\(\frac{x^2+3x+9}{2x+10}.\frac{x+5}{x^3-27}\)

\(=\frac{x^2+3x+9}{2\left(x+5\right)}.\frac{x+5}{\left(x-3\right)\left(x^2+3x+9\right)}\)

\(=\frac{\left(x+5\right)\left(x^2+3x+9\right)}{2\left(x+5\right)\left(x-3\right)\left(x^2+3x+9\right)}\)

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

\(\left(\frac{6x+1}{x^2-6x}+\frac{6x-1}{x^2+6x}\right)\left(\frac{x^2-36}{x^2+1}\right)\)

\(=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right]\left[\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\right]\)

\(=\frac{\left(6x+1\right)\left(x+6\right)+\left(6x-1\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(=\frac{12x^2+12}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(=\frac{12\left(x^2+1\right).\left(x-6\right)\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)\left(x^2+1\right)}\)

\(=\frac{12}{x}\)

để A nhỏ nhất => x²+1 nhỏ nhất và lớn hơn 0 (vì 2>0 và không đổi)

ta có: \(x^2+1\ge1\)

dấu = xảy ra khi x²=0

=> x=0

Vậy Min A=\(\frac{1}{2}\)khi x=0

Gọi thương là \(cx^2+dx+e\)

\(\left(cx^2+dx+e\right)\left(x^2-2x+2\right)=cx^4-2cx^3+2cx^2+dx^3-2dx^2+2dx+ex^2-2ex+2e\)

\(=cx^4+x^3\left(d-2c\right)+x^2\left(2c-2d+e\right)+x\left(2d-2e\right)+2e\)

Đồng nhất hệ số

\(\hept{\begin{cases}c=1;d-2c=1\Leftrightarrow d=3\\2d-2e=4\Leftrightarrow e=1;b=2e\Leftrightarrow b=2\\2c-2d+e=a\Leftrightarrow a=-3\end{cases}}\)

Vậy a=-3;b=2

\(\hept{\begin{cases}xyz=12\\x^3+y^3+z^3=36\end{cases}}\Leftrightarrow x^3+y^3+z^3=3xyz\)

\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)

\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)-3xyz+z^3=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)

\(\Leftrightarrow x=y=z\left(x+y+z>0\right)\)

Thay x=y=z vào r tính thôi bạn

\(5x^2-30x-20y^2+45\)

\(=5\left(x^2-6x-4y^2+9\right)\)

\(=5\left[\left(x-3\right)^2-4y^2\right]\)

\(=5\left(x-3-2y\right)\left(x-3+2y\right)\)

A = \(x^3-y^3-3xy\left(x-y\right)-\left(x^2-2xy+y^2\right)+3xy\left(x-y\right)\)

= \(\left(x-y\right)^3-\left(x-y\right)^2+3xy\left(x-y\right)\)

= \(5^3-5^2+3.\left(-6\right).5\)

= \(125-25-90=10\)

Áp dụng BĐT Svác - xơ.

\(F=\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\)

\(=\frac{a^2}{ba+ca}+\frac{b^2}{cb+db}+\frac{c^2}{dc+ac}+\frac{d^2}{ad+bd}\)

\(\ge\frac{\left(a+b+c+d\right)^2}{ba+ca+bd+db+dc+ac+ad+bd}\)(1)

Xét:  \(\left(a+b+c+d\right)^2-2\left(ba+ca+bd+db+dc+ac+ad+bd\right)\)

\(=a^2+b^2+c^2+d^2-2bd-2ac\)

\(=\left(a-c\right)^2+\left(b-d\right)^2\ge0\)

=> \(\left(a+b+c+d\right)^2\ge2\left(ba+ca+bd+db+dc+ac+ad+bd\right)\)

=> \(\frac{\left(a+b+c+d\right)^2}{ba+ca+bd+db+dc+ac+ad+bd}\ge2\)(2)

Từ ( 1); (2) => \(F\ge2\)

Dấu "=" xảy ra <=> a = b = c = d.