a, \(A=\left(\frac{4}{2x+1}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)

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

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

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

\(P=3\left(x^2+\frac{7}{3}x-\frac{5}{3}\right)\)

\(P=3\left(x^2+2.x.\frac{7}{6}+\frac{49}{36}-\frac{109}{36}\right)\)

\(P=3\left(x+\frac{7}{6}\right)^2-\frac{109}{12}\)

\(P_{min}=-\frac{109}{12}\Leftrightarrow x==-\frac{7}{6}\)

tại sao lại có -109/12 vậy bạn?

\(a)\)\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)

\(\Leftrightarrow-4x+7=0\)

\(\Leftrightarrow x=\frac{7}{4}\)

Vậy\(x=\frac{7}{4}\)

\(b)\)\(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)

\(\Leftrightarrow4\left(x^2-6x+9\right)-4x^2+1-10=0\)

\(\Leftrightarrow4x^2-24x+36-4x^2+1-10=0\)

\(\Leftrightarrow-24x+27=0\)

\(\Leftrightarrow x=\frac{9}{8}\)

Vậy\(x=\frac{9}{8}\)

\(c)\)\(\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=6\)

\(\Leftrightarrow x^2-8x+16-x^2+4-6=0\)

\(\Leftrightarrow-8x+14=0\)

\(\Leftrightarrow x=\frac{7}{4}\)

Vậy\(x=\frac{7}{4}\)

Linz

\(a.\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(x^2-4x+4-x^2+9-6=0\)

\(-4x=-7\)

\(x=\frac{7}{4}\)

\(b.4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)

\(4\left(x^2-6x+9\right)-4x^2+1=10\)

\(-24x=-27\Leftrightarrow x=\frac{9}{8}\)

à nêu cảm nhận thì mình nêu đc nhưng dùng biện pháp nói quá thì mình ko quen

không cần biện pháp nói quá cũng được