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分解因式求完整 x²-xy+四分之一y² a&#17...

y=3-x 所以xy=3x-x²=-2 x²-3x-2=0 x=(3±√17)/2 y=3-x 所以 x=(3-√17)/2,y=(3+√17)/2 x=(3+√17)/2,y=(3-√17)/2

答: (y²-5y-2)(y²-5y-4)-16 =(y²-5y)²-6(y²-5y)+8-16 =(y²-5y)²-6(y²-5y)-8 =(y²-5y-3+√17)(y²-5y-3-√17) (y²-5y+2)(y²-5y-4)-16 =(y²-5y)²-2(y²-5y)-8-16 =(y...

2x²+xy-2y² =2(x+y)(x-y)+xy

杀鸡不用牛刀,初数问题初数解决: 以待求式代入约束条件得, x²+4(z-2x)²=17 →17x²-16zx+4z²-17=0. ∴△=(-6z)²-4×17×(4z²-17)≥0 →-17/2≤z≤17/2. 以z=17/2代回原式得x=4,y=1/2; 以z=-17/2代回原式得x=-4,y=-1/2. ...

解: x¹⁷-x =x(x¹⁶-1) =x(x⁸-1)(x⁸+1) =x(x⁴-1)(x⁴+1)(x⁸+1) =x(x²-1)(x²+1)(x⁴+1)(x⁸+1) =x(x-1)(x+1)(x²+1)(x⁴+1)(x⁸+1)

2x² + 5x + 1 =2(x² + 5x/2) + 1 =2[x² + 5x/2 + (5/4)² - (5/4)²]+1 =2(x + 5/4)² - 17/8 =2[(x + 5/4)² - 17/16] =2(x + 5/4 + √17/4)(x + 5/4 - √17/4)

余弦定理: cosx=(a²+b²-c²)/(2ab) =(17²+18²-30²)/(2*17*18) =-287/(2*17*18)≈-0.469 x...

f(x)=ax²+x-a=a(x-1/(2a))²-(4a²+1)/(4a) 有最大值,a<0 最大值17/8,-(4a²+1)/(4a)=17/8 8a²+17a+2=0 (a+2)(8a+1)=0 a=-2,或-1/8

设f(x) = x³-3x²+1. 可算得f(3) = 1 > 0, f(1) = -1 < 0, f(0) = 1 > 0, f(-1) = -3 < 0. 于是f(x) = 0在区间(1,3), (0,1), (-1,0)内分别存在实根. 而f(x) = 0至多只有3个实根, 因此在上述区间内各有一个. 分别记为a, b, c, 有-1 < c ...

5x5-15x3y-20xy2 =5x(x^4-3x²y-4y²) =5x(x²-4y)(x²+y) 3a2b2-17abxy+10x2y2 =(3ab-2xy)(ab-5xy)

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