a, \(2\frac{7}{9}-\frac{12}{13}x=\frac{7}{9}\)

\(\Leftrightarrow\frac{25}{9}-\frac{12}{13}x=\frac{7}{9}\Leftrightarrow\frac{12}{13}x=2\Leftrightarrow x=\frac{13}{6}\)

b, \(\frac{x-12}{4}=\frac{9-3x}{x}\)

\(\Leftrightarrow x^2-12x=36-12x\Leftrightarrow x^2-12x-36+12x=0\)

\(\Leftrightarrow x^2-36=0\Leftrightarrow x^2=36\Leftrightarrow x=\pm6\)

\(x^2-2\left(-3-1\right)x+\left(-3\right)^2-1=0\)

\(\Leftrightarrow x^2-2\left(-4\right)x+9-1=0\)

\(\Leftrightarrow x^2+8x+8=0\)

Ta có : \(\Delta=8^2-4.1.8=84-32=52>0\)

Nên phương trình có 2 nghiệm phân biệt 

\(x_1=\frac{-8-\sqrt{52}}{2};x_2=\frac{-8+\sqrt{52}}{2}\)

Ta chứng minh:\(\left(a+b\right)^2\le2\left(a^2+b^2\right)\Leftrightarrow\left(a-b\right)^2\ge0\) ( luôn đúng )

Khi đó:\(\left(a+b\right)^2\le2\left(a^2+b^2\right)\le16\)

\(\Rightarrow\left(a+b\right)^2\le16\Rightarrow-4\le a+b\le4\Rightarrowđpcm\)

a, \(\frac{x}{3}-\frac{1}{4}=-\frac{5}{6}\Leftrightarrow\frac{x}{3}+\frac{7}{12}=0\Leftrightarrow\frac{4x}{12}+\frac{7}{12}=0\)

Khử mẫu ta đc : \(4x+7=0\Leftrightarrow4x=-7\Leftrightarrow x=-\frac{7}{4}\)

b, \(\frac{x+3}{15}=\frac{1}{3}\Leftrightarrow\frac{x+3}{15}=\frac{5}{15}\)

Khử mẫu ta đc : \(x+3=5\Leftrightarrow x=2\)

\(P=\frac{1}{x\left(x+1\right)}+\frac{1}{y\left(y+1\right)}+\frac{1}{z\left(z+1\right)}\)

\(\ge3\sqrt[3]{\frac{1}{xyz\left(x+1\right)\left(y+1\right)\left(z+1\right)}}\)

Mà theo BĐT AM - GM ta có tiếp:

\(xyz\le\left(\frac{x+y+z}{3}\right)^3=1\)

\(\left(x+1\right)\left(y+1\right)\left(z+1\right)\le\left(\frac{x+y+z+3}{3}\right)^3=8\)

\(\Rightarrow P\le\frac{3}{2}\)

Đẳng thức xảy ra tại x=y=z=1

Vậy..................