a.

Ta co:

\(\orbr{\begin{cases}x^2-2x-3=0\left(1\right)\left(x\ge0\right)\\x^2+2x-3=0\left(2\right)\left(x< 0\right)\end{cases}}\)

(1)\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\left(l\right)\\x=3\left(n\right)\end{cases}}\)

(2)\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(l\right)\\x=-3\left(n\right)\end{cases}}\)

b.

Ta lai co:

\(\orbr{\begin{cases}x^2-2x+1-4a^2=0\left(3\right)\left(x\ge0\right)\\x^2+2x+1-4a^2=0\left(4\right)\left(x< 0\right)\end{cases}}\)

Xet (3)

De phuong trinh dau co 4 nghiem thi PT(3) co nghiem

\(\Rightarrow\Delta^`>0\)

\(\Leftrightarrow4a^2>0\)

\(\Leftrightarrow a>0\)

\(\Rightarrow x_1=1+2a;x_2=1-2a\)

Tuong tu

(4)

\(a>0\)

\(\Rightarrow x_3=-1+2a;x_4=-1-2a\)

\(\Rightarrow S=\left(1+2a\right)^2+\left(1-2a\right)^2+\left(-1+2a\right)^2+\left(-1-2a\right)^2\)

\(=2\left(1+2a\right)^2+2\left(1-2a\right)^2\)

\(\Rightarrow S< +\infty\)

bài này hay đấy

Áp dụng BĐT Cô-si cho 3 số không âm, ta có :

\(\frac{1+\sqrt{a}}{1+\sqrt{b}}+\frac{1+\sqrt{b}}{1+\sqrt{c}}+\frac{1+\sqrt{c}}{1+\sqrt{a}}\ge3\sqrt[3]{\frac{1+\sqrt{a}}{1+\sqrt{b}}.\frac{1+\sqrt{b}}{1+\sqrt{c}}.\frac{1+\sqrt{c}}{1+\sqrt{a}}}=3\)

Chứng minh \(\frac{1+\sqrt{a}}{1+\sqrt{b}}+\frac{1+\sqrt{b}}{1+\sqrt{c}}+\frac{1+\sqrt{c}}{1+\sqrt{a}}\le3+a+b+c\)( 1 )

đặt \(\sqrt{a}=x;\sqrt{b}=y;\sqrt{c}=z\)( x,y,z \(\ge\)0 )

do a,b,c là số nguyên 

Nếu a = b = c = 0 thì x = y = z = 0 nên ( 1 ) đúng

Nếu a,b,c không đồng thời bằng 0 \(\Rightarrow\)x+ y + z \(\ge\)1

Ta có : VT ( 1 ) 

\(\Leftrightarrow\frac{\left(1+x\right)\left(1+y\right)-\left(1+x\right)y}{1+y}+\frac{\left(1+y\right)\left(1+z\right)-\left(1+y\right)z}{1+z}+\frac{\left(1+z\right)\left(1+x\right)-\left(1+z\right)x}{1+z}\)

\(=3+x+y+z-\left[\frac{\left(1+x\right)y}{1+y}+\frac{\left(1+y\right)z}{1+z}+\frac{\left(1+z\right)x}{1+x}\right]\)

\(\le3+x+y+z-\frac{\left(1+x\right)y+\left(1+y\right)z+\left(1+z\right)x}{1+x+y+z}=3+x+y+z-\frac{x+y+z+xy+yz+xz}{1+x+y+z}\)

\(=3+\frac{x^2+y^2+z^2+xy+yz+xz}{1+x+y+z}\le3+x^2+y^2+z^2\)

Cần chứng minh : \(\frac{x^2+y^2+z^2+xy+yz+xz}{1+x+y+z}\le x^2+y^2+z^2\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2\right)\ge xy+yz+xz\)

Mà \(\left(x+y+z\right)\left(x^2+y^2+z^2\right)\ge1.\left(x^2+y^2+z^2\right)\ge xy+yz+xz\)

suy ra đpcm

Áp dụng bđt AM-GM ta có

\(x^4+y^2\ge2x^2y\)

\(x^2+y^4\ge2xy^2\)

\(\Rightarrow M\le\frac{x}{2x^2y}+\frac{y}{2xy^2}=\frac{1}{2xy}+\frac{1}{2xy}=\frac{1}{xy}=1\)

Dấu "=" xảy ra khi \(x=y=1\)

Vậy..........

Sao không nói x , y , z thuộc N cho nhanh bạn

Áp dụng BĐT Bunhiacopski ta có:

\(\sqrt{x^2+\frac{1}{x^2}}=\frac{1}{\sqrt{17}}\sqrt{\left(x^2+\frac{1}{x^2}\right)\left(4^2+1^2\right)}\ge\frac{1}{\sqrt{17}}\left(4x+\frac{1}{x}\right)\)

Tương tự:

\(\sqrt{y^2+\frac{1}{y^2}}\ge\frac{1}{\sqrt{17}}\left(4y+\frac{1}{y}\right)\)

Cộng lại ta được:

\(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}\ge\frac{1}{\sqrt{17}}\left(4x+4y+\frac{1}{x}+\frac{1}{y}\right)\)

\(\ge\frac{1}{\sqrt{17}}\left[4\left(x+y\right)+\frac{4}{x+y}\right]=\frac{1}{\sqrt{17}}\left(16+1\right)=\sqrt{17}\)

Dấu "=" xảy ra tại x=y=2

A.2

......

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