Question

a, 4(5x-3)-3(2x+1)=9

b, |x-9|=2x+5

c, \(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{x^2-9}\)

Answer 1

a,\(4\left(5x-3\right)-3\left(2x+1\right)=9\)

\(\Leftrightarrow20-12-6x-3=9\)

\(\Leftrightarrow14x=9+12+3\)

\(\Leftrightarrow14x=24\)

\(\Leftrightarrow x=\dfrac{12}{7}\)

Vậy.....

Answer 2

b, |x-9|=2x+5

* Với x ≥ 9 thì |x – 9| = x – 9 ta có PT: x – 9 = 2x + 5

\(\Leftrightarrow\) x = - 14 ( loại)

*Với x < 9 thì |x – 9| = 9 – x ta có PT: 9 – x = 2x + 5

\(\Leftrightarrow\) x = 4/3(thỏa mãn)

=>Vậy phương trình có tập nghiệm là S = \(\left\{\dfrac{4}{3}\right\}\)

Answer 3

c, \(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{x^2+9}\)

ĐKXĐ: x ≠ ±3

\(\Leftrightarrow\)\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\)\(\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x-3\right)}+\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+5}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow\)\(2\left(x+3\right)+3\left(x-3\right)=3x+5\)

\(\Leftrightarrow\)\(2x+6+3x-9=3x+5\)

\(\Leftrightarrow\)\(2x+3x-3x=5-6+9\)

\(\Leftrightarrow\)\(2x=8\)

\(\Leftrightarrow x=4\)(thỏa mãn ĐKXĐ)

Vậy.....