Lỗi r, xem lại nha

a: Hàm số nghịch biến trên R

b: \(\dfrac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\dfrac{x_1^2-4x_1+5-x_2^2+4x_2-5}{x_1-x_2}\)

\(=x_1+x_2-4\)

Trường hợp 1: x<=2

\(\Leftrightarrow x_1+x_2-4< =0\)

Vậy: Hàm số nghịch biến khi x<=2

\(a,x+\dfrac{1}{2}=\dfrac{3}{4}\\ x=\dfrac{3}{4}-\dfrac{1}{2}\\ x=\dfrac{1}{2}\\ b,-\dfrac{2}{3}-x=1\\x=-\dfrac{2}{3}-1\\ x=-\dfrac{5}{3}\\ d,\dfrac{1}{4}+\dfrac{3}{4}:x=\dfrac{5}{2}\\ \dfrac{3}{4}:x=\dfrac{5}{2}-\dfrac{1}{4}\\ \dfrac{3}{4}:x=\dfrac{9}{4}\\ x=\dfrac{3}{4}:\dfrac{9}{4}\\ x=\dfrac{1}{3}\\ e,\left(x+\dfrac{1}{4}\right)\cdot\dfrac{3}{4}=-\dfrac{5}{8}\\ x+\dfrac{1}{4}=-\dfrac{5}{8}:\dfrac{3}{4}\\ x+\dfrac{1}{4}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{1}{4}\\ x=\dfrac{7}{12}\)

\(g,\dfrac{x-3}{15}=\dfrac{-2}{5}\\ 5\left(x-3\right)=-30\\ x-3=-6\\ x=-6+3\\ x=-3\\ h,\dfrac{x}{-2}=\dfrac{-8}{x}\\ x^2=16\\ x=\pm\sqrt{16}\\ x=\pm4\\ k,\dfrac{x+2}{3}=\dfrac{x-4}{5}\\ 5\left(x+2\right)=3\left(x-4\right)\\ 5x+10=3x-12\\ 5x-3x=-12-10\\ 2x=-22\\ x=-11\)

\(m,\left(2x-1\right)^2=4\\ \Rightarrow\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Đánh giá giúp nhá cảm ơn

\(a^3+b^3+c^3=3abc\)

=>\(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)

=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

=>\(a^2+b^2+c^2-ab-ac-bc=0\)

=>\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

=>\(\left(a^2-2ba+b^2\right)+\left(b^2-2cb+c^2\right)+\left(a^2-2ac+c^2\right)=0\)

=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

=>\(\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow a=b=c\)

\(A=\dfrac{a^{2023}}{b^{2023}}+\dfrac{b^{2023}}{c^{2023}}+\dfrac{c^{2023}}{a^{2023}}\)

\(=\dfrac{a^{2023}}{a^{2023}}+\dfrac{b^{2023}}{b^{2023}}+\dfrac{c^{2023}}{c^{2023}}\)

=1+1+1

=3