B2:GIẢI PT:
A)(X-6)(X^2-4)=0
B)(2X+5)(4X^2-9)=0
C)(X-2)^2.(X-9)=0
D)X^2=2X
E)X^2-2X+1=4
F)(X^2+1)(X-1)=0
G)4X^2+4X+1=0
H)X^2-5X+6
i)2X^2+3X+1=0
J)(2X-3)(X+1)+X(X-2)=3(X+2)^2
B2:GIẢI PT:
A)(X-6)(X^2-4)=0
B)(2X+5)(4X^2-9)=0
C)(X-2)^2.(X-9)=0
D)X^2=2X
E)X^2-2X+1=4
F)(X^2+1)(X-1)=0
G)4X^2+4X+1=0
H)X^2-5X+6
i)2X^2+3X+1=0
J)(2X-3)(X+1)+X(X-2)=3(X+2)^2
B2: GIẢI PT:
$\text{A) (x-6)(x$^2$-4)=0}$
$⇔$\(\left[ \begin{array}{l}x-6=0\\x^2-4=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=6\\x=2; -2\end{array} \right.\)
$\text{Vậy pt có tập n$^o$ S = {6; 2; -2}}$
$\text{B) (2x+5)(4x$^2$ – 9)=0}$
$⇔$\(\left[ \begin{array}{l}2x+5=0\\4x^2-9=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=\frac{5}{2}\\x=\frac{3}{2}; -\frac{3}{2}\end{array} \right.\)
$\text{Vậy pt có tập n$^o$ S = {$\frac{5}{2}$; $\frac{3}{2}$; $\frac{-3}{2}$}}$
$\text{C)(x-2)$^2$(x-9)=0}$
⇔ \(\left[ \begin{array}{l}(x-2)^2=0\\x-9=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x^2-4x-4=0\\x=9\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=2\\x=9\end{array} \right.\)
$\text{Vậy pt có tập n$^o$ S = {2; 9}}$
a)7x+14=0 => 7x=-14 =>x=-2
b)16-8x=0 => 8x=16 =>x=2
c)3x-5=7 => 3x=12 => x=4
d)5(x-6)=0 => 5x-30=0 => 5x=30 => x=6
e)1/3x+2/3=0 => 1/3x=-2/3 =>x=-2
g)8-3x=6 => 3x=2 => x=2/3
bài 4:
5x-6 = 6+2x => 5x-2x= 6+6 => 3x=12 => x=4
b)7-(2x+4)=-(x+4) => 7-2x-4=-x-4 => -2x+x=-4+4-7 => -x=-7 => x=7
c) 4(3x-2)-3(x-4)=7x+10 =>12x-8-3x+12=7x+10 => 12x-3x-7x=10+8-12 => 2x=6 =>x=3
d)-x^2-12x+21=(3-x)(x+11) => -x^2-12x+21=-x^2-8x+33 =>-12x+8x=33-21 => -4x=12 =>x=-3
e)9x+5x^2+1=5x^2-22+13x => 9x-13x=-22-1 => -4x=-23 => x=23/4
f)(3x-1/2)(-2/3x+1)=0
th1: 3x-1/2=0 => 3x=1/2 => x=1/6
th2: -2/3x+1=0 => -2/3x=-1 =>x=3/2
g) (3x-5)(x+3)=0
th1: 3x-5=0 => 3x=5 => x=5/3
th2: x+3=0 => x=-3
h) 7x-2/12 + 2-x/4 = 3x+1/3 =>7x-2 +3(2-x)=4(3x+1) => 7x-2 +6-3x=12x+4 => 7x-3x-12x=4+2-6 =>-8x=0 => x=0