Bài 1: tìm x
a, 4x.(x-5)-5x(x+1)+x^2=10
b,(x+3).(x-4)-(x-1).(x+1)=10
Bài 1: tìm x a, 4x.(x-5)-5x(x+1)+x^2=10 b,(x+3).(x-4)-(x-1).(x+1)=10
By Raelynn
By Raelynn
Bài 1: tìm x
a, 4x.(x-5)-5x(x+1)+x^2=10
b,(x+3).(x-4)-(x-1).(x+1)=10
a, 4x.(x-5)-5x(x+1)+x^2=10
4x.x – 4x.5 – 5x.x – 5x.1 + x² = 10
4x² – 20x – 5x² -5x + x² = 10
( 4x² + x²- 5x² ) – ( 20x + 5x ) =10
0 – 25x =10
-25x = 10
x= 10 : (-25)
x= -2/5
b) (x+3).(x-4)-(x-1).(x+1)=10
x.x – x.4 + 3.x – 3.4 -( x.x + x.1 – 1.x – 1.1 ) = 10
x.x – x.4 + 3.x – 3.4 -x.x – x.1 + 1.x + 1.1 = 10
x² – 4x + 3x – 12 – x² – x + x + 1 = 10
( x² – x² ) + ( 3x + x -x -4x ) – ( 12 – 1 ) = 10
-x -11 = 10
-x = 10 + 11
-x = 21
x=-21
Đáp án:
Giải thích các bước giải:
Bài 1:
a)
$ 4x(x – 5) – 5x(x + 1) + x^2 = 10$
$⇒ 4x^2 – 20x – 5x^2 – 5x + x^2 = 10$
$⇒ (4x^2 – 5x^2 + x^2) – (20x + 5x) = 10$
$⇒ -25x = 10$
$⇒ x = \frac{-2}{5}$
Vậy $ x = \frac{-2}{5}$
b)
$ (x + 3)(x – 4) – (x – 1)(x + 1) = 10$
$⇒ x^2 + 3x – 4x – 12 – (x^2 – x + x – 1) = 10$
$⇒ (x^2 – x^2) + (3x – 4x + x – x) + (-12 + 1) = 10$
$⇒ -x – 11 = 10$
$⇒ -x = 21$
$⇒ x = -21$
Vậy $x = -21$