例题:解一元二次不等式x² -9>0 ∵x²-9=(x+3)(x-3),∴(x+3)(x-3)>0,由于有理数法则“两号相乘,同号得正”有(1)(X+3)大于0 ( X-3)大于0 (2)(X+3)小于0 (X-3)小于0解得:
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/08 13:01:17
![例题:解一元二次不等式x² -9>0 ∵x²-9=(x+3)(x-3),∴(x+3)(x-3)>0,由于有理数法则“两号相乘,同号得正”有(1)(X+3)大于0 ( X-3)大于0 (2)(X+3)小于0 (X-3)小于0解得:](/uploads/image/z/1098860-68-0.jpg?t=%E4%BE%8B%E9%A2%98%EF%BC%9A%E8%A7%A3%E4%B8%80%E5%85%83%E4%BA%8C%E6%AC%A1%E4%B8%8D%E7%AD%89%E5%BC%8Fx%26%23178%3B+-9%EF%BC%9E0+%E2%88%B5x%26%23178%3B-9%3D%EF%BC%88x%2B3%EF%BC%89%EF%BC%88x-3%EF%BC%89%2C%E2%88%B4%EF%BC%88x%2B3%EF%BC%89%EF%BC%88x-3%EF%BC%89%EF%BC%9E0%2C%E7%94%B1%E4%BA%8E%E6%9C%89%E7%90%86%E6%95%B0%E6%B3%95%E5%88%99%E2%80%9C%E4%B8%A4%E5%8F%B7%E7%9B%B8%E4%B9%98%2C%E5%90%8C%E5%8F%B7%E5%BE%97%E6%AD%A3%E2%80%9D%E6%9C%89%EF%BC%881%EF%BC%89%EF%BC%88X%2B3%EF%BC%89%E5%A4%A7%E4%BA%8E0+%EF%BC%88+X-3%EF%BC%89%E5%A4%A7%E4%BA%8E0+%EF%BC%882%EF%BC%89%EF%BC%88X%2B3%29%E5%B0%8F%E4%BA%8E0+%EF%BC%88X-3%EF%BC%89%E5%B0%8F%E4%BA%8E0%E8%A7%A3%E5%BE%97%EF%BC%9A)
xN@_ FG>aIHtabmҍm*Ð]
.R;cܸaf9? ]x}¥o%3|5 D&_
$mJ:O7\d?WN.%;qˢ)}}쏈|t8r@7t2TDYZ%1`DF(rYCއZ<FG^"ihyBk-/ΈmAe$2`@ARAZ,$*ة$"åjZ䃲Ca|@
.i51~a1ӈVf}3
O0O
例题:解一元二次不等式x² -9>0 ∵x²-9=(x+3)(x-3),∴(x+3)(x-3)>0,由于有理数法则“两号相乘,同号得正”有(1)(X+3)大于0 ( X-3)大于0 (2)(X+3)小于0 (X-3)小于0解得:
例题:解一元二次不等式x² -9>0 ∵x²-9=(x+3)(x-3),∴(x+3)(x-3)>0,
由于有理数法则“两号相乘,同号得正”有
(1)(X+3)大于0 ( X-3)大于0 (2)(X+3)小于0 (X-3)小于0
解得:X大于3或X小于-3
求分式不等式2X-3分之5X+1小于0的解集
例题:解一元二次不等式x² -9>0 ∵x²-9=(x+3)(x-3),∴(x+3)(x-3)>0,由于有理数法则“两号相乘,同号得正”有(1)(X+3)大于0 ( X-3)大于0 (2)(X+3)小于0 (X-3)小于0解得:
∵5x+1/2x-30 ,2x-3
x²>9,即X的绝对值大于3,so,X大于3或者X小于-3