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

Lời giải:
a.

\(A=\frac{\sqrt{x}(\sqrt{x^3}-1)}{x+\sqrt{x}+1}-\frac{\sqrt{x}(2\sqrt{x}+1)}{\sqrt{x}}+\frac{2(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}-1}\)

\(=\frac{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1}-(2\sqrt{x}+1)+2(\sqrt{x}+1)\)

\(=\sqrt{x}(\sqrt{x}-1)-2\sqrt{x}-1+2\sqrt{x}+2\\ =x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\\ =x-\sqrt{x}+1\)

b.

$A=x-\sqrt{x}+1=(x-\sqrt{x}+\frac{1}{4})+\frac{3}{4}$

$=(\sqrt{x}-\frac{1}{2})^2+\frac{3}{4}\geq 0+\frac{3}{4}=\frac{3}{4}$

$\Rightarrow A_{\min}=\frac{3}{4}$

Giá trị này đạt tại $\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}$

\(A=\dfrac{2x^2-8x+17}{x^2-2x+1}\left(x\ne1\right)\)

\(\Leftrightarrow A\left(x^2-2x+1\right)=2x^2-8x+17\)

\(\Leftrightarrow Ax^2-2Ax+A=2x^2-8x+17\)

\(\Leftrightarrow x^2\left(A-2\right)-2x\left(A-4\right)+A-17=0\left(1\right)\)

\(A-2=0\Leftrightarrow A=2\Leftrightarrow x=3,75\left(tm\right)\left(2\right)\)

\(A-2\ne0\Leftrightarrow A\ne2\Rightarrow\Delta'\ge0\Leftrightarrow\left(A-4\right)^2-\left(A-17\right)\left(A-2\right)\ge0\Leftrightarrow A\ge\dfrac{18}{11}\Rightarrow A_{min}=\dfrac{18}{11}\Leftrightarrow x=\dfrac{13}{2}\left(tm\right)\left(3\right)\)

\(\left(2\right)và\left(3\right)\Rightarrow A_{min}=\dfrac{18}{11}\Leftrightarrow x=\dfrac{13}{2}\)

ĐKXĐ : \(x\ne0;x\ne\pm1\)

a) Bạn ghi lại rõ đề.

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

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

c) \(P=A.B=\dfrac{x^2+x-2}{x.\left(x-1\right)}=\dfrac{\left(x-1\right).\left(x+2\right)}{x\left(x-1\right)}=\dfrac{x+2}{x}=1+\dfrac{2}{x}\)

Không tồn tại Min P \(\forall x\inℝ\)

Bài 1: 

Ta có: \(D=\sqrt{16x^4}-2x^2+1\)

\(=4x^2-2x^2+1\)

\(=2x^2+1\)

1:

ĐKXĐ: \(x\notin\left\{3;-2;1\right\}\)

 \(A=\left(\dfrac{x\left(x+2\right)-x+1}{\left(x-3\right)\left(x+2\right)}\right):\left(\dfrac{x\left(x-3\right)+5x+1}{\left(x+2\right)\left(x-3\right)}\right)\)

\(=\dfrac{x^2+2x-x+1}{\left(x-3\right)\left(x+2\right)}\cdot\dfrac{\left(x+2\right)\left(x-3\right)}{x^2-3x+5x+1}\)

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