Tải Qanda về

\(\frac{x-1}{x^2+5x};\frac{x+1}{x^2-25}\)

ta có: \(\frac{x-1}{x^2+5x}=\frac{x-1}{x\left(x+5\right)}\)

          \(\frac{x+1}{x^2-25}=\frac{x+1}{\left(x-5\right)\left(x+5\right)}\)

MTC: x(x-5).(x+5)

\(\frac{x-1}{x\left(x+5\right)_{\left(x-5\right)}}=\frac{\left(x-1\right)\left(x-5\right)}{x\left(x+5\right)\left(x-5\right)}=\frac{x^2-5x-x+5}{x\left(x+5\right)\left(x-5\right)}=\frac{x^2-4x+5}{x\left(x+5\right)\left(x-5\right)}\)

\(\frac{x+1}{\left(x-5\right)\left(x+5\right)_{\left(x\right)}}=\frac{x^2+x}{x\left(x-5\right)\left(x+5\right)}\)

Vậy ....

Để \(A\inℤ\)thì \(7⋮x^2-x+1\)(1)

Vì \(x^2-x+1=x^2-2x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)

Mà \(\left(x-\frac{1}{2}\right)^2\ge0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\Leftrightarrow x^2-x+1\ge\frac{3}{4}>0\)(2)

Từ (1) và (2) \(\Rightarrow\) \(x^2-x+1\in\left\{1;7\right\}\)

Trường hợp \(x^2-x+1=1\Leftrightarrow x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Trường hợp \(x^2-x+1=7\Leftrightarrow x^2-x-6=0\Leftrightarrow x^2-3x+2x-6=0\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

Vậy để \(A\inℤ\)thì \(x\in\left\{-2;0;1;3\right\}\)

Answer:

\(a^2+b^2+c^2\ge\frac{1}{3}\)

\(\Rightarrow3.\left(a^2+b^2+c^2\right)\ge1\)

\(\Rightarrow3.\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow3.\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2ab+2bc+2ac\)

\(\Rightarrow3.a^2+3.b^2+3.c^2\ge a^2+b^2+c^2+2ab+2bc+2ac\)

\(\Rightarrow2.a^2+2.b^2+2.c^2\ge2ab+2bc+2ac\)

\(\Rightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2\ge\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (Luôn đúng)

Vậy ta có điều cần phải chứng minh.

À thôi mk ko nx