tìm x biết 1.(x+3)^2-(x+2)(x-2)=4x+17 2.(2x+1)^2-(4x-1)(x-3)-15=0 3.(2x+3)(x-1)+(2x-3)(1-x)=0 14/08/2021 Bởi Melody tìm x biết 1.(x+3)^2-(x+2)(x-2)=4x+17 2.(2x+1)^2-(4x-1)(x-3)-15=0 3.(2x+3)(x-1)+(2x-3)(1-x)=0
Đáp án: Giải thích các bước giải: 1) $ (x + 3)^2 – (x – 2)(x + 2) = 4x + 17$ $=>x^2 + 6x + 9 – (x^2 – 4) – 4x = 17$ $=>(x^2 – x^2) + (6x – 4x) + (9 + 4) = 17$ $=> 2x + 13 = 17$ $=> 2x = 4$ $=> x = 2$ $ Vậy x = 2$ 2) $ (2x + 1)^2 – (4x – 1)(x – 3) – 15 = 0$ $=> 4x^2 + 4x + 1 – (4x^2 – 12x – x + 3) – 15 = 0$ $=> (4x^2 – 4x^2) + (4x + 12x + x) + (1 – 3 – 15) = 0$ $=> 17x – 17 = 0$ $=> 17x = 17$ $=> x = 1$ $ Vậy x = 1$ 3) $ (2x + 3)(x – 1) + (2x – 3)(1 – x) = 0$ $=> (2x + 3)(x – 1) – (2x – 3)(x – 1) = 0$ $=> (2x + 3 – 2x + 3)(x – 1) = 0$ $=> 6(x – 1) = 0$ $=> x – 1 = 0 $ $=> x = 1$ $ Vậy x = 1$ Bình luận
Đáp án:
1. x=2
2. x=1
3. x=1
Giải thích các bước giải:
Đáp án:
Giải thích các bước giải:
1)
$ (x + 3)^2 – (x – 2)(x + 2) = 4x + 17$
$=>x^2 + 6x + 9 – (x^2 – 4) – 4x = 17$
$=>(x^2 – x^2) + (6x – 4x) + (9 + 4) = 17$
$=> 2x + 13 = 17$
$=> 2x = 4$
$=> x = 2$
$ Vậy x = 2$
2)
$ (2x + 1)^2 – (4x – 1)(x – 3) – 15 = 0$
$=> 4x^2 + 4x + 1 – (4x^2 – 12x – x + 3) – 15 = 0$
$=> (4x^2 – 4x^2) + (4x + 12x + x) + (1 – 3 – 15) = 0$
$=> 17x – 17 = 0$
$=> 17x = 17$
$=> x = 1$
$ Vậy x = 1$
3)
$ (2x + 3)(x – 1) + (2x – 3)(1 – x) = 0$
$=> (2x + 3)(x – 1) – (2x – 3)(x – 1) = 0$
$=> (2x + 3 – 2x + 3)(x – 1) = 0$
$=> 6(x – 1) = 0$
$=> x – 1 = 0 $
$=> x = 1$
$ Vậy x = 1$