A, 3x+9=0
3x=-9
x=-3
Vậy tập nghiệm S = {-3}
a. 3x+9=0
<=> 3x= -9
<=> x= -3
Vậy S=\(\left\{-3\right\}\)
b. 5x-5=3x2-3x
<=> 5x-3x2+3x-5=0
<=> 8x-3x2-5=0
<=> 8x-8x2+5x2-5=0
<=> 8x(1-x)+5(x2-1)=0
<=> -8x(x-1)+5(x-1)(x+1)=0
<=> (x-1)(-8x+5x+5)=0
<=> (x-1)(-3x+5)=0
<=>\(\left[{}\begin{matrix}x-1=0\\-3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)
Vậy S=\(\left\{1;\dfrac{5}{3}\right\}\)
c. \(\dfrac{x-5}{x-1}+\dfrac{2}{x-3}=1\)
ĐKXĐ: x\(\ne1\), x\(\ne3\)
Ta có: \(\dfrac{x-5}{x-1}+\dfrac{2}{x-3}=1\)
<=> \(\dfrac{\left(x-5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}+\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}\)
<=> \(\dfrac{\left(x-5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}+\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=0\)
=> x2-3x-5x+15+2x-2-x2+3x+x-3=0
<=> -2x+10=0
<=> -2(x-5)=0
<=> x-5 = 0
<=> x=5 (thỏa mãn ĐKXĐ)
a. 3x + 9 =0
<=> 3x = -9
\(< =>\dfrac{3x}{3}=\dfrac{-9}{3}\)
<=> x= -3
vậy nghiệm của phương trình là x= -3
b.5x - 5 = 3x2 - 3x
<=> 5x -5 - 3x2 + 3x = 0
<=> 5(x - 1) - 3x(x - 1) =0
<=> (x - 1)(5 - 3x)= 0
=> x - 1 =0 hoặc 5 - 3x =0
1) x -1 =0
<=> x= 1
2) 5 - 3x =0
<=> -3x = -5
<=> \(\dfrac{-3x}{-3}=\dfrac{-5}{-3}\)
<=> x= \(\dfrac{5}{3}\)
vậy tập nghiệm của phương trình là S= \(\left\{1;\dfrac{5}{3}\right\}\)
c.\(\dfrac{x-5}{x-1}+\dfrac{2}{x-3}=1\) (đk: x< > 1; x< > 3 )
<=> \(\dfrac{\left(x-5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}+\dfrac{2\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}=\dfrac{\left(x-3\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}\)
<=> (x - 5)(x - 3) + 2(x - 1) = (x - 3)(x - 1)
<=> x2 -3x - 5x + 15 + 2x -2 = x2 -x - 3x +3
<=> x2 - 6x + 13 = x2 - 4x +3
<=> x2 -6x - x2 + 4x = 3 - 13
<=> -2x = -10
<=> \(\dfrac{-2x}{-2}=\dfrac{-10}{-2}\)
<=> x= 5 ( thỏa mãn đk)
vậy nghiệm của phương trình là: x=5
