Tìm x bằng ba đẳng thức ở đầu, của bảy hằng đảng thức
Bài 20
a, (-x+5)(x-2)+(x-7)(x+7)=(3x+1)^2-(3x-2)(3x+2)
b, (5x-1)(x+1)-2(x-3)^2=(x+2)(3x-1)-(x+4)^2+(X^2-x)
c, (4x-1)^2-(3x+2)(3x-2)=(7x-1).(x+2)+(2x+1)^2-(4x^2+7)
d, (2x+3)^2-(5x-4)(5x-4)=(x+5)^2-(3x-1)(7x+2)-(x^2-x+1)
e, (1-3x^2)-(x-2)(9x+1)=(3x-4).(3x+4)-9(x+3)^2
g, (3x+4)(3x-4)-(2x+5)^2=(x-5)^2+(2x+1)^2-(x^2-2x)+(x-1)^2
h, (x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x^2
i,...
Đọc tiếp
Tìm x bằng ba đẳng thức ở đầu, của bảy hằng đảng thức
Bài 20
a, (-x+5)(x-2)+(x-7)(x+7)=(3x+1)^2-(3x-2)(3x+2)
b, (5x-1)(x+1)-2(x-3)^2=(x+2)(3x-1)-(x+4)^2+(X^2-x)
c, (4x-1)^2-(3x+2)(3x-2)=(7x-1).(x+2)+(2x+1)^2-(4x^2+7)
d, (2x+3)^2-(5x-4)(5x-4)=(x+5)^2-(3x-1)(7x+2)-(x^2-x+1)
e, (1-3x^2)-(x-2)(9x+1)=(3x-4).(3x+4)-9(x+3)^2
g, (3x+4)(3x-4)-(2x+5)^2=(x-5)^2+(2x+1)^2-(x^2-2x)+(x-1)^2
h, (x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x^2
i, -5(x+3)^2+(x-1)(x+1)+(2x-3)^2=(5x-2)^2-5x(5x+3)
xy – 1 với đa thức x3 – 2x – 6.
g) (3x + 4)(3x - 4) - (2x + 5)² = (x - 5)² + (2x + 1)² - (x² - 2x) + (x - 1)²
9x² - 16 - 4x² - 20x - 25 = x² - 10x + 25 + 4x² + 4x + 1 - x² + 2x + x² - 2x + 1
5x² - 20x - 41 = 5x² - 6x + 27
5x² - 5x² - 20x + 6x = 27 + 41
-14x = 68
i) -5(x + 3)² + (x - 1)(x + 1) + (2x - 3)² = (5x - 2)² - 5x(5x + 3)
-5(x² + 6x + 9) + x² - 1 + 4x² - 12x + 9 = 25x² - 20x + 4 - 25x² - 15x
-5x² - 30x - 45 + x² - 1 + 4x² - 12x + 9 = -35x + 4
-42x - 37 = -35x + 4
-42x + 35x = 4 + 37
-7x = 41
a) (-x + 5)(x - 2) + (x - 7)(x + 7) = (3x + 1)² - (3x - 2)(3x + 2)
-x² + 2x + 5x - 10 + x² - 49 = 9x² + 6x + 1 - 9x² + 4
7x - 59 = 6x + 5
7x - 6x = 5 + 59
x = 64
b) (5x - 1)(x + 1) - 2(x - 3)² = (x + 2)(3x - 1) - (x + 4)² + (x² - x)
5x² + 5x - x - 1 - 2(x² - 6x + 9) = 3x² - x + 6x - 2 - x² - 8x - 16 + x² - x
5x² + 4x - 1 - 2x² + 12x - 18 = 3x² - 4x - 18
3x² + 16x - 19 = 3x² - 4x - 18
3x² + 16x - 3x² + 4x = -18 + 19
20x = 1