\(i< 3u\)

Ta sẽ chứng minh: \(\sqrt{\frac{x^4+1}{2}}+\frac{4x^2}{x^2+1}\ge3x\)

Thật vậy: \(\Leftrightarrow\left(\sqrt{\frac{x^4+1}{2}}-x\right)+2\left(\frac{2x^2}{x^2+1}-x\right)\ge0\)

\(\Leftrightarrow\left(x-1\right)^2\left[\frac{\left(x+1\right)^2}{2\sqrt{\frac{x^4+1}{2}}+2x}-\frac{2x}{x^2+1}\right]\ge0\)

Bây giờ ta quy về chứng minh: \(\frac{\left(x+1\right)^2}{2\sqrt{\frac{x^4+1}{2}}}\ge\frac{2x}{x^2+1}\Leftrightarrow\left(x^2+1\right)\left(x+1\right)^2\ge4x\left(\sqrt{\frac{x^4+1}{2}+x}\right)\)

\(\Leftrightarrow x^4+1+2x^3+2x\ge2x^2+4x\sqrt{\frac{x^4+1}{2}}\)

\(\Leftrightarrow\frac{x^4+1}{2}+x^3+x\ge x^2+2x\sqrt{\frac{x^4+1}{2}}\)

Bất đẳng thức trên đúng theo AM - GM:

\(\frac{x^4+1}{2}+x^3+x\ge\left(\frac{x^4+1}{2}+x^2\right)+x^2\ge2x\sqrt{\frac{x^4+1}{2}}+x^2\)

Vậy hoàn tất chứng minh trên nên ta có:

\(\sqrt{\frac{a^2+1}{2}}+\frac{4a}{a+1}\ge3\sqrt{a}\);\(\sqrt{\frac{b^2+1}{2}}+\frac{4b}{b+1}\ge3\sqrt{b}\)

\(\sqrt{\frac{c^2+1}{2}}+\frac{4c}{c+1}\ge3\sqrt{c}\); \(\sqrt{\frac{d^2+1}{2}}+\frac{4c}{d+1}\ge3\sqrt{d}\)

Cộng từng vế của các bđt trên. ta được: \(\text{Σ}_{cyc}\sqrt{\frac{a^2+1}{2}}\ge3\left(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\right)\)

\(-4\left(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}\right)\)\(=3\left(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\right)-8\)

Dấu "=" xảy ra khi a = b = c = 1 

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

\(\Leftrightarrow\left(x^3+2x+1\right)+\left(x+2\right)-1-\left(x+2\right).\sqrt{x^3+2x+1}=0\)

Đặt \(\sqrt{x^3+2x+1}=a\left(a\ge0\right)\), \(x+2=b\)Ta có:

\(PT\Leftrightarrow a^2+b-ab-1=0\)

\(\Leftrightarrow\left(a^2-1\right)-\left(ab-b\right)=0\)

\(\Leftrightarrow\left(a+1\right)\left(a-1\right)-b\left(a-1\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(a-b+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a-1=0\\a-b+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=1\\a-b+1=0\end{cases}}}\)

Th1: \(a=1\Leftrightarrow\sqrt{x^2+2x+1}=1\Leftrightarrow x^3+2x+1=1\)\(\Leftrightarrow x^3+2x=0\Rightarrow x=0\)

Th2: \(a-b+1=0\Leftrightarrow\sqrt{x^3+2x+1}-x-2+1=0\)

\(\Leftrightarrow\sqrt{x^3+2x+1}=x+1\)

\(\Leftrightarrow\hept{\begin{cases}x\ge-1\\x^3+2x+1=x^2+2x+1\end{cases}\Rightarrow\orbr{\begin{cases}x=0\left(TM\right)\\x=1\left(TM\right)\end{cases}}}\)

Vậy pt có 2 nghiệm x=0 và x=1

91,35898385

\(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-2\sqrt{25}+\sqrt{300}\right)\)

\(=2\sqrt{3}.\left(\sqrt{9.3}+2\sqrt{16.3}-10+\sqrt{100.3}\right)\)

\(=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-10+10\sqrt{3}\right)\)

\(=2\sqrt{3}.\left(21\sqrt{3}-10\right)\)

\(=126-20\sqrt{3}\)

=> x^2 +6x-8=4x+7

=> x^2+2x-15=0

=> x^2+2.x.1+1-16=0

=> (x+1)^2=4^2

=> x+1=4 hoặc x+1=-4

=> x=3 hoặc x=-5

+, x=3 => y=-17

+, x=-5=> y= -13

ban oi

ĐK: \(x\ge-3\)

Đặt \(t=\sqrt{x+3}\)  \(\left(t\ge0\right)\)  \(\Rightarrow t^2=x+3\)

\(x^2+2x+\sqrt{x+3}+2x\sqrt{x+3}=9\)

\(x^2+x+\left(x+3\right)+t+2xt=12\)

\(t^2+t\left(2x+1\right)+\left(x^2+x-12\right)=0\)

Goi phương trình trên là phương trình bậc 2 ẩn t 

\(\Delta=\left(2x+1\right)^2-4\cdot1\cdot\left(x^2+x-12\right)\)

\(=4x^2+4x+1-4x^2-4x+48=49>0\)

\(\Rightarrow\)Phương trình có hai nghiệm phân biệt 

\(t_1=\frac{-2x-1-\sqrt{49}}{2\cdot1}=\frac{-2x-8}{2}=-x-4\)

\(t_2=\frac{-2x-1+\sqrt{49}}{2}=3-x\)

+) \(t=-x-4\)

\(\Rightarrow\sqrt{x+3}=-x-4\)

ĐK : \(x\le-4\)

Bình phương 2 vế \(\Rightarrow x+3=x^2+8x+16\)

\(x^2+7x+13=0\)

\(\Delta=-3< 0\Rightarrow x\in\varnothing\)

+) \(t=3-x\)

\(\Rightarrow\sqrt{x+3}=3-x\)

ĐK : \(x\le3\)

BÌnh phương 2 vế \(\Rightarrow x+3=9-6x+x^2\)

\(x^2+7x-6=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-7+\sqrt{73}}{2}\left(tm\right)\\x=\frac{-7-\sqrt{73}}{2}\left(ktm\right)\end{cases}}\)

Vậy \(S=\left\{\frac{-7+\sqrt{73}}{2}\right\}\)

ĐK:  \(x\ge0;\)\(x\le-2\)

Đặt: \(\sqrt{x^2+2x}=t>0\)  

Khi đó ta có: 

\(t=-2t^2+3\)

<=>  \(2t^2+t-3=0\)

<=> \(\left(2t+3\right)\left(t-1\right)=0\)

<=> \(\orbr{\begin{cases}t=-1,5\left(L\right)\\t=1\left(t/m\right)\end{cases}}\)

suy ra:  \(\sqrt{x^2+2x}=1\)

<=>  \(x^2+2x-1=0\)

Xét    \(\Delta'=2\) 

=>  \(\orbr{\begin{cases}x=-1+\sqrt{2}\\x=-1-\sqrt{2}\end{cases}}\)(t.m)

Vậy...

\(\sqrt{3-2\sqrt{2}}-\sqrt{6+4\sqrt{2}}\)

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

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

\(=-3\)