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解方程:(1)2x2+4x%1=0;(2)4(2x%1)2=9(x+4)2

(1)a=2,b=4,c=-1,b2-4ac=16-4×2×(-1)=24>0,x=?2±62,x1=?2?62,x2=?2+62;(2)4(2x-1)2=9(x+4)2,(2)4(2x-1)2-9(x+4)2=0,(4x-2+3x+12)(4x-2-3x-12)=0,(7x+10)(x-14)=0,x1=-107,x2=14.

(1)由原方程,得(2x-8+6)(2x-8-6)=0,即(2x-2)(2x-14)=0,解得x1=1,x2=7;(2)由原方程,得2(x2-2x+1)=9+2,即(x-1)2=112,开方,得x-1=±222,解得x1=1+222,x2=1?222;(3)由原方程,得(x+2)(x-1-2)=0,即(x+2)(x-3)=...

(1)方程开方得:2x-1=3或2x-1=-3,解得:x1=2,x2=-1;(2)分解因式得:(x-1)(x+4)=0,解得:x1=1,x2=-4;(3)方程变形得:(x+4)2-5(x+4)=0,分解因式得:(x+4)(x+4-5)=0,解得:x1=-4,x2=1;(4)方程变形得:x2+4x+4=6,即...

(1)3/4x+1/2x=9/4 5x/4=9/4 x=9/5 (2)4/5x×〔1/2+0.4〕=4.32 4x/5=4.32÷0.9 x=4.8X5/4 x=6

(1)x 2 -2x+1=9,移项得:x 2 -2x-8=0,即(x-4)(x+2)=0,可得x-4=0或x+2=0,解得:x 1 =4,x 2 =-2;(2)去分母得:4x 2 +2x-1=0,这里a=4,b=2,c=-1,∵b 2 -4ac=2 2 -4×4×(-1)=20>0,∴x= -2± 20 8 = -1± 5 4 ,则x 1 = -1+ 5 4 ,x...

(1)开方得x-1=±2即x-1=2或x-1=-2.解得x 1 =3,x 2 =-1.(2)∵(x+2)(x-1)=0∴x+2=0或x-1=0∴x 1 =-2,x 2 =1.(3)∵x 2 -2x-3=0∴(x+1)(x-3)=0,即x+1=0或x-3=0解得x 1 =-1,x 2 =3.(4)∵a=1,b=4,c=2∴b 2 -4ac=16-8=8.∴x= -4± 8 ...

(1)(2x-1) 2 -9=0,(2x-1+3)(2x-1-3)=0,2x+2=0或2x-4=0,∴x 1 =-1,x 2 =2;(2)(x-4)(x+2)=0,x-4=0或x+2=0,∴x 1 =4,x 2 =-2;(3)(x+4)(x-1)=0,(x+4)=0或x-1=0,∴x 1 =-4,x 2 =1;(4)△=2 2 -4×4<0,∴原方程无解.

①方程两边开平方,得2x-1=±3,解得x1=2,x2=-1;②移项,得x2+3x=4配方,得x2+3x+94=4+94(x+32)2=254,开平方,得x+32=±52解得x1=1,x2=-4;③∵a=1,b=-2,c=-8,△=(-2)2-4×1×(-8)=36∴x=2±362,解得x1=4,x2=-2;④移项,化系数为“1”,得x2-2...

(1)x(x+1)=0,x=0或x+1=0,∴x1=0,x2=-1;(2)(2x+11)(2x-11)=0,2x+11=0或2x-11=0,∴x1=-112,x2=112;(3)(2x+1)(3x-2)=0,2x+1=0或3x-2=0,∴x1=-12,x2=23;(4)(x-4+5-2x)(x-4-5+2x)=0,(1-x)(3x-9)=0,1-x=0或3x-9=...

(1)因式分解得,(2x+1)(x+3)=0,解得,x1=12,x2=3;(2)移项得,(x-1)2+2(x-1)=0,提公因式得,(x-1)(x-1+2)=0,解得,x1=1,x2=-1;(3)移项得,x(2x+3)-(4x+6)=0,提公因式得,(2x+3)(x-2)=0,解得,x1=-32;x2=2;...

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