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

\(x^4\)-2x\(^3\)+3x\(^2\)-2x+2

=(\(x^4\)-2x\(^3\)+x\(^2\))+(2x\(^2\)-2x)+2

=(x\(^2\)-x)\(^2\)+2(x\(^2\)-x)+2

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

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

=[x\(^2\)-2.x.\(\dfrac{1}{2}\)+\(\left(\dfrac{1}{2}\right)^2\)+\(\dfrac{3}{4}\)]\(^2\)+1

=[(x-\(\dfrac{1}{2}\))\(^2\)+\(\dfrac{3}{4}\)]²+1

Ta có:(x-\(\dfrac{1}{2}\))\(^2\)\(\ge0\)

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

=>[(x-\(\dfrac{1}{2}\))\(^2\)+\(\dfrac{3}{4}\)]²\(\ge\dfrac{9}{16}\)

=>[(x-\(\dfrac{1}{2}\))\(^2\)+\(\dfrac{3}{4}\)]²+1\(\ge\dfrac{9}{16}+1\)=\(\dfrac{25}{16}\)

Vậy Min F(x)=\(\dfrac{25}{16}\)khi x-\(\dfrac{1}{2}\)=0=>x=\(\dfrac{1}{2}\)

thắc mắc j hỏi mik nha

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

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

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

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

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  \(x=-\frac{1}{2}\)

bài này lm kiểu j z các bn

Với các số thực không âm a; b ta luôn có BĐT sau:

\(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\) (bình phương 2 vế được \(2\sqrt{ab}\ge0\) luôn đúng)

Áp dụng:

a. 

\(A\ge\sqrt{x-4+5-x}=1\)

\(\Rightarrow A_{min}=1\) khi \(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(A\le\sqrt{\left(1+1\right)\left(x-4+5-x\right)}=\sqrt{2}\) (Bunhiacopxki)

\(A_{max}=\sqrt{2}\) khi \(x-4=5-x\Leftrightarrow x=\dfrac{9}{2}\)

b.

\(B\ge\sqrt{3-2x+3x+4}=\sqrt{x+7}=\sqrt{\dfrac{1}{3}\left(3x+4\right)+\dfrac{17}{3}}\ge\sqrt{\dfrac{17}{3}}=\dfrac{\sqrt{51}}{3}\)

\(B_{min}=\dfrac{\sqrt{51}}{3}\) khi \(x=-\dfrac{4}{3}\)

\(B=\sqrt{3-2x}+\sqrt{\dfrac{3}{2}}.\sqrt{2x+\dfrac{8}{3}}\le\sqrt{\left(1+\dfrac{3}{2}\right)\left(3-2x+2x+\dfrac{8}{3}\right)}=\dfrac{\sqrt{510}}{6}\)

\(B_{max}=\dfrac{\sqrt{510}}{6}\) khi \(x=\dfrac{11}{30}\)

a)Ta có:A=\(\sqrt{x-4}+\sqrt{5-x}\)

        =>A²=\(x-4+2\sqrt{\left(x-4\right)\left(5-x\right)}+5-x\)

        =>A²= 1+\(2\sqrt{\left(x-4\right)\left(5-x\right)}\ge1\)

        =>A\(\ge\)1

Dấu '=' xảy ra <=> x=4 hoặc x=5

Vậy,Min A=1 <=>x=4 hoặc x=5

Còn câu b tương tự nhé

Ta có : \(\left|2x-5\right|+\left|7-2x\right|\ge\left|2x-5+7-2x\right|\forall x\)

\(\Leftrightarrow\left|2x-5\right|+\left|7-2x\right|\ge2\forall x\)

\(\Rightarrow A_{min}=2\)

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