\(1,a+b+c=0\Leftrightarrow a=-b-c\Leftrightarrow a^2=b^2+2bc+c^2\Leftrightarrow b^2+c^2=a^2-2bc\)

Tương tự: \(\left\{{}\begin{matrix}a^2+b^2=c^2-2ab\\c^2+a^2=b^2-2ac\end{matrix}\right.\)

\(\Leftrightarrow N=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ca}+\dfrac{c^2}{c^2-c^2+2ac}\\ \Leftrightarrow N=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{a^3+b^3+c^3-3abc+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)

a) \(A=\dfrac{x\sqrt{y}+y\sqrt{x}}{x+2\sqrt{xy}+y}\)

\(A=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)

\(A=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

b) \(B=\dfrac{x\sqrt{y}-y\sqrt{x}}{x-2\sqrt{xy}+y}\)

\(B=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)

\(B=\dfrac{\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\)

c) \(C=\dfrac{3\sqrt{a}-2a-1}{4a-4\sqrt{a}+1}\)

\(C=\dfrac{-\left(2a-3\sqrt{a}+1\right)}{\left(2\sqrt{a}\right)^2-2\sqrt{a}\cdot2\cdot1+1^2}\)

\(C=\dfrac{-\left(\sqrt{a}-1\right)\left(2\sqrt{a}-1\right)}{\left(2\sqrt{a}-1\right)^2}\)

\(C=\dfrac{-\sqrt{a}+1}{2\sqrt{a}-1}\)

d) \(D=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)

\(D=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}\)

\(D=\sqrt{a}+2-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)

\(D=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)\)

\(D=0\)

\(a,A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\left(x\ge0;x\ne1\right)\\ A=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\\ A=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)

\(b,x+\sqrt{x}+1=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\\ \Rightarrow\dfrac{2}{x+\sqrt{x}+1}>0\left(1\right)\)

\(\sqrt{x}+\dfrac{1}{2}\ge\dfrac{1}{2}\\ \Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2\ge\dfrac{1}{4}\\ \Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge1\\ \Leftrightarrow\dfrac{2}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{2}{1}=2\\ \Rightarrow A< 2\left(2\right)\)

\(\left(1\right)\left(2\right)\Leftrightarrow0< A< 2\)

\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{x^3+1}\)

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

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

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

P = (\(\dfrac{1}{\sqrt{x}-1}\) - \(\dfrac{1}{\sqrt{x}}\)) : (\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\) - \(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)) với  0 < \(x\) ≠ 1; 4

P = \(\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}\): (\(\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right).\left(\sqrt{x-2}\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}\))

P = \(\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\): \(\dfrac{x-1-\left(x-4\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}\)

P = \(\dfrac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\) : \(\dfrac{3}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}\)

P = \(\dfrac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\) \(\times\) \(\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}{3}\)

P = \(\dfrac{\sqrt{x}-2}{3.\sqrt{x}}\)

P = \(\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{3x}\) 

b, P = \(\dfrac{1}{4}\)

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

⇒4\(x\) - 8\(\sqrt{x}\) = 3\(x\)

⇒ 4\(x\) - 8\(\sqrt{x}\) - 3\(x\) = 0

     \(x\) - 8\(\sqrt{x}\)   = 0

      \(\sqrt{x}\).(\(\sqrt{x}\) - 8) = 0

       \(\left[{}\begin{matrix}x=0\\\sqrt{x}=8\end{matrix}\right.\)

      \(\left[{}\begin{matrix}x=0\\x=64\end{matrix}\right.\)

      \(x=0\) (loại)

      \(x\) = 64

Lời giải:

a. \(P=\frac{\sqrt{x}-(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-1)}: \frac{(\sqrt{x}+1)(\sqrt{x}-1)-(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}-1)}\)

\(=\frac{1}{\sqrt{x}(\sqrt{x}-1)}: \frac{x-1-(x-4)}{(\sqrt{x}-2)(\sqrt{x}-1)}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{3}{(\sqrt{x}-1)(\sqrt{x}-2)}\\ =\frac{1}{\sqrt{x}(\sqrt{x}-1)}.\frac{(\sqrt{x}-1)(\sqrt{x}-2)}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)

b.

\(P=\frac{\sqrt{x}-2}{3\sqrt{x}}=\frac{1}{4}\\ \Rightarrow 4(\sqrt{x}-2)=3\sqrt{x}\\ \Leftrightarrow \sqrt{x}=8\Leftrightarrow x=64\) 

(thỏa mãn) 

c.

