a)

<=> f(x) = .

Xét dấu của f(x) ta được tập nghiệm của bất phương trình:

T = ∪ [3; +∞).

b)

<=> f(x) = = .

f(x) không xác định với x = ± 1.

Xét dấu của f(x) cho tập nghiệm của bất phương trình:

T = (-∞; - 1) ∪ (0; 1) ∪ (1; 3).

c) <=> f(x) =

= .

Tập nghiệm: \(\left(-12;-4\right)\cup\left(-3;0\right)\).

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

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

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

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

\(\Leftrightarrow Q=\dfrac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}:\left(\dfrac{\sqrt{1-x}}{\sqrt{x}}-\dfrac{1}{x}\right)\)

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

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

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

a/ \(\left(x+1\right)\left(x-2\right)< 0\)

TH1:\(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\) (vô lý)

TH2:\(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\)\(\Rightarrow-1< x< 2\)

Vậy.........

b/ \(\left(x-3\right)\left(x-4\right)>0\)

TH1:\(\left\{{}\begin{matrix}x-3>0\\x-4>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>3\\x>4\end{matrix}\right.\)\(\Rightarrow x>4\)

TH2:\(\left\{{}\begin{matrix}x-3< 0\\x-4< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< 3\\x< 4\end{matrix}\right.\)\(\Rightarrow x< 3\)

Vậy...............

c/ \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}-\dfrac{1}{8}\)

\(\Rightarrow\dfrac{-1}{12}< x< -\dfrac{5}{48}\)

Vậy...............

Để ( x + 1 ) ( x - 2 ) < 0

=> x + 1 và x - 2 phải khác dấu mà x + 1 > x + 2

=> x + 1 dương x + 2 âm

Tức là x + 1 > 0 => x > - 1 và x - 2 < 0 => x < 2

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\sqrt{3}\)

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

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xy}+\dfrac{2}{yz}+\dfrac{2}{xz}=3\)

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2.\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)=3\)

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

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2.1=3\) ( Do x+y+z=xyz )

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=3-2=1\)

Vậy P = 1

Lời giải:

Vì $0< x< 1$ nên $x; 1-x>0$

Áp dụng BĐT Bunhiacopxky ta có:

\(\left(\frac{1}{x}+\frac{2}{1-x}\right)[x+(1-x)]\geq (1+\sqrt{2})^2\)

\(\Leftrightarrow A.1\geq (1+\sqrt{2})^2\)

\(\Leftrightarrow A\geq (1+\sqrt{2})^2\)

Vậy GTNN của $A$ là \((1+\sqrt{2})^2\). Dấu "=" xảy ra khi \(\frac{1}{x}=\frac{\sqrt{2}}{1-x}\Leftrightarrow x=\sqrt{2}-1\)

Vì \(0< x< \dfrac{1}{2}\) áp dụng BĐT Cauchy-Schwarz dạng Engel

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

\(\ge2.\dfrac{\left(1+1\right)^2}{2x+1-2x}=\dfrac{2.4}{1}=8\)

Đẳng thức xảy ra \(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{1-2x}\Leftrightarrow x=\dfrac{1}{4}\)

a) \(A=\dfrac{1}{\sqrt{x}+1}-\dfrac{3}{x\sqrt{x}+1}+\dfrac{2}{x-\sqrt{x}+1}\)

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

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

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

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

\(A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)

b) Chứng minh \(A\ge0\)

Ta có \(A=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)

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

Mà \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\) và \(\sqrt{x}\ge0\)

\(\Rightarrow A=\dfrac{\sqrt{x}}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\ge0\) (1)

Chứng minh \(A\le1\)

Ta có \(A=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)

\(\Leftrightarrow\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\le1\)

\(\Leftrightarrow\sqrt{x}\le x-\sqrt{x}+1\)

\(\Leftrightarrow2\sqrt{x}\le x+1\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow x+1\ge2\sqrt{x}\) ( luôn đúng với mọi \(x\ge0\) )

Vậy \(A\le1\) (2)

Từ (1) và (2)

\(\Rightarrow0\le A\le1\) ( đpcm )