標題:
急急急!中4quadrativ formula 聽日交!
發問:
1.using quadrativ formula, solve the quadrativ formula (1-x)(4+x)+9x-2=02.if the equation(x+1)(x-4)+kx-4k=0 has a repeated root,fing the possible values of k.3.suppose f(x)=x/(1+x0,find f(1/2)f(2)4.suppose f(x)=x^2-4x,if f(k-2)=f(k),find the value of k.5.if g(x)=5x^2,express g(x+2)-g(x-2) in terms... 顯示更多 1.using quadrativ formula, solve the quadrativ formula (1-x)(4+x)+9x-2=0 2.if the equation(x+1)(x-4)+kx-4k=0 has a repeated root,fing the possible values of k. 3.suppose f(x)=x/(1+x0,find f(1/2)f(2) 4.suppose f(x)=x^2-4x,if f(k-2)=f(k),find the value of k. 5.if g(x)=5x^2,express g(x+2)-g(x-2) in terms of x. 6,consider f(x)=6K-(k+2)x such that f(5)=-5. a) find the value of k b) hence find the value of x such that f(x)=86. 7.find the values of x such that (4x-1)^2-3(4x-1)-4=-4 8.if f(x)=(2^x)-(k*2^-x and 2f(1)=3,find a)the value of k b)the value of x such that f(x)=63/8 9.solve the equation 2^(2x+3) - 4^x=35
1. (1-x)(4+x)+9x-2=0 4 + x - 4x - x^2 + 9x - 2 = 0 -x^2 + 6x + 2 = 0 x^2 - 6x - 2 = 0 x = [6 +/- √(36 + 8)]/2 x = 3 + √11 or 3 - √11 2. (x + 1)(x - 4) + kx - 4k = 0 (x + 1)(x - 4) + k(x - 4) = 0 (x - 4)(x + 1 + k) = 0 x = 4 or x = -1 - k For repeated roots, 4 = -1 - k k = -5 3. f(x)=x/(1+x) f(1/2) = (1/2) / (3/2) = 1/3 f(2) = 2/3 f(1/2)f(2) = 1/3 * 2/3 = 2/9 4. f(x) = x^2 - 4x f(k-2) = (k-2)^2 - 4(k-2) = k^2 - 4k + 4 - 4k + 8 = k^2 - 8k + 12 f(k) = k^2 - 4k f(k-2) = f(k) => k^2 - 8k + 12 = k^2 - 4k 4k = 12 k = 3 5. g(x)=5x^2 g(x + 2) - g(x - 2) = 5(x+2)^2 - 5(x-2)^2 = 5[(x+2) + (x-2)][(x+2) - (x-2)] = 5(2x)(4) = 40x 6. f(x)=6k-(k+2)x a) f(5) = 6k-(k+2)(5) = -5 6k - 5k - 10 = -5 k = 5 b) f(x) becomes 30 - 7x f(x) = 86 => 30 - 7x = 86 7x = -56 x = -8 7. (4x-1)^2-3(4x-1)-4=-4 (4x-1)^2 - 3(4x-1) = 0 (4x-1)[(4x-1) - 3] = 0 (4x - 1)(4x - 4) = 0 x = 1/4 or x = 1 8. a) f(x) = 2^x - k*2^(-x) f(1) = 2 - k/2 2f(1) = 4 - k = 3 k = 1 b) f(x) becomes 2^x - 1/2^x f(x) = 63/8 => 2^x - 1/2^x = 63/8 Let u = 2^x u - 1/u = 63/8 8u - 8/u = 63 8u^2 - 63u - 8 = 0 (8u + 1)(u - 8) = 0 u = -1/8 or u = 8 2^x = -1/8 (rejected) or 2^x = 8 x = 3 9. 2^(2x+3) - 4^x=35 (2^3)(2^2x) - 4^x = 35 8(4^x) - 4^x = 35 7(4^x) = 35 4^x = 5 xlog4 = log5 x = log5/log4 = 1.161
其他解答:
急急急!中4quadrativ formula 聽日交!
