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Đặt a + 2b = x
Ta có:
\(A=15x^2-3x\left(x+19\right)+12\left(1-x\right)\)
\(=15x^2-3x^2-57x+12x-12x^2\)
\(=-45x\)
\(=-45\left(a+2b\right)\)
Đặt a + 2b = x
Ta có:
\(A=15x^2-3x=\left(x+19\right)+12\left(1-x\right)\)
\(=15x^2-3x^2-57x+12x-12x^2\)
\(=-45x\)
\(=-45\left(a+2b\right)\)
a) (x - 1)(x + 1)(x2 + 1)(x4 + 1)(x8 + 1)
= (x2 - 1)(x2 + 1)(x4 + 1)(x8 + 1)
= (x4 - 1)(x4 + 1)(x8 + 1)
= (x8 - 1)(x8 + 1)
= x16 - 1
b) (a2 - 2b)(a2 + 2b)(a4 + 4b2)(a8 + 16b4)
= (a4 - 4b2)(a4 + 4b2)(a8 + 16b4)
= (a8 - 16b4)(a8 + 16b4)
= a16 - 256b8
\(1.a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)\)
\(=\left(b-c\right)\left(a^2-b^2\right)-\left(a-b\right)\left(b^2-c^2\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(b+c\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
\(2.\dfrac{a^2-b^2+4b-4}{2a-2b+4}\)
\(=\dfrac{a^2-\left(b-2\right)^2}{2\left(a-b+2\right)}\)
\(=\dfrac{\left(a-b+2\right)\left(a+b-2\right)}{2\left(a-b+2\right)}\)
\(=\dfrac{a+b-2}{2}\)
A = x(x - 3)(x + 3) - (x + 1)3
=> A = x(x2 - 9) - x3 - 3x2 - 3x - 1
=> A = x3 - 9x - x3 - 3x2 - 3x - 1
=> A = -3x2 - 12x - 1
B = (a - 2b)2 - (a + 2b)2
=> B = (a - 2b + a + 2b)(a - 2b - a - 2b)
=> B = 2a(-2a - 4b)
=> B = -4a2 - 8ab
a)(a+2b-3c-d)(a+2b+3c+d)=[(a+2b)-(3c+d)][(a+2b)+(3c-d)]
=(a+2b)2-(3c-d)2=a2+4ab+4b2-9c2+6cd-d2
câu b tương tự
Lời giải:
$(a+2b-c)(a+2b+c)-(a^2+4b^2-c^2)=(a+2b)^2-c^2-a^2-4b^2+c^2$
$=(a+2b)^2-a^2-4b^2$
$=a^2+4ab+4b^2-a^2-4b^2=4ab$
\(=\left[\left(a+2b\right)^2-c^2\right]-\left(a^2+4b^2-c^2\right)\)
\(=a^2+4ab+4b^2-c^2-a^2-4b^2+c^2\)
\(=4ab\)