ta có :

\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}\)

\(=\dfrac{2}{2\sqrt{1}}+\dfrac{2}{2\sqrt{2}}+\dfrac{2}{2\sqrt{3}}+...+\dfrac{2}{2\sqrt{100}}\)

\(>\dfrac{2}{\sqrt{1}+\sqrt{2}}+\dfrac{2}{\sqrt{2}+\sqrt{3}}+\dfrac{2}{\sqrt{3}+\sqrt{4}}+...+\dfrac{2}{\sqrt{100}+\sqrt{101}}\)

\(=2\left(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{100}+\sqrt{101}}\right)\)

\(=2\left(\dfrac{\sqrt{1}-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+\dfrac{\sqrt{3}+\sqrt{4}}{3-4}+...+\dfrac{\sqrt{100}-\sqrt{101}}{100-101}\right)\)

\(=2\left(\dfrac{\sqrt{1}-\sqrt{101}}{-1}\right)=2\left(\sqrt{101}-\sqrt{1}\right)=18,1\)

\(>18\)

\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>18\)

a,

\(A=\dfrac{2x}{x+3}-\dfrac{x+1}{3-x}-\dfrac{3-11x}{x^2-9}\)

\(A=\dfrac{2x^2-6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x^2+4x+3}{\left(x-3\right)\left(x+3\right)}-\dfrac{3-11x}{\left(x-3\right)\left(x+3\right)}\)

\(A=\dfrac{3x^2+9x+1}{\left(x-3\right)\left(x+3\right)}\)

vẽ lại hình : (mk k bt vẽ hình trực tuyến nên chỉ hg dẫn bn vẽ thui nha)

vẽ R₁ nối tiếp R₃

R₂ // R₁₃

R₄ // R123

vì R₁ nt R₃

=> R₁₃=2R

vì R₁₃ // R₂

\(\Rightarrow R_{123}=\dfrac{R^2}{2R}=\dfrac{R}{2}\)

vì R₄ // R₁₂₃

\(\Rightarrow R_{tđ}=\dfrac{R.\dfrac{R}{2}}{R+\dfrac{R}{2}}=\dfrac{\dfrac{R^2}{2}}{\dfrac{3R}{2}}=\dfrac{R^2}{3R}=\dfrac{R}{3}\)

mk k chắc là đúng đâu bn nhé !

\(A=\sin^275+\sin^215-\cos^250-\cos^240+\cot40.\cot50\)

\(A=\sin^275+\cos^275-\cos^250-\sin^250+\cot40.\tan40\)

\(A=1-1+1\)

\(A=1\)

\(\sqrt{11-4\sqrt{7}}-\sqrt{2}\sqrt{8+3\sqrt{7}}\)

\(=\sqrt{7-2\sqrt{4}\sqrt{7}+4}-\sqrt{16+6\sqrt{7}}\)

\(=\sqrt{\left(\sqrt{7}-\sqrt{4}\right)^2}-\sqrt{9+2\sqrt{9}\sqrt{7}+7}\)

\(=\sqrt{7}-\sqrt{4}-\sqrt{9}-\sqrt{7}\)

\(=-2-3=-5\)

chúc bn hc giỏi

ta có :

\(S=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{25}}\)\(\Leftrightarrow S=\dfrac{2}{2\sqrt{2}}+\dfrac{2}{2\sqrt{3}}+\dfrac{2}{2\sqrt{4}}+...+\dfrac{2}{2\sqrt{25}}\)

\(>\dfrac{2}{\sqrt{2}+\sqrt{3}}+\dfrac{2}{\sqrt{3}+\sqrt{4}}+\dfrac{2}{\sqrt{4}+\sqrt{5}}+...+\dfrac{2}{\sqrt{25}+\sqrt{26}}\)

\(\Leftrightarrow S>2.\left(\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+\dfrac{\sqrt{3}-\sqrt{4}}{3-4}+\dfrac{\sqrt{4}+\sqrt{5}}{4-5}+...+\dfrac{\sqrt{25}-\sqrt{26}}{25-26}\right)\)\(\Leftrightarrow S>2\left(\dfrac{\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+\sqrt{4}-\sqrt{5}+...+\sqrt{25}-\sqrt{26}}{-1}\right)\)

\(\Leftrightarrow S>2\left(\dfrac{\sqrt{2}-\sqrt{26}}{-1}\right)\)

\(\Leftrightarrow S>2.\left(\sqrt{26}-\sqrt{2}\right)\)

\(\Leftrightarrow S>7,4\)

\(\Leftrightarrow S>7\)

chúc bn hc tốt