Question

（2y－1/3－y）²≤1∴（y＋2）（3y－4）≤0这步怎么来的？？？

Answer 1

（2y－1/3－y）²-1≤1-1

（y-1/3）²-1²≤0

（y-1/3+1）(y-1/3-1)≤0

(y+2/3)(y-4/3)≤0

1/3×1/3×(3y+2)(3y-4)≤0

9×1/3×1/3×(3y+2)(3y-4)≤0×9

(3y+2)(3y-4)≤0

Answer 2

解由（2y－1/3－y）²≤1
得(2y-1)^2≤(3-y)^2
则4y^2-4y+1≤y^2-6y+9
即3y^2+2y-8≤0
即(y+2)(3y-4)≤0