Câu 6:
1: \(\left(2x-3\right)^2\ge0\forall x\)
=>\(\left(2x-3\right)^2+15\ge15\forall x\)
Dấu '=' xảy ra khi 2x-3=0
=>2x=3
=>\(x=\frac32\)
2: \(\left(5x+7\right)^8\ge0\forall x\)
=>\(\left(5x+7\right)^8-2020\ge-2020\forall x\)
Dấu '=' xảy ra khi 5x+7=0
=>5x=-7
=>\(x=-\frac75\)
3: \(\left|1-2019x\right|\ge0\forall x\)
=>\(\left|1-2019x\right|+2016\ge2016\forall x\)
Dấu '=' xảy ra khi 1-2019x=0
=>2019x=1
=>\(x=\frac{1}{2019}\)
4: \(\left|4x+1\right|\ge0\forall x\)
=>\(\left|4x+1\right|-1\ge-1\forall x\)
Dấu '=' xảy ra khi 4x+1=0
=>4x=-1
=>\(x=-\frac14\)
5: \(\left|x-1\right|+\left|x-2\right|\ge\left|x-1-x+2\right|=1\forall x\)
Dấu '=' xảy ra khi (x-1)(x-2)<=0
=>1<=x<=2
6: \(\left|x-2021\right|\ge0\forall x\)
=>\(\left|x-2021\right|+3\ge3\forall x\)
=>\(\frac{15}{\left|x-2021\right|+3}\le\frac{15}{3}=5\forall x\)
=>\(-\frac{15}{\left|x-2021\right|+3}\ge-5\forall x\)
=>\(-\frac{15}{\left|x-2021\right|+3}+2021\ge-5+2021=2016\forall x\)
Dấu '=' xảy ra khi x-2021=0
=>x=2021
Câu 7:
1: \(\left(4x-7\right)^2\ge0\forall x\)
=>\(-\left(4x-7\right)^2\le0\forall x\)
=>\(-\left(4x-7\right)^2+8\le8\forall x\)
Dấu '=' xảy ra khi 4x-7=0
=>4x=7
=>x=7/4
2: \(\left|6x-1\right|\ge0\forall x\)
=>\(-\left|6x-1\right|\le0\forall x\)
=>\(-\left|6x-1\right|+7\le7\forall x\)
Dấu '=' xảy ra khi 6x-1=0
=>6x=1
=>x=1/6
3: \(x^2+5\ge5\forall x\)
=>\(\left|x^2+5\right|\ge5\forall x\)
=>\(-\left|x^2+5\right|\le-5\forall x\)
=>\(-\left|x^2+5\right|+6\le-5+6=1\forall x\)
Dấu '=' xảy ra khi x=0
4: \(\left(5x-6\right)^2\ge0\forall x\)
=>\(\left(5x-6\right)^2+2\ge2\forall x\)
=>\(\frac{3}{\left(5x-6\right)^2+2}\le\frac32\forall x\)
=>\(\frac{3}{\left(5x-6\right)^2+2}+14\le14+\frac32=\frac{31}{2}\forall x\)
Dấu '=' xảy ra khi 5x-6=0
=>x=6/5
5: \(\left|7x+4\right|>=0\forall x\)
=>\(\left|7x+4\right|+5>=5\forall x\)
=>\(\frac{15}{5+\left|7x+4\right|}\le\frac{15}{5}=3\forall x\)
=>\(\frac{15}{5+\left|7x+4\right|}-6\le3-6=-3\forall x\)
Dấu '=' xảy ra khi 7x+4=0
=>x=-4/7
6: \(\frac{4x^2+9}{x^2+1}=\frac{4x^2+4+5}{x^2+1}=4+\frac{5}{x^2+1}\)
\(x^2+1\ge1\forall x\)
=>\(\frac{5}{x^2+1}\le\frac51=5\forall x\)
=>\(\frac{5}{x^2+1}+4\le5+4=9\forall x\)
Dấu '=' xảy ra khi x=0
giúp mik với ạ mik cần gấp. GIẢI CỤ THỂ GIÚP MIK vs ạ
