\(M=\left(100-1\right)\left(100-2^2\right)\left(100-3^2\right).....\left(100-50^2\right)\\ =\left(100-1\right)\left(100-2^2\right)\left(100-3^2\right).....\left(100-10^2\right).....\left(100-50^2\right)\\ =\left(100-1\right)\left(100-2^2\right)\left(100-3^2\right).....\left(100-100\right).....\left(100-50^2\right)\\ =\left(100-1\right)\left(100-2^2\right)\left(100-3^2\right)....0....\left(100-50^2\right)=0\)

dạng toán này được gọi là gì v ạ?

a: Xét ΔABD vuông tại A và ΔEBD vuông tại E có

BD chung

\(\widehat{ABD}=\widehat{EBD}\)

Do đó: ΔBAD=ΔBED

=>DA=DE

b: DA=DE

=>D nằm trên đường trung trực của AE(1)

Ta có: BA=BE

=>B nằm trên đường trung trực của AE(2)

Từ (1),(2) suy ra BD là đường trung trực của AE

mà BD cắt AE tại F

nên F là trung điểm của AE

=>CF là đường trung tuyến của ΔAEC

nhanh giúp với ạ

\(\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+.\dfrac{x-64}{9}=10\\ \Leftrightarrow\left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0\\ \Leftrightarrow\dfrac{x-100}{15}+\dfrac{x-100}{13}+\dfrac{x-100}{11}+\dfrac{x-100}{9}=0\\ \Leftrightarrow\left(x-100\right)\left(\dfrac{1}{15}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{9}\right)=0\\ \Leftrightarrow x-100=0\\ \Leftrightarrow x=100\)

\(\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}=10\\\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}-10=0\\ \left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0 \\ \dfrac{x-85-15}{15}+\dfrac{x-74-26}{13}+\dfrac{x-67-33}{11}+\dfrac{x-64-36}{9}=0\\ \dfrac{x-100}{15}+\dfrac{x-100}{13}+\dfrac{x-100}{11}+\dfrac{x-100}{9}=0\\ \left(x-100\right)\cdot\left(\dfrac{1}{15}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{9}\right)=0\)

Vì \(\dfrac{1}{15}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{9}\ne0\)

\(A=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\\ 3A=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\\ 3A+A=\left(3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\right)+\left(3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\right)\\ 4A=3^{101}+\left(3^{100}-3^{100}\right)+\left(3^{99}-3^{99}\right)+...+\left(3^2-3^2\right)+\left(3-3\right)-1\\ 4A=3^{101}-1\\ A=\dfrac{3^{101}-1}{4}\)

Nếu đề bài là: (\(\dfrac{2}{3}\))⁵ x \(\dfrac{1}{2^5}\) thì làm như này em nhé.

(\(\dfrac{2}{3}\))⁵ x \(\dfrac{1}{2^5}\) = (\(\dfrac{2}{3}\) x \(\dfrac{1}{2}\))⁵ = (\(\dfrac{1}{3}\))⁵  

\(\dfrac{11}{8}\left[\left(\dfrac{-5}{11}:\dfrac{13}{8}-\dfrac{5}{11}:\dfrac{13}{5}\right)+\dfrac{-6}{33}\right]+\dfrac{3}{4}\\ =\dfrac{11}{8}\left[\left(\dfrac{-5}{11}\cdot\dfrac{8}{13}+\dfrac{-5}{11}\cdot\dfrac{5}{13}\right)+\dfrac{-2}{11}\right]+\dfrac{3}{4}\\ =\dfrac{11}{8}\left[\dfrac{-5}{11}\cdot\left(\dfrac{8}{13}+\dfrac{5}{13}\right)+\dfrac{-2}{11}\right]+\dfrac{3}{4}\\ =\dfrac{11}{8}\left(\dfrac{-5}{11}+\dfrac{-2}{11}\right)+\dfrac{3}{4}\\ =\dfrac{11}{8}\cdot\dfrac{-7}{11}+\dfrac{3}{4}\\ =\dfrac{-7}{8}+\dfrac{6}{8}\\ =\dfrac{-1}{8}\)

Bài 3:

\(a.\dfrac{3x+2}{x-3}=\dfrac{3x-9+11}{x-3}=\dfrac{3\left(x-3\right)+11}{x-3}=3+\dfrac{11}{x-3}\)

Để bt nguyên thì 11 ⋮ x - 3

=> x - 3 ∈ Ư(11) = {1; -1; 11; -11} 

=> x ∈ {4; 2; 14; -8} 

\(b.\dfrac{x-2}{3x-2}\) nguyên \(\Rightarrow\dfrac{3\left(x-2\right)}{3x-2}=\dfrac{3x-6}{3x-2}=\dfrac{3x-2-4}{3x-2}=1-\dfrac{4}{3x-2}\)

Để bt nguyên thì: 4 ⋮ 3x - 2

=> 3x - 2 ∈ Ư(4) = {1; -1; 2; -2; 4; -4}

=> 3x ∈ {3; 1; 4; 0; 5; -2} 

=> x ∈ {1; `1/3`; `4/3`; 0;`5/3`;`-2/3`} 

Mà: x nguyên => x ∈ {1;0}

Bài 4:

\(A=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)

=>\(3A=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\)

=>\(3A+A=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3+3^{100}-3^{99}+...+3^2-3+1\)

=>\(4A=3^{101}+1\)

=>\(A=\dfrac{3^{101}+1}{4}\)