Cho x,y,z là các sô dương.Chứng minh rằng x/2x+y+z+y/2y+z+x+z/2z+x+y<=3/4

a)\(\left(x^2+1\right)\left(x^2-4x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2-4x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-1\left(vn\right)\\\left(x-2\right)^2=0\end{cases}\Rightarrow}x=2}\)

b)\(\left(3x-2\right)\left(\frac{2x+6}{7}-\frac{4x-3}{5}\right)=0\\ \Rightarrow\left(3x-2\right)\left(\frac{10x+30-28x+21}{35}\right)=0\)

\(\Rightarrow\left(3x-2\right)\left(\frac{-18x+51}{35}\right)=0\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{17}{6}\end{cases}}\)

c)\(\left(3,3-11x\right)\left(\frac{21x+6+10-30x}{15}\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{16}{9}\end{cases}}\)

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

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

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

a)\(ĐKXĐ:x\ne1\)

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

b)\(\frac{\left(x-1\right)^3}{x^3-1}-\frac{1-2x^2+4x}{x^3-1}-\frac{x^2+x+1}{x^3-1}=0\)

\(\Rightarrow\left(x-1\right)^3-\left(1-2x^2+4x\right)-\left(x^2+x+1\right)=0\)

\(\Leftrightarrow x^3-3x^2+27x-1-1+2x^2-4x-x^2-x-1=0\)

\(\Leftrightarrow x^3-2x^2+22x-3=0\)

ĐẾN ĐÂY THÌ BÍ RỒI T_T

\(\Leftrightarrow\left(x^2-6x+9\right)^2-1-15\left(x^2-6x+10\right)=0\)

\(\Leftrightarrow\left(x^2-6x+8\right)\left(x^2-6x+10\right)-15\left(x^2-6x+10\right)=0\)

\(\Leftrightarrow\left(x^2-6x+10\right)\left(x^2-6x-7\right)=0\)

\(\Leftrightarrow\left(x^2-6x+10\right)\left(x^2+x-7x-7\right)=0\)

\(\Leftrightarrow\left(x^2-6x+10\right)\left(x+1\right)\left(x-7\right)=0\)

\(Vi:x^2-6x+10=0\Leftrightarrow\left(x-3\right)^2+1>0,\forall x\)

\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

\(hay:x-7=0\Leftrightarrow x=7\)

\(V...\)

\(:)\)