Question

giúp mình câu này với,ghi rõ luôn ạ

Answer 1

cái này cơ bản lắm mà b vẫn chx bt làm luôn?

Answer 2

đkxđ: x khác 3 ; x khác -3 

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

\(\Leftrightarrow\dfrac{..}{..}-\dfrac{...}{...}=-\dfrac{3x-1}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{...}{...}-\dfrac{....}{....}+\dfrac{3x-1}{....}=0\)

\(\Leftrightarrow x^2+6x+9-x^2+6x-9+3x-1=0\)

\(\Leftrightarrow15x-1=0\)

=> x = 1/15 ( thỏa mãn)

Answer 3

\(ĐK:x\ne\pm3\)

\(\Rightarrow\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=-\dfrac{3x-1}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+3\right)-\left(x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-1}{\left(x-3\right)\left(x+3\right)}\)

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

\(\Leftrightarrow x^2+6x+9-x^2+6x-9+3x-1=0\)

\(\Leftrightarrow15x-1=0\)

\(\Leftrightarrow15x=1\)

\(\Leftrightarrow x=\dfrac{1}{15}\)

Answer 4

đkxđ: x khác 3 ; x khác -3 

⇔....−......=−3x−1(x−3)(x+3)⇔....−......=−3x−1(x−3)(x+3)