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(1)用配方法解方程:2x2-3x-1=0 (2)解...

(1)x2-3x-2=0, x2-3x=2, x2-3x+94=2+94,(x-32)2=174, x-32=±172,x1=32+172=3+172,x2=32-172=3?172; (2)x2=4(2-x),x2+4x=8,x2+4x+4=12,(x+2)2=12,x+2=±2

(1)2x2-3x-1=0,2x2-3x=1,x2-32x=12,x2-32x+(34)2=12+(34)2,(x-34)2=1716x=3±174,x1=3+174,x2=3?174;(2)2x?3≥x+1 ①x?2>12(x+1) ②,由①得x≥4,由②得x>5,则原不等式组的解集为x>5;

题目应该是:2x² - 3x - 1 = 0 用配方法解方程 2x² - 3x - 1 = 0 x² - 3x/2 - 1/2 = 0 x² - 3x/2 + (3/4)² =1/2 +(3/4)² x² - 3x/2 + (3/4)² = 17/16 (x -3/4)² = 17 / 16 x -3/4 = ±√17 / 4 x = ...

(1)2x2+3x-1=0,2x2+3x=1,x2+32x=12,平配方得:x2+32x+(34)2=12+(34)2,(x+34)2=1716,开方得:x+34=±174,x1=?3+174,x2=-3+174;(2)x2+4x-2=0,b2-4ac=42-4×1×(-2)=24,x=?4±242,x1=-2+6,x2=-2-6;(3)(2x+1)2=(x-3)2,...

∵2x 2 +3x+1=0∴2x 2 +3x=-12(x 2 + 3 2 x)=-12(x 2 + 3 2 x+ 9 16 )=-1+ 9 8 ∴2(x+ 3 4 ) 2 = 1 8 即2(x+ 3 4 ) 2 - 1 8 =0故选B.

(1)方程变形得:x2-2x=-12,配方得:x2-2x+1=12,即(x-1)2=12,开方得:x-1=±22,则x1=1+22,x2=1-22;(2)方程变形得:2x2-3x+1=0,分解因式得:(2x-1)(x-1)=0,解得:x1=12,x2=1;(3)方程变形得:x2-11x-24=0,分解因式得:(x-3...

(1)2x 2 -5x-3=0(用配方法) x 2 - 5 2 x-3=0 ∴ (x- 5 4 ) 2 = 49 16 ∴ x 1 =3, x 2 =- 1 2 ;(2)3x(x-1)=2-2x∴3(x-1)+2(x-1)=0∴(x-1)( 3x+2)=0 ∴ x 1 =1, x 2 =- 2 3 .

(1)x=3±9?4×1×(?1)2×1=3±132,∴x1=3+132,x2=3?132.(2)x=4±16+4×3×12×3=2±73,∴x1=2+73,x2=

(1)移项,得2x2-6x=-1,化二次项系数为1,得x2-3x=-12,配方,得x2-3x+(32)2=-12+(32)2,即(x-32)2=74,开方,得x-32=±72,解得,x1=3+72,x2=3?72;(2)移项,得(2x-1)2-x2=0,所以(2x-1-x)(2x-1+x)=0,即(x-1)(3x-1)=0,解得,...

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