lê ngoc long

a,(x-3)ngũ2+2x-6=0

b,x+3/x-3+48/9-x ngũ2=x-3/x+3

 

HT.Phong (9A5)
7 tháng 8 2023 lúc 15:13

a) \(\left(x-3\right)^2+2x-6=0\)

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

\(\Leftrightarrow x^2-4x+3=0\)

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

\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

b) \(\dfrac{x+3}{x-3}+\dfrac{48}{9-x^2}=\dfrac{x-3}{x+3}\) (ĐKXĐ: \(x\ne\pm3\))

\(\Leftrightarrow\dfrac{x+3}{x-3}-\dfrac{48}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x+3}\)

\(\Leftrightarrow\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}-\dfrac{48}{\left(x+3\right)\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}\)

\(\Leftrightarrow x^2+6x+9-48=x^2-6x+9\)

\(\Leftrightarrow x^2-x^2+6x+6x+9-9-48=0\)

\(\Leftrightarrow12x-48=0\)

\(\Leftrightarrow12x=48\)

\(\Leftrightarrow x=\dfrac{48}{12}\)

\(\Leftrightarrow x=4\left(tm\right)\)

Nguyễn Lê Phước Thịnh
7 tháng 8 2023 lúc 15:10

a: (x-3)^2+2x-6=0

=>(x-3)^2+2(x-3)=0

=>(x-3)(x-3+2)=0

=>(x-3)(x-1)=0

=>x=3 hoặc x=1

b:

ĐKXĐ: x<>3; x<>-3

 \(\dfrac{x+3}{x-3}+\dfrac{48}{9-x^2}=\dfrac{x-3}{x+3}\)

=>\(\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{48}{\left(x-3\right)\cdot\left(x+3\right)}=\dfrac{\left(x-3\right)^2}{\left(x+3\right)^2}\)

=>(x+3)^2-48=(x-3)^2

=>x^2+6x+9-48=x^2-6x+9

=>6x-39=-6x+9

=>12x=48

=>x=4(nhận)

\(a,\left(x-3\right)^2-2x+6=0\\ \Leftrightarrow x^2-6x+9-2x+6=0\\ \Leftrightarrow x^2-8x+15=0\\ \Leftrightarrow x^2-3x-5x+15=0\\ \Leftrightarrow x\left(x-3\right)-5\left(x-3\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\\ Vậy:S=\left\{5;3\right\}\\ b,\dfrac{x+3}{x-3}+\dfrac{48}{9-x^2}=\dfrac{x-3}{x+3}\left(ĐKXĐ:x\ne\pm3\right)\\ \Leftrightarrow\dfrac{\left(x+3\right)^2-48-\left(x-3\right)^2}{x^2-9}=0\\ \Leftrightarrow\left[\left(x+3-x+3\right)\left(x+3+x-3\right)\right]-48=0\\ \Leftrightarrow6.2x-48=0\\ \Leftrightarrow12x=48\\ \Leftrightarrow x=4\left(TM\right)\\ Vậy:S=\left\{4\right\}\)


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