c) \(\dfrac{1}{\left(1-x\right)\left(x-2\right)}+\dfrac{1}{\left(2-x\right)\left(x-3\right)}+...+\dfrac{1}{\left(99-x\right)\left(x-100\right)}\)

\(=-\dfrac{1}{\left(x-1\right)\left(x-2\right)}-\dfrac{1}{\left(x-2\right)\left(x-3\right)}-...-\dfrac{1}{\left(x-99\right)\left(x-100\right)}\)

\(=-\left[\dfrac{\left(x-1\right)-\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}+\dfrac{\left(x-2\right)-\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+...+\dfrac{\left(x-99\right)-\left(x-100\right)}{\left(x-99\right)\left(x-100\right)}\right]\)

\(=-\left(\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+...+\dfrac{1}{x-100}-\dfrac{1}{x-99}\right)\)

\(=-\left(\dfrac{1}{x-100}-\dfrac{1}{x-1}\right)\)

\(=\dfrac{1}{x-1}-\dfrac{1}{x-100}\)

Bài 5: 

\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)

TH1: \(x=-y\): 

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{-y}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{z}=\dfrac{1}{0+z}=\dfrac{1}{x+y+z}\)

Ta có đpcm. 

Các trường hợp \(y=-z\) và \(z=-x\) tương tự. 

a) \(-5x^2+3x=-\left(5x^2-3x\right)=-\left[\left(\sqrt{5}x\right)^2-2.\sqrt{5}x.\dfrac{3}{2\sqrt{5}}+\left(\dfrac{3}{2\sqrt{5}}\right)^2-\dfrac{9}{20}\right]\)

\(=-\left(\sqrt{5}x-\dfrac{3}{2\sqrt{5}}\right)^2+\dfrac{9}{20}\le\dfrac{9}{20}\)

Dấu \(=\)xảy ra khi \(\sqrt{5}x-\dfrac{3}{2\sqrt{5}}=0\Leftrightarrow x=\dfrac{3}{10}\).

b) \(4x^2-3x=\left(2x\right)^2-2.\left(2x\right).\dfrac{3}{4}+\left(\dfrac{3}{4}\right)^2-\dfrac{9}{16}=\left(2x-\dfrac{3}{4}\right)^2-\dfrac{9}{16}\ge-\dfrac{9}{16}\)

Dấu \(=\) xảy ra khi \(2x-\dfrac{3}{4}=0\Leftrightarrow x=\dfrac{3}{8}\). 

c) Bạn làm tương tự ý b).

Bài này là bài khảo sát chất lượng HSG mong mn giúp bọn mình với ạ

Ta có 3x ( x² - 9 ) = 0 ⇒ 3x = 0 hoặc x² - 9 = 0

Nếu 3x = 0 ⇒ x = 0

Nếu x² - 9 = 0 ⇒ x ² = 0 + 9 = 9 = 3²

Vậy x = 3

Vậy x ϵ { 0; 3 } để 3x ( x² - 9 ) = 0 

là thế nào

Em nhập lại câu hỏi nha em!