\(a,4x-6< 7x-12\)

\(\Leftrightarrow6< 3x\Leftrightarrow x>2\)

\(b,\frac{3x-7}{4}\ge2-\frac{x+5}{3}\)

\(\Leftrightarrow3\left(3x-7\right)\ge24-4\left(x+5\right)\)

\(\Leftrightarrow13x\ge25\Leftrightarrow x\ge\frac{25}{13}\)

\(c,\frac{3x-8}{-7}\ge1-\frac{x+2}{-3}\)

\(\Leftrightarrow-3\left(3x-8\right)\ge21+7\left(x+2\right)\)

\(\Leftrightarrow-16x\ge11\)

\(\Leftrightarrow x\le-\frac{11}{16}\)

\(d,-12-8x>3+2x-\left(5-7x\right)\)

\(\Leftrightarrow14>17x\Leftrightarrow x< \frac{14}{17}\)

\(e,-1+\frac{x-1}{-3}\le\frac{x+2}{-9}\)

\(\Leftrightarrow-9-3\left(x-1\right)\le-\left(x+2\right)\)

\(\Leftrightarrow-2x\le4\Leftrightarrow x\ge-2\)

nhiều quá bạn ơi , bạn k biết câu nào mình giải zúp cho 

hết luôn đó bạn Ngọc Vi ... nhưng bạn giúp được câu nào thì mình cảm ơn

ừm , vậy thì chờ mk xíu

a/ \(\left\{{}\begin{matrix}2x>-1\\4x\ge3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x>-\frac{1}{2}\\x\ge\frac{3}{4}\end{matrix}\right.\) \(\Rightarrow x\ge\frac{3}{4}\)

b/ \(x-1-\frac{1}{x-1}\ge0\)

\(\Leftrightarrow\frac{\left(x-1\right)^2-1}{x-1}\ge0\)

\(\Leftrightarrow\frac{x\left(x-2\right)}{x-1}\ge0\Rightarrow\left[{}\begin{matrix}x\ge2\\0\le x< 1\end{matrix}\right.\)

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

Tương tự \(y^2-1=\frac{1}{4}\left(b-\frac{1}{b}\right)^2\)

\(P=\frac{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)-\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)+\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}\)

\(=\frac{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}-ab+\frac{a}{b}+\frac{b}{a}-\frac{1}{ab}}{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}+ab-\frac{a}{b}-\frac{b}{a}+\frac{1}{ab}}=\frac{\frac{a}{b}+\frac{b}{a}}{ab+\frac{1}{ab}}=\frac{a^2+b^2}{a^2b^2+1}\)

Ta có : \(\frac{3x-1}{x+2}\ge1\)

=> \(3x-1\ge x-2\)

=> \(3x-x\ge-2+1\)

=> \(x\ge-\frac{1}{2}\)