25(x-1)²-9(2x+3)²=0 用因式分解

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25(x-1)²-9(2x+3)²=0 用因式分解
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25(x-1)²-9(2x+3)²=0 用因式分解
25(x-1)²-9(2x+3)²=0 用因式分解

25(x-1)²-9(2x+3)²=0 用因式分解
25(x-1)²-9(2x+3)²=0
[5(x-1)]²-[3(2x+3)]²=0
[5(x-1)+3(2x+3)][5(x-1)-3(2x+3)]=0
(11x+4)(-x-14)=0
11x+4=0 -x-14=0
x1=-4/11 x2=-14

(5(x-1))²-(3(2x+3))²=0
((5z-5)+(6x+9))((5x-5)-(6x+9))=0
∴11x-4=0或x+14=0
x1=4/11
x2=-14

[5(x-1)]²-[3(2x+3)]²=0
[5(x-1)-3(2x+3)][5(x-1)+3(2x+3)]=0
-(x+14)(11x+4)=0
x1=-14, x2=-4/11

25(x-1)²-9(2x+3)²=0 用因式分解 1.(ax+by)²+(bx-ay)²2.(x²-x)²-4(x²-x-1)3.4(x+y)²-9(x-y)²4.(a+b)(a-b)+4(b-1)5.x²+4xy+4y²-2x-4y-36.(x+1)(x+2)(x+3)(x+4)+17,3x²-11x+10 求几道因式分解1.(ax+by)²+(bx-ay)²2.(x²-x)²-4(x²-x-1)3.4(x+y)²-9(x-y)²4.(a+b)(a-b)+4(b-1)5.x²+4xy+4y²-2x-4y-36.(x+1)(x+2)(x+3)(x+4)+17,3x²-11x+10 因式分解:(x²-2x)²-3x²+6x 2.(2X²-3X+1)²-22X²+33X-1 1、解关于x的方程:a(x²+1)=4x²+22、√(x+11)=1-x3、(5x/x²-9)+(5/x+3)=1+(x/x-3)4、(x²+2/2x²-1)-(6x²-3/x²+2)+2=0x+2y=125、{x&s 3y/2x+2y+2xy/x²+xy 2x/x²-64y²-1/x-8y (1/a+1/b)²÷(1/a²-1/b²) (3x&su3y/2x+2y+2xy/x²+xy 2x/x²-64y²-1/x-8y (1/a+1/b)²÷(1/a²-1/b²) (3x²/4y)²×2y/3x+x²/2y²÷2y²/ 因式分解:(x²+5x+9)(x²-3x+7)-3(4x+1)²要详解求人不如求己,看看对还是错。设y=[(x²+5x+9)+(x²-3x+7)]/2=x²+x+8 再设x²+x+8 =t,则原式 化简~(x²+3x+9/x²-27)+(6x/9x-x²)-(x-1/6+2x)(x²+3x+9/x²-27)+(6x/9x-x²)-(x-1/6+2x)注意了!是三个分式!(x²+3x+9)/(x²-27)+(6x)/(9x-x²)-(x-1)/(6+2x) xy²/(x²y-y) × x²/(x²+x)=(x²-3x)/(x²-5x) × 2x-10/(x²-6x+9)=化简求植x²-1/(x²-x-2)除以x/2x-4,其中x=1/2[x-x/(x+1)]除以[x/(2x-4)] ,其中x=√2+1 1、(3x²-2x²+5x-2)-( )=4x³-x²+52、如果xy²-2x²y²+3x²Y是x²y²-2x²y减去另一个多项式所得到的差,求另一个多项式.3、已知三角形的周长是(3x&sup 因式分解(x²-2x)²-9 因式分解4x²-13x²+9 计算:(B-1)(1+B)(B²-1),(a-2b+3c)(a+2b-3c)分解因式:a²x²+16ax +6425(x+y)²-16(x-y)²x²-6x+9-Y²(a²+b²-1)²-4a²b² 求值:【(x+2y-3∕2)(x-2y+3∕2)+2y(2y-3)+(2ˆ-2∕3)】 x²-4x+1=0(公式法)2(x-3)²=x²-9 (2)(x-2)(x+2)-(x+1)(x-3)(3) (ab+1)²-(ab-1)²(4)(2x-y)²-4(x-y)(x+2y) 计算极限x→2,limx²-3x+2/x²-4, 1(x²-9)/(x²+2x²+x)*(3x³+9x²)/(x³-3x),其中x=-1/3