Ta có : \(D=3x^2+2x+1=3\left(x^2+\frac{2}{3}x+\frac{1}{3}\right)=3\left(x^2+\frac{2}{3}x+\frac{1}{9}+\frac{2}{9}\right)=3\left(x+\frac{1}{3}\right)^2+\frac{2}{3}\ge\frac{2}{3}\)

\(\Rightarrow\)Min D = 2/3

Dấu "=" xảy ra khi x + 1/3 = 0 

\(\Rightarrow x=-\frac{1}{3}\)

Vậy Min D = 2/3 khi x = -1/3 

D = 3x² + 2x + 1 = 3( x² + 2/3x + 1/9 ) + 2/3 = 3( x + 1/3 )² + 2/3 ≥ 2/3 ∀ x

Dấu "=" xảy ra <=> x = -1/3 . Vậy MinD = 2/3

`S=3x^2+2x+1`

`=(3x^2+2x+1/3)+2/3`

`=[(\sqrt3 x)^2+ 2.\sqrt3 x . 1/\sqrt3 + (1/\sqrt3)^2]+2/3`

`=(\sqrt3 x+1/\sqrt3)^2 + 2/3`

`=(\sqrt3x+\sqrt3/3)^2+2/3`

`=> D_(min) =2/3 <=> \sqrt3x+\sqrt3/3=0 <=>x=-1/3`

\(D=3\left(x^2+\dfrac{2}{3}x+\dfrac{1}{3}\right)=3\left(x^2+\dfrac{2}{3}x+\dfrac{1}{9}+\dfrac{2}{9}\right)=3\left(x+\dfrac{1}{3}\right)^2+\dfrac{2}{3}\ge\dfrac{2}{3}\)

dấu"=" xảy ra<=>x=-1/3

Ta có:D=3x²+2x+1

            = \(3\left(x^2+2.\dfrac{1}{3}x+\dfrac{1}{9}\right)-\dfrac{1}{3}\)

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

Vì \(3\left(x-\dfrac{1}{3}\right)^2\ge0\forall x\)

\(\Rightarrow D\ge-\dfrac{1}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)

b: \(A=\dfrac{2-1}{3\cdot2}=\dfrac{1}{6}\)

a: Để \(\dfrac{3x-2}{4}\) không nhỏ hơn \(\dfrac{3x+3}{6}\) thì \(\dfrac{3x-2}{4}>=\dfrac{3x+3}{6}\)

=>\(\dfrac{6\left(3x-2\right)}{24}>=\dfrac{4\left(3x+3\right)}{24}\)

=>18x-12>=12x+12

=>6x>=24

=>x>=4

b: Để \(\left(x+1\right)^2\) nhỏ hơn \(\left(x-1\right)^2\) thì \(\left(x+1\right)^2< \left(x-1\right)^2\)

=>\(x^2+2x+1< x^2-2x+1\)

=>4x<0

=>x<0

c: Để \(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}\) không lớn hơn \(\dfrac{x^2}{7}-\dfrac{2x-3}{5}\) thì

\(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}< =\dfrac{x^2}{7}-\dfrac{2x-3}{5}\)

=>\(\dfrac{2x-3+5x\left(x-2\right)}{35}< =\dfrac{5x^2-7\cdot\left(2x-3\right)}{35}\)

=>\(2x-3+5x^2-10x< =5x^2-14x+21\)

=>-8x-3<=-14x+21

=>6x<=24

=>x<=4

3) Ta có: \(C=x^2-4x+7=\left(x-2\right)^2+3\ge3\)

Dấu "=" xảy ra khi x = 2

4) Ta có: \(D=2x^2+3x+4=2\left(x^2+\dfrac{3}{2}x+\dfrac{9}{16}\right)+\dfrac{23}{8}=2\left(x+\dfrac{3}{4}\right)^2+\dfrac{23}{8}\ge\dfrac{23}{8}\)

Dấu "=" xảy ra khi \(x=-\dfrac{3}{4}\)

