解方程9x²-8=0

是不是没学过数学,先看△≥0还是＜0,前一种情况按实数公式求解,后一种情况只有虚数解,估计你还是初中生,公式x=(-b±√(b^2-4ac))/2a,过程x=(-(-9)±√(〖(-9)〗^2-4×2

原方程可变为：2x²-18x+16=0即,x²-9x+8=0把方程左边因式分解的,（x-1）（x-8）=0从而的,x-1=0或x-8=0解得,x=1或x=8

[1-1/(x-4)]+[1-1/(x-8)]=[1-1/(x-7)]+[1-1/(x-5)]1/(x-4)+1/(x-8)=1/(x-7)+1/(x-5)(x-4+x-8)/(x-4)(x-8)=(

x=-0.5这题很简单好好加油吧

1、X²-2X-8=0x²-2x+1-9=0x²-2x+1=9（x-1）²=9x-1=±3x1=4,x2=-22、X²+6X=-9x²+6x

/>因为：x-5/x-6+x-8/x-9=x-7/x-8+x-6/x-7,所以：1+1/(x-6)+1+1/(x-9)=1+1/(x-8)+1+1/(x-7)即：1/(x-6）+1/(x-9)=1/(

(1)(X+7)/(X+6)+(X+9)/(X+8)=(X+10)/(X+9)+(X+6)/(X+5)这类方程重在运算的技巧性,观察分母7+9=10+6;6+8=5+9所以先移项:(X+7)/(X+6

4x^2(x+2)-9(x+2)=0(x+2)(4x^2-9)=0(x+2)(2x-3)(2x+3)=0x=-2,1.5,-1.5

通分得[（x-8）(x-4)-(x-9)(x-3)]/(x-3)(x-4)=[(x+7)(x+3)-(x+2)(x+8)]/(x+8)(x+3)解得5/(x-3)(x-4)=5/(x+8)(x+3),

(x^2+x)(x^2+x-3)-3(x^2+x)+8=0（x^2+x)^2-6(x^2+x)+8=0(x^2+x-4)(x^2+x-2)=0x^2+x-4=0x^2+x-2=0(x+1/2)^2=1

(x+8)/(x-3)-(x-9)/(x-4)=(x+7)/(x+8)-(x+2)/(x+3)[（x-8）(x-4)-(x-9)(x-3)]/(x-3)(x-4)=[(x+7)(x+3)-(x+2)(

这道题不用那么复杂.先移项.(x-4)/(x-5)-(x-5)/(x-6)=(x-7)/(x-8)-(x-8)/( x-9)再通分.（x-4)(x-6)/(x-5)(x-6)-(x-5)(x

用换元法设x*x+11x-8＝A然后再做

x³-4x²+16x+192=0x³+4x²-8x²+16x+192=0x²(x+4)-8(x²-2x-24)=0x²(

左右两边同乘以(X-5)(X-6)(X-8)(X-9)得到(X-4)(X-6)(X-8)(X-9)-(X-5)(X-5)(X-8)(X-9)=(X-5)(X-6)(X-7)(X-9)-(X-5)(X-

(x-4)/(x-5)-(x-5)/(x-6)=(x-7)/(x-8)-(x-8)/(x-9)1+1/(x-5)-1-1/(x-6)=1+1/(x-8)-1-1/(x-9)1/(x-5)-1/(x-6

x-9=4*8x-9=32x=32+9x=41再问：检验？再答：检验：把x=41代入原方程，左边=41-9=32右边=32因为：左边=右边，所以：x=41是原方程的解

令a=x^2-2x则(a+9)(a-8)+72=0a^2+a-72+72=0a(a+1)=0a=0,a=-1x^2-2x=0,x(x-2)=0,x=0,x=2x^2-2x=-1,x^2-2x+1=0,

令X^2+X=YY(Y-3)-3Y+8=0Y^2-6Y+8=0(Y-2)(Y-4)=0y=2Y=4X^2+X=2x^2+x=4x^2+x-2=0x^2+x-4=0(x+2)(x-1)=0X=1/2(-