{3}+\frac{1}{4}\)

\(=\frac{7}{12}\)

\(\sqrt{4+36+81}\)

\(=\sqrt{121}\)

\(=\pm11\)

\(\sqrt{1^3}+2^3\)

\(=\sqrt{1}+8\)

\(=1+8\)

\(=9\)

\(=20-12+5-6\)

\(=7\)

= 20 - 12 + 5 - 6

= 7

\(E=5.4-4.3+5-0,3.20=20-12+5-6=7\)

=5x4-4x3+5-6

=20-12-1

=7

Xét phân thức phụ sau:

Ta có: \(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+1\right)}}\cdot\left(\frac{1}{\sqrt{n}+\sqrt{n+1}}\right)\)

\(=\frac{1}{\sqrt{n\left(n+1\right)}}\cdot\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\frac{1}{\sqrt{n\left(n+1\right)}}\cdot\left(\sqrt{n+1}-\sqrt{n}\right)\)

\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Thay vào ta được:

\(BT=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)

\(BT=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)

Đặt biểu thức đã cho là A

Tổng quát ta có: Với \(a\inℕ^∗\)ta có:

\(\frac{1}{\left(a+1\right)\sqrt{a}+a.\sqrt{a+1}}=\frac{\left(a+1\right)-a}{\sqrt{a}.\sqrt{a+1}.\left(\sqrt{a}+\sqrt{a+1}\right)}\)

\(=\frac{\left(\sqrt{a+1}-\sqrt{a}\right)\left(\sqrt{a+1}+\sqrt{a}\right)}{\sqrt{a}.\sqrt{a+1}.\left(\sqrt{a}+\sqrt{a+1}\right)}=\frac{\sqrt{a+1}-\sqrt{a}}{\sqrt{a}.\sqrt{a+1}}\)

\(=\frac{\sqrt{a+1}}{\sqrt{a}.\sqrt{a+1}}-\frac{\sqrt{a}}{\sqrt{a}.\sqrt{a+1}}=\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a+1}}\)

Áp dụng kết quả trên ta có:

Với \(n=1\)\(\Rightarrow\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)

Với \(n=2\)\(\Rightarrow\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)

Với \(n=3\)\(\Rightarrow\frac{1}{4\sqrt{3}+3\sqrt{4}}=\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\)

.....................

Với \(n=399\)\(\Rightarrow\frac{1}{400\sqrt{399}+399\sqrt{400}}=\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)

\(\Rightarrow A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+......+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)

\(5\sqrt{16}-4\sqrt{9}+\sqrt{25}-\sqrt{400}\)

\(=5.4-4.3+5-20\)

\(=20-12+5-20\)

\(=-7\)

5√16−4√9+√25−√400

=5.4−4.3+5−20

=20−12+5−20

=8+5-20

=13-20

=-7

\(5\sqrt{16}-4\sqrt{9}-\sqrt{25}-0,3\sqrt{400}\)

\(=5.4-4.3+5-0,3.20\)

\(=20-12+5-6\)

\(=7\)

\(5.\sqrt{16}-4.\sqrt{9}+\sqrt{25}-0,3.\sqrt{400}\)

\(=5.4-4.3+5-\frac{3}{10}.20\)

\(=20-12+5-6\)

\(=8+5-6\)

\(=13-6\)

\(=7\)

=5*4-4*3+5-0.3*20

=20-12+5-6

=8+5-6

=13-6

=7

a) \(\sqrt{16}+\sqrt{225}.\sqrt{9}=4+15.3=4+45=49\)

b) \(\sqrt{\dfrac{10000}{400}}+\sqrt{\left(-3\right)^2}.\sqrt{6^4}=\dfrac{100}{20}+\sqrt{9}.\sqrt{36^2}=5+3.36=5+108=113\)

b. \(\sqrt{\dfrac{10000}{400}}+\sqrt{\left(-3\right)^2}.\sqrt{6^4}=\dfrac{100}{20}+3.6^2=5+3.36=5+108=113\)