3) \(C=x^2-4x+7\)

\(=\left(x-2\right)^2+3\text{≥}3\) ∀x (vì \(\left(x-2\right)^2\text{≥}0\))

MinC=3 ⇔ x=2

4) \(D=2x^2+3x+4\)

\(=2\left(x+\dfrac{3}{4}\right)^2+\dfrac{23}{8}\text{≥}\dfrac{23}{8}\) ∀x (vì \(2\left(x+\dfrac{3}{4}\right)^2\text{≥}0\))

MinD= \(\dfrac{23}{8}\) ⇔ \(x=-\dfrac{3}{4}\)

\(P=x^4+2x^3+3x^2+2x+1\)

\(=\left(x^4+2x^2+1\right)+\left(2x^3+2x\right)+x^2\)

\(=\left(x^2+1\right)^2+2x\left(x^2+1\right)+x^2\)

\(=\left(x^2+x+1\right)^2\)

giải tiếp : 

Vì \(x^2+x+1=\left(x^2+2x.\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}\)

                            \(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Nên  \(P\ge\left(\frac{3}{4}\right)^2=\frac{9}{16}\)

Dấu "=" xảy ra khi và chỉ khi  \(x=-\frac{1}{2}\)

bạn kacura làm thiếu rồi mình bổ sung tiếp bạn ấy nha:

ta có :P là giá trị nhỏ nhất khi \(x^2+x+1\)là nhỏ nhất( giá trị nhỏ nhất của biểu thức)

\(\Leftrightarrow x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

đẳng thức xảy ra khi : \(x=\frac{-1}{2}\)

\(\Leftrightarrow P\ge\left(\frac{3}{4}\right)^2\)

\(\Leftrightarrow P\ge\frac{9}{16}\)

=> GIÁ TRỊ NHỎ NHẤT CỦA BIỂU THỨC LÀ:\(\frac{1}{2}\)

bài này học nhóm nên mk biết làm:

# chúc bạn học tốt #

Answer:

a) \(\frac{5x}{2x+2}+1=\frac{6}{x+1}\)

\(\Rightarrow\frac{5x}{2\left(x+1\right)}+\frac{2\left(x+1\right)}{2\left(x+1\right)}=\frac{12}{2\left(x+1\right)}\)

\(\Rightarrow5x+2x+2-12=0\)

\(\Rightarrow7x-10=0\)

\(\Rightarrow x=\frac{10}{7}\)

b) \(\frac{x^2-6}{x}=x+\frac{3}{2}\left(ĐK:x\ne0\right)\)

\(\Rightarrow x^2-6=x^2+\frac{3}{2}x\)

\(\Rightarrow\frac{3}{2}x=-6\)

\(\Rightarrow x=-4\)

c) \(\frac{3x-2}{4}\ge\frac{3x+3}{6}\)

\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\ge0\)

\(\Rightarrow9x-6-6x-6\ge0\)

\(\Rightarrow3x-12\ge0\)

\(\Rightarrow x\ge4\)

d) \(\left(x+1\right)^2< \left(x-1\right)^2\)

\(\Rightarrow x^2+2x+1< x^2-2x+1\)

\(\Rightarrow4x< 0\)

\(\Rightarrow x< 0\)

e) \(\frac{2x-3}{35}+\frac{x\left(x-2\right)}{7}\le\frac{x^2}{7}-\frac{2x-3}{5}\)

\(\Rightarrow\frac{2x-3+5\left(x^2-2x\right)}{35}\le\frac{5x^2-7\left(2x-3\right)}{35}\)

\(\Rightarrow2x-3+5x^2-10x\le5x^2-14x+21\)

\(\Rightarrow6x\le24\)

\(\Rightarrow x\le4\)

f) \(\frac{3x-2}{4}\le\frac{3x+3}{6}\)

\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\le0\)

\(\Rightarrow9x-6-6x-6\le0\)

\(\Rightarrow3x\le12\)

\(\Rightarrow x\le4\)