\(\left|x+1\right|+\left|x+5\right|=4\\ Áp\text{ }dụng\text{ }bất\text{ }đẳng\text{ }thức\text{ }\left|a\right|+\left|b\right|\ge\left|a+b\right|,\text{ }ta\text{ }được:\\ \left|x+1\right|+\left|x+5\right|\ge\left|x+1+x+5\right|\\ \Leftrightarrow\left|2x+6\right|\le4\\ \Leftrightarrow\left[{}\begin{matrix}2x+6\le-4\\2x+6\le4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x\le-10\\2x\le-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le-5\\x\le-1\end{matrix}\right.\\\text{ }Vậy\text{ }x\le-5\text{ }hoặc\text{ }x\le-1\)

\(Ta\text{ }có\text{ }:\text{ }\dfrac{nguyên\text{ }tử\text{ }Fe}{nguyên\text{ }tử\text{ }Si}=\dfrac{56}{28}=2\\ Vậy\text{ }nguyên\text{ }tử\text{ }Fe\text{ }nặng\text{ }gấp\text{ }2\text{ }lần\text{ }nguyên\text{ }tử\text{ }Si\)

\(\text{Ta có : }a+b+c=0\\ \Rightarrow c=-\left(a+b\right)\text{ }\text{ }\text{ }\left(\text{*}\right)\\ \Rightarrow c^3=-\left(a+b\right)^3\\ \Rightarrow a^3+b^3+c^3=a^3+b^3-\left(a+b\right)^3\\ =\left(a^3+b^3\right)-\left(a^3+3a^2b+3ab^2+b^3\right)\\ =-\left(3a^2b+3ab^2\right)\\ =-3ab\left(a+b\right)\text{ }\text{ }\text{ }\text{ }\left(1\right)\\ Thay\text{ }\left(\text{*}\right)\text{ }vào\text{ }\left(1\right),ta\text{ }được:\\ \left(1\right)=\left(-3ab\right)\cdot\left(-c\right)=3abc\left(đpcm\right)\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\)

Vậy \(a^3+b^3+c^3=3abc\)

\(\text{a) }3x^2y^2:x^2=3y^2\)

\(\text{b) }\left(x^5+4x^3-6x^2\right):4x^2\\ =\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)

\(\text{c) }\left(x^3-8\right):\left(x^2+2x+4\right)\\ =\left(x-2\right)\left(x^2+2x+4\right):\left(x^2+2x+4\right)\\ =x-2\)

\(\text{d) }\left(3x^2-6x\right):\left(2-x\right)\\ =3x\left(x-2\right):\left(2-x\right)\\ =-3x\left(2-x\right):\left(2-x\right)\\ =-3x\)

\(\text{e) }\left(x^3+2x^2-2x-1\right):\left(x^2+3x+1\right)\\ =\left(x^3+3x^2-x^2+x-3x-1\right):\left(x^2+3x+1\right)\\ =\left[\left(x^3+3x^2+x\right)-\left(x^2+3x+1\right)\right]:\left(x^2+3x+1\right)\\ =\left[x\left(x^2+3x+1\right)-\left(x^2+3x-1\right)\right]:\left(x^2+3x+1\right)\\ =\left(x-1\right)\left(x^2+3x+1\right):\left(x^2+3x+1\right)\\ =x-1\)

\(\text{a) }\dfrac{x^2+x+1}{x^2+2x+1}\\ =\dfrac{x^2+2x-x+1+1-1}{x^2+2x+1}\\ =\dfrac{\left(x^2+2x+1\right)-\left(x+1\right)+1}{x^2+2x+1}\\ =\dfrac{x^2+2x+1}{x^2+2x+1}-\dfrac{x+1}{\left(x+1\right)^2}+\dfrac{1}{\left(x+1\right)^2}\\ =1-\dfrac{1}{x+1}+\dfrac{1}{\left(x+1\right)^2}\left(1\right)\\ Đặt\text{ }\dfrac{1}{x+1}=y\\ \Rightarrow\left(1\right)=1-y+y^2\\ =y^2-y+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(y^2-y+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\\ Do\text{ }\left(y-\dfrac{1}{2}\right)^2\ge0\forall x\\ \Rightarrow\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\\ Dấu\text{ }"="\text{ }xảy\text{ }ra\text{ }khi:\\ \left(y-\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow y-\dfrac{1}{2}=0\\ \Leftrightarrow y=\dfrac{1}{2}\\ \Leftrightarrow\dfrac{ 1}{x+1}=\dfrac{1}{2}\\ \Leftrightarrow x+1=2\\ \Leftrightarrow x=1\\ Vậy\text{ }GTNN\text{ }của\text{ }phân\text{ }thức\text{ }là\text{ }\dfrac{3}{4}\text{ }khi\text{ }x=1\)

\(\text{b) }\dfrac{4x^2-6x+1}{\left(2x-1\right)^2}\\ =\dfrac{4x^2-4x-2x+1+1-1}{\left(2x-1\right)^2}\\ =\dfrac{\left(4x^2-4x+1\right)-\left(2x-1\right)-1}{\left(2x-1\right)^2}\\ =\dfrac{\left(2x-1\right)^2}{\left(2x-1\right)^2}-\dfrac{2x-1}{\left(2x-1\right)^2}-\dfrac{1}{\left(2x-1\right)^2}\\ =1-\dfrac{1}{2x-1}-\dfrac{1}{\left(2x-1\right)^2}\left(1\right)\\ Đặt\text{ }-\dfrac{1}{2x-1}=y\\ \Rightarrow\left(1\right)=1+y+y^2\\ =y^2+y+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\\ Do\text{ }\left(y+\dfrac{1}{2}\right)^2\ge0\forall x\\ \Rightarrow\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\\ Dấu\text{ }"="\text{ }xảy\text{ }ra\text{ }khi:\\ \left(y+\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow y+\dfrac{1}{2}=0\\ \Leftrightarrow y=-\dfrac{1}{2}\\ \Leftrightarrow-\dfrac{1}{2x-1}=-\dfrac{1}{2}\\ \Leftrightarrow2x-1=2\\ \Leftrightarrow2x=3\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\text{ }GTNN\text{ }của\text{ }biểu\text{ }thức\text{ }là\text{ }\dfrac{3}{4}\text{ }khi\text{ }x=\dfrac{3}{2}\)