Tại $x=4+2\sqrt{3}=(\sqrt{3}+1)^2\Rightarrow \sqrt{x}=\sqrt{3}+1$
Khi đó:

$P=\frac{\sqrt{3}+1-2}{3(\sqrt{3}+1)}=\frac{2-\sqrt{3}}{3}$

a) \(C=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\cdot\left(x^2-2x+1\right)\) (ĐK: \(x\ne-\dfrac{5}{3};x\ne3;x\ne-2;x\ne1\))

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

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

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

\(C=\dfrac{2x^2+4x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)

\(C=\dfrac{2\left(x+1\right)^2}{\left(3x+5\right)\left(x-3\right)\left(x+2\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)

\(C=\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}\)

b) Thay x = 2003 ta có: 

\(C=\dfrac{2}{\left(2003-1\right)^4\left(2003-3\right)}=\dfrac{2}{2002^4\cdot2000}=\dfrac{1}{2002^4\cdot1000}\)

c) \(C>0\) khi: 

\(\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}>0\) mà: \(\left\{{}\begin{matrix}2>0\\\left(x-1\right)^4>0\end{matrix}\right.\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\) (đpcm) 

Bài 3:

a) Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{xy}+\frac{2}{x^2+y^2}=2\left(\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)\) \(\geq 2.\frac{(1+1)^2}{2xy+x^2+y^2}=\frac{8}{(x+y)^2}=8\)

Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)

b) Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{xy}+\frac{1}{x^2+y^2}=\frac{1}{2xy}+\left (\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)\geq \frac{1}{2xy}+\frac{(1+1)^2}{2xy+x^2+y^2}\)

\(=\frac{1}{2xy}+\frac{4}{(x+y)^2}\)

Theo BĐT AM-GM:

\(xy\leq \frac{(x+y)^2}{4}=\frac{1}{4}\Rightarrow \frac{1}{2xy}\geq 2\)

Do đó \(\frac{1}{xy}+\frac{1}{x^2+y^2}\geq 2+4=6\)

Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)

Bài 1: Thiếu đề.

Bài 2: Sai đề, thử với \(x=\frac{1}{6}\)

Bài 4 a) Sai đề với \(x<0\)

b) Áp dụng BĐT AM-GM:

\(x^4-x+\frac{1}{2}=\left (x^4+\frac{1}{4}\right)-x+\frac{1}{4}\geq x^2-x+\frac{1}{4}=(x-\frac{1}{2})^2\geq 0\)

Dấu bằng xảy ra khi \(\left\{\begin{matrix} x^4=\frac{1}{4}\\ x=\frac{1}{2}\end{matrix}\right.\) (vô lý)

Do đó dấu bằng không xảy ra , nên \(x^4-x+\frac{1}{2}>0\)

Bài 6: Áp dụng BĐT AM-GM cho $6$ số:

\(a^2+b^2+c^2+d^2+ab+cd\geq 6\sqrt[6]{a^3b^3c^3d^3}=6\)

Do đó ta có đpcm

Dấu bằng xảy ra khi \(a=b=c=d=1\)

Bài 5:

a) Áp dụng BĐT Cauchy-Schwarz:

\(\text{VT}=\frac{a^2}{ab+ac-a^2}+\frac{b^2}{ab+bc-b^2}+\frac{c^2}{ac+bc-c^2}\geq \frac{(a+b+c)^2}{2(ab+bc+ac)-(a^2+b^2+c^2)}\)

Theo hệ quả của BĐT AM-GM ta có:

\(a^2+b^2+c^2\geq ab+bc+ac\Rightarrow \text{VT}\geq \frac{(a+b+c)^2}{ab+bc+ac}\) \((1)\)

Lại có:

\(a^2+b^2+c^2\geq ab+bc+ac\)

\(\Rightarrow (a+b+c)^2\geq 3(ab+bc+ac)\Rightarrow ab+bc+ac\leq \frac{(a+b+c)^2}{3}\) \((2)\)

Từ \((1),(2)\Rightarrow \text{VT}\geq 3\) (đpcm)

Dấu bằng xảy ra khi \(a=b=c\)

b) Để CM \(\frac{1}{a+b},\frac{1}{b+c},\frac{1}{c+a}\) ta cần chỉ ra:

\(\frac{1}{a+b}+\frac{1}{b+c}>\frac{1}{c+a}\), \(\frac{1}{a+b}+\frac{1}{c+a}>\frac{1}{b+c},\frac{1}{a+c}+\frac{1}{b+c}>\frac{1}{a+b}\)

Xét hiệu \(\frac{1}{a+b}+\frac{1}{b+c}-\frac{1}{c+a}=\frac{2b+a+c}{(a+b)(b+c)}-\frac{1}{a+c}=\frac{b(a+c-b)+a^2+c^2}{(a+b)(b+c)(c+a)}\)

Vì \(a,b,c\) là độ dài ba cạnh tam giác nên hiệu trên luôn lớn hơn $0$

Do đó \(\frac{1}{a+b}+\frac{1}{b+c}>\frac{1}{a+c}\)

Hoàn toàn tương tự với các hiệu còn lại, ta thu được đpcm.