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證明下列各方程是恆等式..列埋式拉吾該=] 1. x(y-z)+y(z-y)=(x-y)(y-z) 2. (x+y)^2-(x-y)^2=4xy 求下列各恆等式中A.B.C的值 3. Cx^2+A(x+1)-B恆等2(x+3)+3(x-2) 4. Ax(x-1)+B(x+1)(x-1)+C(x-1)+C(x+1)x恆等x^2+4 ^代表次方
最佳解答:
1. LHS :xy-xz+yz-y^2 RHS: xy-xz-y^2+yz LHS=RHS, 方程是恆等 2. LHS: x^2+2xy+y^2-X^2+2xy-y^2=4xy=RHS , 方程是恆等 3. LHS: Cx^2+Ax+A-B RHS: 2x+6+3x-6=5x so C=0 A=5 B=5 4.LHS:Ax^2-Ax+Bx^2-B+Cx-C+Cx^2+Cx= (A+C)x^2+(-A+C)x-(B+C) =x^2+4 (A+C)=1 (-A+C)=0 A=C (B+C)=-4 A=C=1/2 B=7/2 2007-01-11 20:02:43 補充： 第3題的後半部LHS: Cx^2 Ax (A-B)RHS: 0x^2 5x 0C=0A=5A-B=0B=5 2007-01-11 22:11:50 補充： 樓下的joshsin1996做乜要抄我的答案..........
其他解答:
1. x(y - z)+ y(z - y) = x(y - z) - y(y - z) = (x - y)(y - z) (By factorization, grouping) 2. (x + y)^2 - (x - y)^2 = [(x + y) - (x - y)] * [(x + y) - (x - y)] (By factorization, using identity a^2 - b^2 = (a - b)(a + b)) = (2x)(2y) = 4xy 3. Since RHS is a linear polynomial, so the coefficient of x^2 should be 0, so C = 0 Next, 2(x + 3) + 3(x - 2) = 5x = 5x + 5 - 5 = 5(x + 1) - 5 which gives (by conparing LHS) A = 5, B = -5 4. Expanding LHS will be very tedious!! instead, we can put some particular values into the identity put x = 1, we have C(1 + 1) * 1 = 1^2+4 gives C = 5/2 put x = 0, we have B(0 - 1)(0 + 1) + C(0 - 1) = 4, i.e. B + C = -4, B = - 13/4 put x = -1, we have A(-1)(-1 - 1) + C(-1 - 1) = (-1)^2 + 4, i.e. 2A - 2C = 5, A = 5
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恆等式!help!發問:
證明下列各方程是恆等式..列埋式拉吾該=] 1. x(y-z)+y(z-y)=(x-y)(y-z) 2. (x+y)^2-(x-y)^2=4xy 求下列各恆等式中A.B.C的值 3. Cx^2+A(x+1)-B恆等2(x+3)+3(x-2) 4. Ax(x-1)+B(x+1)(x-1)+C(x-1)+C(x+1)x恆等x^2+4 ^代表次方
最佳解答:
1. LHS :xy-xz+yz-y^2 RHS: xy-xz-y^2+yz LHS=RHS, 方程是恆等 2. LHS: x^2+2xy+y^2-X^2+2xy-y^2=4xy=RHS , 方程是恆等 3. LHS: Cx^2+Ax+A-B RHS: 2x+6+3x-6=5x so C=0 A=5 B=5 4.LHS:Ax^2-Ax+Bx^2-B+Cx-C+Cx^2+Cx= (A+C)x^2+(-A+C)x-(B+C) =x^2+4 (A+C)=1 (-A+C)=0 A=C (B+C)=-4 A=C=1/2 B=7/2 2007-01-11 20:02:43 補充： 第3題的後半部LHS: Cx^2 Ax (A-B)RHS: 0x^2 5x 0C=0A=5A-B=0B=5 2007-01-11 22:11:50 補充： 樓下的joshsin1996做乜要抄我的答案..........
其他解答:
1. x(y - z)+ y(z - y) = x(y - z) - y(y - z) = (x - y)(y - z) (By factorization, grouping) 2. (x + y)^2 - (x - y)^2 = [(x + y) - (x - y)] * [(x + y) - (x - y)] (By factorization, using identity a^2 - b^2 = (a - b)(a + b)) = (2x)(2y) = 4xy 3. Since RHS is a linear polynomial, so the coefficient of x^2 should be 0, so C = 0 Next, 2(x + 3) + 3(x - 2) = 5x = 5x + 5 - 5 = 5(x + 1) - 5 which gives (by conparing LHS) A = 5, B = -5 4. Expanding LHS will be very tedious!! instead, we can put some particular values into the identity put x = 1, we have C(1 + 1) * 1 = 1^2+4 gives C = 5/2 put x = 0, we have B(0 - 1)(0 + 1) + C(0 - 1) = 4, i.e. B + C = -4, B = - 13/4 put x = -1, we have A(-1)(-1 - 1) + C(-1 - 1) = (-1)^2 + 4, i.e. 2A - 2C = 5, A = 5
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