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如何解一元四次方程x^4+2x^3+9x=0

用单变量求解,无实解

(1)解:4(x-1)2=9,开方得:2(x-1)=±3,即2(x-1)=3,2(x-1)=-3,解方程得:x1=52,x2=-12,即方程的解是x1=52,x2=-12.(2)解:3x2+10x+3=0,∴(3x+1)(x+3)=0,即:3x+1=0,x+3=0,解方程得:x1=-13,x2=-3,∴方程的解是x1=-13,...

(1) (2)x 1 =1 x 2 =9(3)x 1 =-3 x 2 =- (4)x 1 =-2 x 2 =3 试题分析:(1)找出a,b,c的值,代入求根公式即可求出解;(2)方程左边分解因式后,利用两数相乘积为0,两因式中至少有一个为0转化为两个一元一次方程来求解;(3)利用平...

(1)移项,得4(x-1) 2 -9(x-5) 2 =0,分解因式,得[2(x-1)+3(x-5)][2(x-1)-3(x-5)]=0,即(5x-17)(-x+13)=0,所以5x-17=0或-x+13=0,解得x 1 = 17 5 ,x 2 =13;(2)移项,得2x(x-3)+x-3=0,分解因式,得(x-3)(2x+1)=0...

(1)(x+1+3)(x+1-3)=0(x+4)(x-2)=0∴x 1 =-4,x 2 =2.(2)(2x-5)(x+1)=0∴x 1 = 5 2 ,x 2 =-1.(3)(x-5)[3(x-5)+2]=0(x-5)(3x-13)=0∴x 1 =5,x 2 = 13 3 .(4)(x+1) 2 +2(x+1)+1=0(x+1+1) 2 =0(x+2) 2 =0∴x ...

是想求x等于多少还是怎么的?每个方程是一个题目么?

(1)x^2-9x+8=0 答案:x1=8 x2=1  (2)x^2+6x-27=0 答案:x1=3 x2=-9  (3)x^2-2x-80=0 答案:x1=-8 x2=10  (4)x^2+10x-200=0 答案:x1=-20 x2=10  (5)x^2-20x+96=0 答案:x1=12 x2=8  (6)x^2+23x+76=0 答案:x1=-19 x2=-...

(x+3)(x-1)=5 x²+2x-3-5=0 x²+2x-8=0 (x-2)(x+4)=0 x=2或x =-4

(-2)²;-4*(-k)>0 即4+4k>0,k>-1 又一元二次方程x²-6x+k=0与x²-mx-12=0有一个相同的根 则(-6)²-4*k≥0,k≤9 又K是符合条件的最大整数 则K=9 x²-6x+9=0 x=3 3*3-3m-12=0 m=-1

(3-x)^2+4x(x-3)=0 ,5x^2 -18x+9=0, (5x-3)(x-3)=0, x1= 3/5, x2=3 9(x-2)^=4(x+1)^ 2, 5x^2 -44x +32=0, (5x-4)(x-8)=0, x1=4/5,x2=8 (2x-1)^-x^-4x-4=0, 3x^2 -8x -3=0, (3x+1)(x-3)=0, x1= -1/3, x2=3 x^-13x+12=42, x^2 -13x - 30=0, (x -1...

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