\(M=1-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)

Đặt \(N=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\)

\(2N=1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\)

\(\Rightarrow2N-N=1-\dfrac{1}{2^{10}}\)

\(\Rightarrow N=1-\dfrac{1}{2^{10}}\)

\(\Rightarrow M=1-\left(1-\dfrac{1}{2^{10}}\right)=\dfrac{1}{2^{10}}>\dfrac{1}{2^{11}}\)

Vậy \(M>\dfrac{1}{2^{11}}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}=4\)

\(\Rightarrow2+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}=4\)

\(\Rightarrow\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}=2\)

\(\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)

\(\Rightarrow\dfrac{c+a+b}{abc}=1\)

\(\Rightarrow a+b+c=abc\) 

\(a,\left(\dfrac{1}{x-1}-\dfrac{x}{x-1^2}.\dfrac{x^2+1+x}{x+1}\right):\dfrac{1}{x^2-1}\\ =\left(\dfrac{1}{x-1}-\dfrac{x\left(x^2+1+x\right)}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1}{x^2-1}\\ =\left(\dfrac{1\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x^3+x+x^2}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1}{x^2-1}\)

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

b,

thay x=\(\dfrac{1}{2}\) vào bt M ta được:

\(1-\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^2=\dfrac{5}{8}\)

ĐKXĐ: \(x\ge-2;x\ne-1\)

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

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

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

\(M=-\dfrac{1}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\ge-\dfrac{1}{\dfrac{3}{4}}=-\dfrac{4}{3}\)

\(M_{min}=-\dfrac{4}{3}\) khi \(x=\dfrac{1}{2}\)

Help me!!!

Bài này giải ra dài lắm;

Gợi ý : với câu a) cm 1<A<2

với câ u b) 0<B<1

với câu c) áp dụng bài toán của ông gao í; cách tỉnh tổng từ 1->100 trong sách GK 6 có nhé

Mong bạn giải ra

Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

=100

Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)

\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{8}{\dfrac{1}{5}}=40\)

\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)

Bạn ơi thiếu đề rồi, cái biểu thức này không tính được đâu , mình nghĩ thế

(1/m+1/n+1/p)^2=25

=>1/m^2+1/n^2+1/p^2+2(1/mn+1/pn+1/mp)=25

=>\(5+2\cdot\dfrac{m+n+p}{mnp}=25\)

=>\(2\cdot\dfrac{m+n+p}{mnp}=20\)

=>\(\dfrac{m+n+p}{mnp}=10\)

=>m+n+p=10mnp