A = 2 + 3\(\sqrt[]{x^2+1}\) 

Ta có: x² \(\ge\) 0, \(\forall\) x => x² \(\ge\) 1, \(\forall\) x

=> \(\sqrt[]{x^2+1}\) \(\ge\) \(\sqrt[]{1}\) 

=> 3\(\sqrt[]{x^2+1}\) \(\ge\) 3

=> 2 + 3\(\sqrt[]{x^2+1}\) \(\ge\) 5

Vậy A đạt GTNN khi bằng 5

Dấu "=" xảy ra khi x = 0

2: (3x-4)^2+2>=2

=>5/(3x-4)^2+2<=5/2

=>B>=-5/2

Dấu = xảy ra khi x=4/3

4: D=(3x^2+7-4)/(3x^2+7)=1-4/3x^2+7

3x^2+7>=7

=>4/3x^2+7<=4/7

=>-4/3x^2+7>=-4/7

=>D>=3/7

Dấu = xảy ra khi x=0

2) B = \(\dfrac{-5}{\left(3x-4\right)^2+2}\) 

Ta có: ( 3x-4)² \(\ge\) 0 , \(\forall\) x

=> ( 3x-4)² +2 \(\ge\) 2, \(\forall\) x

=> \(\dfrac{1}{\left(3x-4\right)^2+2}\) \(\le\) \(\dfrac{1}{2}\) , \(\forall\) x

=> \(\dfrac{-5}{\left(3x-4\right)^2+2}\) \(\ge\) \(\dfrac{-5}{2}\) , \(\forall\) x

=> B \(\ge\) \(\dfrac{-5}{2}\) 

Vậy B đạt GTNN khi bằng \(\dfrac{-5}{2}\) 

Dấu "= " xảy ra khi 3x - 4 = 0

4) D=\(\dfrac{3x^2+3}{3x^2+7}\) 

= 1 - \(\dfrac{4}{3x^2+7}\) 

Ta có: 3x² \(\ge\) 0, \(\forall\) x

=> 3x² +7 \(\ge\) 7, \(\forall\) x

=> \(\dfrac{1}{3x^2+7}\) \(\le\) \(\dfrac{1}{7}\) 

=> \(\dfrac{4}{3x^2+7}\) \(\le\) \(\dfrac{4}{7}\) 

=> 1 - \(\dfrac{4}{3x^2+7}\) \(\ge\) \(\dfrac{3}{7}\) 

Vậy D đạt GTNN khi bằng \(\dfrac{3}{7}\) 

Dấu "=" xảy ra khi x = 0

Câu 8:

ĐK \(\hept{\begin{cases}x\ne0\\x\ne3\end{cases}}\)

\(A=\frac{x^2}{\left(x-3\right)}.\frac{\left(x-3\right)^2}{x}-4=x\left(x-3\right)-4=x^2-3x-4=\left(x-\frac{3}{2}\right)^2-\frac{25}{4}\\ \)

a) \(A< -6\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{1}{4}< 0\) vô nghiệm

b) A>=-25/4 khi x=3/2

\(a,\dfrac{3x+21}{x^2-9}+\dfrac{2}{x+3}-\dfrac{3}{x-3}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}-\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21+2x-6-3x-9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2}{x-3}\)

\(b,\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\\ =\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x+3}{x^2-1}\\ =\dfrac{3x^2+4x+1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x^2-2x+1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{3x^2+4x+1-x^2+2x-1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x^2+2x-3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{2x^2+6x-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x^2+3x\right)+\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\)

\(=\dfrac{x\left(x+3\right)+\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x+3}{\left(x-1\right)^2}\)

GTNN của

+,G=3/2

+,H=-2015

+,K=5

chờ bông băng đi cấp cứu đã

bà kiếm mấy bài cực trị này ở đâu z? chỉ t vs ,cho t đề cx đc

cho a,b,c thực thỏa mãn a²+b²+c²=1.Tìm min Thắng=ab+bc+2ac 

đấy phúc coi thử