Question

Giải các pt sau:

a) \(\sqrt{x^2-7x+10}-3x=-1\)

b) \(\left|3x-5\right|=2x^2+x-3\)

Answer 1

ĐKXĐ:2<x<5

\(\sqrt{x^2-7x+10}=3x-1(x\ge \frac{1}{3}) \)

<=>\(x^2-7x+10=9x^2-6x+1 \)

<=>\(8x^2+x-9=0\)

<=>\((x-1)(8x+9)=0 \)

<=>x=1(x\(\ge\frac{1}{3} \))

Answer 2

b,

\(\left|3x-5\right|=2x^2+x-3\)(1)

Nếu \(3x-5\ge0\Leftrightarrow x\ge\dfrac{5}{3}\)

Thì pt (1) <=> \(3x-5=2x^2+x-3\)

\(\Leftrightarrow2x^2-2x+2=0\)

\(\Leftrightarrow PT\) vô \(n_o\) (vì \(\Delta< 0\))

Nếu 3x - 5 <0 \(\Leftrightarrow x< \dfrac{5}{3}\)

Thì pt (1) <=> \(5-3x=2x^2+x-3\)

\(\Leftrightarrow2x^2+4x-8=0\)

\(\Leftrightarrow x^2+2x-4=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1+\sqrt{5}\\x=-1-\sqrt{5}\end{matrix}\right.\)(t/m)

Vậy....