百度试题 结果1 题目Factorise fully64-4x^2=4(16-x^2) =4(4-x)(4+x) 相关知识点: 试题来源: 解析 =4(16-)=4(4-x)(4+x) 反馈 收藏
Factorise the following fully: (1) x4−1= (2) 32x8−162= (3) 50(2x+1)2−18(1−x)2= 相关知识点: 试题来源: 解析 (1) (x2+1)(x+1)(x−1) (2) 2(4x4+9)(2x2+3)(2x2−3) (3) 2(7x+8)(13x+2) (1)N/A.(2)N/A.(3)N/A....
解析 (x+3)^6[(x+3)+(2x-1)] (x+3)^6(3x+2) 结果一 题目 Factorise fully(x+3)^7+(x+3)^6(2x-1)Do not attempt to expand brackets. 答案 (x+3)^6[(x+3)+(2x-1)] (x+3)^6(3x+2)相关推荐 1Factorise fully(x+3)^7+(x+3)^6(2x-1)Do not attempt to expand bracke...
百度试题 结果1 题目(b) Factorise fully 6x^3-27x^2+30x=3x(2x2-9x+10) [M1]=3x(2x-5)(x-2) 相关知识点: 试题来源: 解析 =3x(2x²-9x+10) [M1] = 3x(2x-5)(x-2) 反馈 收藏
结果1 题目 (a) Factorise fully 15y^4+20uy^35y3(3y+ 4u)(2)5-8x(b) S 01vc4-3x=(5-8x)/4Show clear algebraic working.4.x(4-3.x)-5-8.x or 16-12.r-5-8.r Or1-3r-1/1-2r x=.(3) 相关知识点: 试题来源: 解析 or or 反馈 收藏 ...
百度试题 结果1 题目Factorise fully(x+3)^7+(x+3)^6(2x-1)Do not attempt to expand brackets. 相关知识点: 试题来源: 解析 (x+3)^6[(x+3)+(2x-1)] (x+3)^6(3x+2) 反馈 收藏
百度试题 结果1 题目Hence, factorise fully x^3-6x^2-x+6 相关知识点: 试题来源: 解析 =(x-1)(x²-5x-6)=(x-1)(x+1)(x-6) 反馈 收藏
25 Given that f(x)=2x^3-17x^2-58x+33(a) show that f(11)=0(2)(b)hence fully factorise f(x).(4) 相关知识点: 试题来源: 解析 (a)f(11)= 2(11)^3- 17(11)² -58(11) + 33(subst.)Note: 2662- 2057- 638 + 33 earns M markORattempt at dividing 2x^3- 17x² - 58...