Question

A)9x-2(x+3)=3x+1

B)\(\dfrac{5}{x-3}\)-\(\dfrac{4}{x+3}\)=\(\dfrac{x-13}{x^2-9}\)

C)7x+8≥3(x-1)

 

Giúp mình với

Answer 1

A.

^{_{\(\Leftrightarrow\)}} 9x - 2x - 6 = 3x + 1

\(\Leftrightarrow\) 4x = 7

\(\Leftrightarrow\) x = \(\dfrac{7}{4}\)

B.

\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-13}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\) 5x + 15 - 4x +12 = x - 13

\(\Leftrightarrow\) 0x = -40 ( phương trình vô nghiệm)

C.

\(\Leftrightarrow\) 7x + 8 \(\ge\) 3x -3

\(\Leftrightarrow\) 4x \(\ge\) - 11

\(\Leftrightarrow\)\(x\ge\dfrac{-11}{4}\)