Đáp án:
a.A=|x+4|+2020≥0+2020=2020A=|x+4|+2020≥0+2020=2020
B=|3x−6|−2≥0−2=−2B=|3x−6|−2≥0−2=−2
C=5x2−3≥5.0−3=−3C=5x2−3≥5.0−3=−3
b.M=7−|x−10|≤7−0=7M=7−|x−10|≤7−0=7
P=−|x+5|−3≤−0−3=−3P=−|x+5|−3≤−0−3=−3
Q=20−4(x−3)2≤20−4.0=20
Giải thích các bước giải:
Ta có : |x|≥0∀x,x2≥0∀x|x|≥0∀x,x2≥0∀x
Dấu = xảy ra khi x=0x=0
Nên suy ra :
a.A=|x+4|+2020≥0+2020=2020A=|x+4|+2020≥0+2020=2020
B=|3x−6|−2≥0−2=−2B=|3x−6|−2≥0−2=−2
C=5x2−3≥5.0−3=−3C=5x2−3≥5.0−3=−3
b.M=7−|x−10|≤7−0=7M=7−|x−10|≤7−0=7
P=−|x+5|−3≤−0−3=−3P=−|x+5|−3≤−0−3=−3
Q=20−4(x−3)2≤20−4.0=20