a, : x+2/x-2 – $\frac{1}{x}$ = 2/x(x – 2) b, : 5 – (x-6) = 4 (3-2x) c, : 2x (x-3) + 5 ( x -3 ) = 0 18/09/2021 Bởi Sadie a, : x+2/x-2 – $\frac{1}{x}$ = 2/x(x – 2) b, : 5 – (x-6) = 4 (3-2x) c, : 2x (x-3) + 5 ( x -3 ) = 0
Đáp án: b) x=1/7c)x=3 x=-5/3 Giải thích các bước giải: b)5-x+6=12-8x -x+8x=12-11 7x=1 x=1/7c)(x-3)(2x+5)=0 =>x-3=0 =>x=3 2x+5=0 x=-5/2 Bình luận
a, : x+2/x-2 – 1/x = 2/x(x – 2) ⇔ x(x+2)/x(x-2) – x-2/x(x-2) = 2/x(x-2) ⇒ x²+2x – x+2 = 2 ⇔ x²+x = 0 ⇔ x(x+1)=0 ⇔ x+1=0 ⇔ x=-1 b, : 5 – (x-6) = 4 (3-2x) ⇔ 5-x+6=12-8x ⇔ 11-x = 12-8x ⇔ -1 = -7x ⇔ x = 1/7 c, : 2x (x-3) + 5 ( x -3 ) = 0 ⇔ (x-3)(2x+5)=0 ⇔ x-3=0 ⇔ x=3 2x+5=0 ⇔ x=-5/2 Bình luận
Đáp án:
b) x=1/7
c)x=3
x=-5/3
Giải thích các bước giải:
b)5-x+6=12-8x
-x+8x=12-11
7x=1
x=1/7
c)(x-3)(2x+5)=0
=>x-3=0 =>x=3
2x+5=0 x=-5/2
a, : x+2/x-2 – 1/x = 2/x(x – 2)
⇔ x(x+2)/x(x-2) – x-2/x(x-2) = 2/x(x-2)
⇒ x²+2x – x+2 = 2
⇔ x²+x = 0
⇔ x(x+1)=0
⇔ x+1=0 ⇔ x=-1
b, : 5 – (x-6) = 4 (3-2x)
⇔ 5-x+6=12-8x
⇔ 11-x = 12-8x
⇔ -1 = -7x
⇔ x = 1/7
c, : 2x (x-3) + 5 ( x -3 ) = 0
⇔ (x-3)(2x+5)=0
⇔ x-3=0 ⇔ x=3
2x+5=0 ⇔ x=-5/2