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九年级数学解答题一般
题目
解下列方程:
(1)x224x=9856(1)x^{2}-24x=9856
(2)x26x9991=0(2)x^{2}-6x-9991=0.
知识点:解一元一次方程章节:未标注

答案与解析

答案

(1)移项得x224x9856=0x^{2}-24x-9856=0
因式分解,得(x112)(x+88)=0\left(x-112\right)\left(x+88\right)=0
x1=112\therefore x_{1}=112x2=88x_{2}=-88

(2)x26x9991=0(2)\because x^{2}-6x-9991=0
(x103)(x+97)=0\therefore \left(x-103\right)\left(x+97\right)=0
解得:x1=103x_{1}=103x2=97x_{2}=-97.

解析

(1)移项得x224x9856=0x^{2}-24x-9856=0
因式分解,得(x112)(x+88)=0\left(x-112\right)\left(x+88\right)=0
x1=112\therefore x_{1}=112x2=88x_{2}=-88

(2)x26x9991=0(2)\because x^{2}-6x-9991=0
(x103)(x+97)=0\therefore \left(x-103\right)\left(x+97\right)=0
解得:x1=103x_{1}=103x2=97x_{2}=-97.

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