\(\frac{3x-2}{x+7}=\frac{6x+1}{2x-3}\) (Đkxđ: \(x\ne-7;x\ne\frac{3}{2}\))
\(\Rightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)
\(\Leftrightarrow6x^2-9x-4x-6x^2-42x-x=7-6\)
\(\Leftrightarrow-56x=1\)
\(\Leftrightarrow x=-\frac{1}{56}\) (t/m đkxđ)
Vậy \(S=\left\{-\frac{1}{56}\right\}\)
ĐKXĐ: x khác -7 và 3/2
Từ đề bài <=> (3x-2)(2x-3) = (6x+1)(x+7)
<=> 6x^2-4x-9x+6 = 6x^2+x+42x+7
<=> -13x+6 = 43x+7
<=> 6-7 = 43x+13x
<=> 56x = -1
<=> x = -1/56 (TM)
Vậy ...
ĐKXĐ:x khác -7;x khác 1,5
=>(3x-2)(2x-3)=(6x+1)(x+7)
=>6x2-4x-9x+6=6x2+x+42x+7
=>6x2-13x+6=6x2+43x+7
=>6x2-6x2-13x-43x+6-7=0
=>-56x-1=0
=>-56x=1
=>x=\(\frac{-1}{56}\)
ĐKXĐ: \(x\)\(\ne\)-7; \(x\)\(\ne\)-\(\frac{3}{2}\)
Ta có: \(\frac{3x-2}{x+7}\)= \(\frac{6x+1}{2x-3}\)
\(\Rightarrow\)\(\frac{\left(3x-2\right)\left(2x-3\right)}{\left(x+7\right)\left(2x-3\right)}\)= \(\frac{\left(6x+1\right)\left(x+7\right)}{\left(2x-3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\left(3x-2\right)\left(2x-3\right)\)= \(\left(6x+1\right)\left(x+7\right)\)
\(\Leftrightarrow\)\(6x^2-13x+6\)= \(6x^2+43x+7\)
\(\Leftrightarrow\)\(6x^2-6x^2-13x-43x=-6+7\)
\(\Leftrightarrow\)\(-56x=1\)\(\Rightarrow\)\(x=-\frac{1}{56}\) (nhận)
Vậy PT có tập nghiệm S= \(\left\{-\frac{1}{56}\right\}\)
