ngu

Lời giải:

a.

\(x=\frac{b^2-a^2}{a^2}: \frac{a^3-b^3}{a^4}=\frac{(b-a)(b+a)}{a^2}.\frac{a^4}{(a-b)(a^2+ab+b^2)}=\frac{-a^2(a+b)}{a^2+ab+b^2}\)

b.

\(x=\frac{a-b}{a^3+b^3}: \frac{a^2+b^2-2ab}{a^2+b^2-ab}=\frac{a-b}{(a+b)(a^2-ab+b^2)}: \frac{(a-b)^2}{a^2-ab+b^2}\)

\(x=\frac{a-b}{(a+b)(a^2-ab+b^2)}.\frac{a^2-ab+b^2}{(a-b)^2}=\frac{1}{(a+b)(a-b)}=\frac{1}{a^2-b^2}\)

Để phân số $\frac{a}{b}$ có nghĩa thì $b\neq 0$
a.

 ĐKXĐ: \(\left\{\begin{matrix} 3x-3\neq 0\\ 4-4x\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 3(x-1)\neq 0\\ -4(x-1)\neq 0\end{matrix}\right.\Leftrightarrow x-1\neq 0\Leftrightarrow x\neq 1\)

b.

ĐKXĐ: \(\left\{\begin{matrix} a+3\neq 0\\ 3a^2+14a+15\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+3\neq 0\\ (3a+5)(a+3)\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 3a+5\neq 0\\ a+3\neq 0\end{matrix}\right.\Leftrightarrow a\neq \frac{-5}{3}; a\neq -3\)

c.

ĐKXĐ: \(\left\{\begin{matrix} b^3+1\neq 0\\ b^2-b+1\neq 0\\ b+1\neq 0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix} b^3+1\neq 0\\ (b+1)(b^2-b+1)\neq 0\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} b^3+1\neq 0\\ b^3+1\neq 0\end{matrix}\right.\Leftrightarrow b^3\neq -1\Leftrightarrow b\neq -1\)

e.

ĐKXĐ: \(\left\{\begin{matrix} 4x^2-2xy\neq 0\\ 2y^2-4xy\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 2x(2x-y)\neq 0\\ 2y(y-2x)\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ 2x-y\neq 0\\ y-2x\neq 0\\ y\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ x\neq \frac{y}{2}\\ y\neq 0\\ \end{matrix}\right.\)

h.

ĐKXĐ: \(\left\{\begin{matrix} x-5\neq 0\\ x+5\neq 0\\ 25-x^2\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x-5\neq 0\\\ x+5\neq 0\\ -(x-5)(x+5)\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x-5\neq 0\\ x+5\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 5\\ x\neq -5\end{matrix}\right.\)

Bunhiacopxki:

\(\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)\ge\left(ab+bc+ca\right)^2\)

\(\Rightarrow\dfrac{ab}{a^2+bc+ca}\le\dfrac{ab\left(b^2+bc+ca\right)}{\left(ab+bc+ca\right)^2}\)

Tương tự: \(\dfrac{bc}{b^2+ca+ab}\le\dfrac{bc\left(c^2+ca+ab\right)}{\left(ab+bc+ca\right)^2}\)

\(\dfrac{ca}{c^2+ab+bc}\le\dfrac{ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)

\(\Rightarrow VT\le\dfrac{ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)

Nên ta chỉ cần chứng minh:

\(\dfrac{ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\le\dfrac{a^2+c^2+c^2}{ab+bc+ca}\)

\(\Leftrightarrow ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)\le\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)\)

Nhân phá và rút gọn 2 vế:

\(\Leftrightarrow a^3b+b^3c+c^3a\ge abc\left(a+b+c\right)\)

\(\Leftrightarrow\dfrac{a^3b+b^3c+c^3a}{abc}\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\ge a+b+c\)

Đúng do: \(\dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)

Dấu "=" xảy ra khi \(a=b=c\)

Biểu thức này chỉ có max khi a;b là số thực dương, đề bài thiếu

Bunhiacopxki:

\(\left(a^3+b\right)\left(\dfrac{1}{a}+b\right)\ge\left(a+b\right)^2\)

\(\Rightarrow\dfrac{1}{a^3+b}\le\dfrac{\dfrac{1}{a}+b}{\left(a+b\right)^2}=\dfrac{ab+1}{a\left(a+b\right)^2}\)

Tương tự: \(\dfrac{1}{b^3+a}\le\dfrac{ab+1}{b\left(a+b\right)^2}\)

\(\Rightarrow P\le\left(a+b\right)\left(\dfrac{ab+1}{a\left(a+b\right)^2}+\dfrac{ab+1}{b\left(a+b\right)^2}\right)-\dfrac{1}{ab}\)

\(P\le\left(a+b\right).\dfrac{ab+1}{\left(a+b\right)^2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{1}{ab}=\dfrac{ab+1}{a+b}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{1}{ab}\)

\(P\le\dfrac{ab+1}{a+b}\left(\dfrac{a+b}{ab}\right)-\dfrac{1}{ab}=\dfrac{ab+1}{ab}-\dfrac{1}{ab}=1+\dfrac{1}{ab}-\dfrac{1}{ab}=1\)

Dấu "=" xảy ra khi \(a=b=1\)