Question

1.Giải phương trình
a)|x+2| = 2x-10
b) |-5x| +1= 3x-9
2.a)Với giá trị nào của x thì \(\frac{x+2}{x-3}\) >0
b)Chứng minh; 9² + 4b² -6a +4b+\(\frac{8}{3}\) >0 với mọ a,b ∈ R

Answer 1

a. * \(\left|x+2\right|=x+2\) nếu \(x+2\ge0\Leftrightarrow x\ge-2\)

\(\left|x+2\right|=-x-2\) nếu \(x+2< 0\Leftrightarrow x< -2\)

* TH1: \(x+2=2x-10\Leftrightarrow x-2x=-10-2\)

\(\Leftrightarrow-x=-12\Leftrightarrow x=12\left(tm\right)\)

TH2: \(-x-2=2x-10\Leftrightarrow-x-2x=-10+2\)

\(\Leftrightarrow-3x=-8\Leftrightarrow x=\frac{8}{3}\left(ktm\right)\)

Vậy, \(S=\left\{12\right\}\)

b. * \(\left|-5x\right|=-5x\) nếu \(-5x\ge0\Leftrightarrow x\le0\)

\(\left|-5x\right|=5x\) nếu \(-5x< 0\Leftrightarrow x>0\)

* TH1: \(-5x+1=3x-9\Leftrightarrow-5x-3x=-9-1\)

\(\Leftrightarrow-8x=-10\Leftrightarrow x=\frac{5}{4}\left(ktm\right)\)

TH2: \(5x+1=3x-9\Leftrightarrow5x-3x=-9-1\)

\(\Leftrightarrow2x=-10\Leftrightarrow x=-5\left(ktm\right)\)

Vậy, \(S=\left\{\varnothing\right\}\)