發問:
1.using quadrativ formula, solve the quadrativ formula (1-x)(4+x)+9x-2=02.if the equation(x+1)(x-4)+kx-4k=0 has a repeated root,fing the possible values of k.3.suppose f(x)=x/(1+x0,find f(1/2)f(2)4.suppose f(x)=x^2-4x,if f(k-2)=f(k),find the value of k.5.if g(x)=5x^2,express g(x+2)-g(x-2) in terms... 顯示更多 1.using quadrativ formula, solve the quadrativ formula (1-x)(4+x)+9x-2=0 2.if the equation(x+1)(x-4)+kx-4k=0 has a repeated root,fing the possible values of k. 3.suppose f(x)=x/(1+x0,find f(1/2)f(2) 4.suppose f(x)=x^2-4x,if f(k-2)=f(k),find the value of k. 5.if g(x)=5x^2,express g(x+2)-g(x-2) in terms of x. 6,consider f(x)=6K-(k+2)x such that f(5)=-5. a) find the value of k b) hence find the value of x such that f(x)=86. 7.find the values of x such that (4x-1)^2-3(4x-1)-4=-4 8.if f(x)=(2^x)-(k*2^-x and 2f(1)=3,find a)the value of k b)the value of x such that f(x)=63/8 9.solve the equation 2^(2x+3) - 4^x=35
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最佳解答:1. (1-x)(4+x)+9x-2=0 4 + x - 4x - x^2 + 9x - 2 = 0 -x^2 + 6x + 2 = 0 x^2 - 6x - 2 = 0 x = [6 +/- √(36 + 8)]/2 x = 3 + √11 or 3 - √11 2. (x + 1)(x - 4) + kx - 4k = 0 (x + 1)(x - 4) + k(x - 4) = 0 (x - 4)(x + 1 + k) = 0 x = 4 or x = -1 - k For repeated roots, 4 = -1 - k k = -5 3. f(x)=x/(1+x) f(1/2) = (1/2) / (3/2) = 1/3 f(2) = 2/3 f(1/2)f(2) = 1/3 * 2/3 = 2/9 4. f(x) = x^2 - 4x f(k-2) = (k-2)^2 - 4(k-2) = k^2 - 4k + 4 - 4k + 8 = k^2 - 8k + 12 f(k) = k^2 - 4k f(k-2) = f(k) => k^2 - 8k + 12 = k^2 - 4k 4k = 12 k = 3 5. g(x)=5x^2 g(x + 2) - g(x - 2) = 5(x+2)^2 - 5(x-2)^2 = 5[(x+2) + (x-2)][(x+2) - (x-2)] = 5(2x)(4) = 40x 6. f(x)=6k-(k+2)x a) f(5) = 6k-(k+2)(5) = -5 6k - 5k - 10 = -5 k = 5 b) f(x) becomes 30 - 7x f(x) = 86 => 30 - 7x = 86 7x = -56 x = -8 7. (4x-1)^2-3(4x-1)-4=-4 (4x-1)^2 - 3(4x-1) = 0 (4x-1)[(4x-1) - 3] = 0 (4x - 1)(4x - 4) = 0 x = 1/4 or x = 1 8. a) f(x) = 2^x - k*2^(-x) f(1) = 2 - k/2 2f(1) = 4 - k = 3 k = 1 b) f(x) becomes 2^x - 1/2^x f(x) = 63/8 => 2^x - 1/2^x = 63/8 Let u = 2^x u - 1/u = 63/8 8u - 8/u = 63 8u^2 - 63u - 8 = 0 (8u + 1)(u - 8) = 0 u = -1/8 or u = 8 2^x = -1/8 (rejected) or 2^x = 8 x = 3 9. 2^(2x+3) - 4^x=35 (2^3)(2^2x) - 4^x = 35 8(4^x) - 4^x = 35 7(4^x) = 35 4^x = 5 xlog4 = log5 x = log5/log4 = 1.161
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