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標題:
發問:
factorise and write down all the STEPS and ANSWERS------------------------------------------1. x^2-y^2+2zx+2yz+2y-2z-1 < arrange in powers of x first> ------ (57a)2. x(b^2-c^2)+(c^2-x^2)+c(x^2-b^2) --- (56a)no need to care these little brackets3. x^2(b-c)+b2(c+x)-c^2(x+b) ---- ... 顯示更多 factorise and write down all the STEPS and ANSWERS ------------------------------------------ 1. x^2-y^2+2zx+2yz+2y-2z-1 < arrange in powers of x first> ------ (57a) 2. x(b^2-c^2)+(c^2-x^2)+c(x^2-b^2) --- (56a)no need to care these little brackets 3. x^2(b-c)+b2(c+x)-c^2(x+b) ---- (56a)no need to care thi brackets 4. x^2(y+z)+y^2(z-)+z^2(y-x)-2xyz --- (56a)no need to care this brackets
最佳解答:
1. x^2-y^2+2zx+2yz+2y-2z-1 = x^2 + (-y^2+2y-1) + (2zx+2yz-2z) = x^2 - (y-1)^2 + z(x+y-1) = [x^2 - (y-1)^2] + z(x+y-1) = (x+y-1)(x-y+1) + z(x+y-1) = (x+y-1)(x-y+z+1) 2. x(b^2-c^2)+(c^2-x^2)+c(x^2-b^2) = xb^2 - xc^2 + c^2 - x^2 + cx^2 - cb^2 = b^2(x-c) - (x^2-c^2) + xc(x-c) = b^2(x-c) - (x-c)(x+c) + xc(x-c) = (x-c)(b^2-x-c+xc) 3. x^2(b-c)+b^2(c+x)-c^2(x+b) = x^2(b-c) + cb^2 + xb^2 - xc^2 - bc^2 = x^2(b-c) + x(b^2-c^2) + bc(b-c) = x^2(b-c) + x(b-c)(b+c) +bc(b-c) = (b-c)(x^2+xb+xc+bc) p.s. in 2nd term, there should be b^2 instead of b2 4. x^2(y+z)+y^2(z-x)+z^2(y-x)-2xyz = yx^2 + zx^2 + zy^2 - xy^2 +z^2(y-x) - 2xyz = z^2(y-x) + z(x^2-2xy+y^2) + xy(x-y) = z^2(y-x) + z(y-x)^2 - xy(y-x) = (y-x)[z^2 + z(y-x) - xy] = (y-x)(z-x)(z+y) p.s. in 2nd term, there should be z-x instead of z-
其他解答:
1. (x+y-1)*(x-y+2*z+1) 2. (x-c)*(c*x-x-c+b^2) 3. (mistake in question, b2 should read b^2 -(c-b)*(x+b)*(x+c) 4. (mistake in question, (z-) should read (z-x) (y-x)*(z-x)*(z+y)
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factorization -------------------urgent發問:
factorise and write down all the STEPS and ANSWERS------------------------------------------1. x^2-y^2+2zx+2yz+2y-2z-1 < arrange in powers of x first> ------ (57a)2. x(b^2-c^2)+(c^2-x^2)+c(x^2-b^2) --- (56a)no need to care these little brackets3. x^2(b-c)+b2(c+x)-c^2(x+b) ---- ... 顯示更多 factorise and write down all the STEPS and ANSWERS ------------------------------------------ 1. x^2-y^2+2zx+2yz+2y-2z-1 < arrange in powers of x first> ------ (57a) 2. x(b^2-c^2)+(c^2-x^2)+c(x^2-b^2) --- (56a)no need to care these little brackets 3. x^2(b-c)+b2(c+x)-c^2(x+b) ---- (56a)no need to care thi brackets 4. x^2(y+z)+y^2(z-)+z^2(y-x)-2xyz --- (56a)no need to care this brackets
最佳解答:
1. x^2-y^2+2zx+2yz+2y-2z-1 = x^2 + (-y^2+2y-1) + (2zx+2yz-2z) = x^2 - (y-1)^2 + z(x+y-1) = [x^2 - (y-1)^2] + z(x+y-1) = (x+y-1)(x-y+1) + z(x+y-1) = (x+y-1)(x-y+z+1) 2. x(b^2-c^2)+(c^2-x^2)+c(x^2-b^2) = xb^2 - xc^2 + c^2 - x^2 + cx^2 - cb^2 = b^2(x-c) - (x^2-c^2) + xc(x-c) = b^2(x-c) - (x-c)(x+c) + xc(x-c) = (x-c)(b^2-x-c+xc) 3. x^2(b-c)+b^2(c+x)-c^2(x+b) = x^2(b-c) + cb^2 + xb^2 - xc^2 - bc^2 = x^2(b-c) + x(b^2-c^2) + bc(b-c) = x^2(b-c) + x(b-c)(b+c) +bc(b-c) = (b-c)(x^2+xb+xc+bc) p.s. in 2nd term, there should be b^2 instead of b2 4. x^2(y+z)+y^2(z-x)+z^2(y-x)-2xyz = yx^2 + zx^2 + zy^2 - xy^2 +z^2(y-x) - 2xyz = z^2(y-x) + z(x^2-2xy+y^2) + xy(x-y) = z^2(y-x) + z(y-x)^2 - xy(y-x) = (y-x)[z^2 + z(y-x) - xy] = (y-x)(z-x)(z+y) p.s. in 2nd term, there should be z-x instead of z-
其他解答:
1. (x+y-1)*(x-y+2*z+1) 2. (x-c)*(c*x-x-c+b^2) 3. (mistake in question, b2 should read b^2 -(c-b)*(x+b)*(x+c) 4. (mistake in question, (z-) should read (z-x) (y-x)*(z-x)*(z+y)
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