用因式分解法解方程 (1)x²+x-6=0(2)5x²-3x=x+1 (3)x²-2x+5=0 (4)x²+4x+8=4x+11(5)4x²-6x=0 (6)x(2x-4)=5-8x(7)(4x+4)²=x(2x+1)(8)3x(x+2)=5(x+2) (9)(3x+1)²-5=0
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/14 03:22:54
![用因式分解法解方程 (1)x²+x-6=0(2)5x²-3x=x+1 (3)x²-2x+5=0 (4)x²+4x+8=4x+11(5)4x²-6x=0 (6)x(2x-4)=5-8x(7)(4x+4)²=x(2x+1)(8)3x(x+2)=5(x+2) (9)(3x+1)²-5=0](/uploads/image/z/468718-70-8.jpg?t=%E7%94%A8%E5%9B%A0%E5%BC%8F%E5%88%86%E8%A7%A3%E6%B3%95%E8%A7%A3%E6%96%B9%E7%A8%8B+%EF%BC%881%EF%BC%89x%26%23178%3B%2Bx-6%3D0%EF%BC%882%EF%BC%895x%26%23178%3B-3x%3Dx%2B1++++++%EF%BC%883%EF%BC%89x%26%23178%3B-2x%2B5%3D0+++++%EF%BC%884%EF%BC%89x%26%23178%3B%2B4x%2B8%3D4x%2B11%EF%BC%885%EF%BC%894x%26%23178%3B-6x%3D0+++++++++%EF%BC%886%EF%BC%89x%EF%BC%882x-4%29%3D5-8x%EF%BC%887%EF%BC%89%EF%BC%884x%2B4%29%26%23178%3B%3Dx%282x%2B1%29%288%293x%28x%2B2%29%3D5%28x%2B2%29++++++%289%29%283x%2B1%29%26%23178%3B-5%3D0)
xVKoF+$z/>/pARRE=THh^H/A9~'>>7z~X/.z~zl6'('2E[0Rhgp9
LJ?zX\y2ei@(?!#㨌dBV?QTUz7ЭNp5Q1ؓ!˝-թ7YoBnAU>@o25{*jD_*H0 RRyƋFj%_}|h{r~|ro!~yewge eY,'YIqƛ`mf)3*-h9)fP4dz
WӶTʅqKrrJ\y1edJn\ 5ӹع!"hl))uC]:2buv t)F\sbG9