Câu hỏi của Cỏ dại - Toán lớp 8 - Học toán với OnlineMath

Theo giả thiết,ta có: \(a+b+c=0\Leftrightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\a+c=-b\end{cases}}\)

Ta lại có: \(a+b+c=0\Leftrightarrow\left(a+b+c\right)^3=0\)

\(\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow a^3+b^3+c^3+3\left(-a\right)\left(-b\right)\left(-c\right)=0\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

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

a)\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

A xác định

\(\Leftrightarrow\hept{\begin{cases}x+3\ne0\\x^2+x-6\ne0\\2-x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-3\\\left(x+3\right)\left(x-2\right)\ne0\\x\ne2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)

Vậy A xác định \(\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)

b) \(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{\left(x^2-2x\right)+\left(3x-6\right)}+\frac{1}{2-x}\)

\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{x.\left(x-2\right)+3.\left(x-2\right)}+\frac{1}{2-x}\)

\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}+\frac{1}{2-x}\)

\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{1}{x-2}\)

\(A=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)

\(A=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

\(A=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)

\(A=\frac{\left(x^2+3x\right)-\left(4x+12\right)}{\left(x+3\right)\left(x-2\right)}\)

\(A=\frac{x.\left(x+3\right)-4.\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)

\(A=\frac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}\)

\(A=\frac{x-4}{x-2}\left(x+3\ne0\right)\)

c) \(A=-\frac{3}{4}\)

\(\Leftrightarrow\frac{x-4}{x-2}=-\frac{3}{4}\)

\(\Leftrightarrow4.\left(x-4\right)=-3.\left(x-2\right)\)

\(\Leftrightarrow4x-16=-3x+6\)

\(\Leftrightarrow7x=22\)

\(\Leftrightarrow x=\frac{22}{7}\)

Vậy \(x=\frac{22}{7}\)

Tham khảo nhé~

Ta có:

\(a+b+c=0\)

\(\Leftrightarrow a+b=-c\)

\(\Leftrightarrow\left(a+b\right)^3=\left(-c\right)^3\)

\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3=-c^3\)

\(\Leftrightarrow a^3+b^3+c^3+3ab\left(a+b\right)=0\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

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

                             đpcm

Tham khảo nhé~

\(x^3+2x^2+x+2\)

\(=x.x.x+x.1+2x^2+2.1\)

\(=x\left(x^2+1\right)+2\left(x^2+1\right)\)

\(=\left(x+2\right)\left(x^2+1\right)\)

\(x^3+2x^2+x+2\)

\(=\left(x^3+2x^2\right)+\left(x+2\right)\)

\(=x^2\left(x+2\right)+\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+1\right)\)

\(\left|5-3x\right|=3x-5\)

\(\Rightarrow5-3x\le0\)

\(\Leftrightarrow5\le3x\)

\(\Leftrightarrow x\ge\frac{5}{3}\)

Vậy \(x\ge\frac{5}{3}\)