PT \(\Leftrightarrow\frac{15x-5}{20}-\frac{20\left(2x-1\right)}{20}=\frac{4x-12}{20}+\frac{-2}{20}\)

\(\Rightarrow15x-5-40x+20=4x-12-2\)

\(\Leftrightarrow-25x+15=4x-14\Leftrightarrow-29x=-29\Leftrightarrow x=1\)

Vậy tập nghiệm của pt là S = { 1 } 

câu trả lời là c nha

vậy bạn cho mình biết cách làm đi 

\(P=\frac{a}{ca+4}+\frac{b}{ba+4}+\frac{c}{bc+4}\)

\(=\frac{ab}{abc+4b}+\frac{bc}{abc+4c}+\frac{ca}{abc+4a}\)

\(=\frac{ab}{8+4b}+\frac{bc}{8+4c}+\frac{ca}{8+4a}\)

\(=\frac{1}{4}\left(\frac{ab}{b+2}+\frac{bc}{c+2}+\frac{ca}{a+2}\right)\)

Ta có: \(\frac{ab}{b+2}=\frac{ab}{4}\left(\frac{1}{b+2}\right)\le\frac{ab}{4}\left(\frac{1}{b}+\frac{1}{2}\right)=\frac{1}{4}\left(a+\frac{ab}{2}\right)\)

Tương tự: \(\frac{bc}{c+2}\le\frac{bc}{4}\left(\frac{1}{c}+\frac{1}{2}\right)=\frac{1}{4}\left(b+\frac{bc}{2}\right);\frac{ca}{a+2}\le\frac{ca}{4}\left(\frac{1}{a}+\frac{1}{2}\right)=\frac{1}{4}\left(a+\frac{ac}{2}\right)\)

\(\Rightarrow\frac{ab}{b+2}+\frac{bc}{c+2}+\frac{ca}{a+2}\le\frac{1}{4}\left[\left(a+b+c\right)+\frac{1}{2}\left(ab+bc+ca\right)\right]\)

\(\Rightarrow P\le\frac{1}{16}\left[\left(a+b+c\right)+\frac{1}{2}\left(ab+bc+ca\right)\right]\)

Ta có: \(a\le\frac{a^2+4}{4}\)

\(\Leftrightarrow\left(a-2\right)^2\ge0\)( đúng )

Dấu "=" xảy ra \(\Leftrightarrow a=2\)

Tương tự \(b\le\frac{b^2+4}{4};c\le\frac{b^2+4}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow b=c=2\)

\(\Rightarrow a+b+c\le\frac{a^2+b^2+c^2}{4}+3\)

Mà \(a^2+b^2+c^2\ge3\sqrt[3]{\left(abc\right)^2}=12\)

\(\Rightarrow3\le\frac{a^2+b^2+c^2}{4}\)

\(\Rightarrow a+b+c\le\frac{a^2+b^2+c^2}{2}\)

Lại có BĐT phụ \(\frac{1}{2}\left(ab+bc+ca\right)\le\frac{1}{2}\left(a^2+b^2+c^2\right)\)

\(\Rightarrow P\le\frac{1}{16}\left(a^2+b^2+c^2\right)\left(đpcm\right)\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=2\)

-6m-9<0

<=> -6m < 9

<=> m>\(\frac{-3}{2}\)

\(-6m-9< 0\Leftrightarrow-6m< 9\)

\(\Leftrightarrow m>-\frac{9}{6}=-\frac{3}{2}\)

Vậy tập nghiệm BFT là S = { -3/2 }