\(P=1+\frac{x+3}{x^2+5x+6}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)

\(P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)

\(P=1+\frac{1}{x+2}:\left(\frac{4x^2.2}{4x^2\left(x-2\right)}-\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{1}{x+2}\right)\)

\(P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right)\)

\(P=1+\frac{1}{x+2}:\left(\frac{2x+4-x-x+2}{\left(x+2\right)\left(x-2\right)}\right)\)

\(P=1+\frac{1}{x+2}:\frac{6}{\left(x+2\right)\left(x-2\right)}=1+\frac{\left(x+2\right)\left(x-2\right)}{6\left(x+2\right)}=1+\frac{x-2}{6}\)

\(=\frac{x+4}{6}.P=0\Leftrightarrow x=-4\)

\(P>0\Leftrightarrow x>-4\)

sai lớp :>>>

\(ĐKXĐ:x\ne0;x\ne\pm2\)

a) \(M=\left[\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right]:\left(x-2+\frac{10-x^2}{x+2}\right)\)

\(\Leftrightarrow M=\left[\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right]:\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\)

\(\Leftrightarrow M=\frac{3x^2-6x\left(x+2\right)+3x\left(x-2\right)}{3x\left(x-2\right)\left(x+2\right)}:\frac{x^2-4+10-x^2}{x+2}\)

\(\Leftrightarrow M=\frac{3x^2-6x^2-12x+3x^2-6x}{3x\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)

\(\Leftrightarrow M=\frac{-18x\left(x+2\right)}{18x\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow M=-\frac{1}{x-2}\)

\(\Leftrightarrow M=\frac{1}{2-x}\)

b) Để M đạt giá trị lớn nhất

\(\Leftrightarrow2-x\)đạt giá trị nhỏ nhất

\(\Leftrightarrow x\)đạt giá trị lớn nhất

Vậy để M đạt giá trị lớn nhất thì x phải đạt giá trị lớn nhất \(\left(x\inℤ\right)\)

玉明, bạn làm sai rồi. Dấu ngoặc vuông là dấu phần nguyên không phải dấu ngoặc thường