Có: \(2a=AE+AF+EF=AE+AF+\sqrt{AE^2+AF^2}\ge\sqrt{AE.AF}\left(2+\sqrt{2}\right)\)

\(\Leftrightarrow\)\(\sqrt{AE.AF}\le\frac{2a}{2+\sqrt{2}}\)\(\Leftrightarrow\)\(\frac{1}{2}AE.AF=\frac{a^2}{3+2\sqrt{2}}\) không đổi 

Dấu "=" xảy ra khi \(AE=AF\)\(\Rightarrow\)\(\hept{\begin{cases}2AE+EF=2a\\2AE^2=EF^2\end{cases}}\Leftrightarrow\hept{\begin{cases}EF=2a-2AE\\EF=\sqrt{2}AE\end{cases}}\)

\(\Rightarrow\)\(2a-2AE=\sqrt{2}AE\)\(\Leftrightarrow\)\(AE=\frac{2a}{2+\sqrt{2}}\)

\(\left(x^2-4x\right)^2-\left(x-2\right)^2+10=0.\)

\(\left(x-2\right)=t\Leftrightarrow x=t+2\)

thay x=t+2 ta được

\(\left\{\left(t+2\right)^2-4\left(t+2\right)\right\}^2-t^2+10=0\)

\(\left(t^2-4\right)^2-t^2+10=0\)

\(t^4-8t^2+16-t^2+10=0\)

\(t^4-9t^2+26=0\)

\(t^2=m\) 

\(m^2-9m+26=0\)

\(\left(m^2-9m+\frac{81}{4}\right)-\left(\frac{81}{4}+26\right)=0\)

\(\left(m-\frac{9}{2}\right)^2-\left(\frac{81}{4}+26\right)=0\)

\(\left(m-\frac{9}{2}+\sqrt{\frac{81}{4}+26}\right)\left(m-\frac{9}{2}-\sqrt{\frac{81}{4}+26}\right)=0\)

1.     \(\frac{x^3-10x^2+25x}{x^2-5x}\)\(=0\)  ( đkxđ: \(x\ne0;5\))

<=> \(\frac{x\left(x-5\right)^2}{x\left(x-5\right)}=0\)<=>   \(x-5=0\)<=> vô n_o

_{2.       \(A=\)\(\frac{2x^2-2}{x^3-x^2-4x+4}\)\(=\frac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x-2\right)\left(x+2\right)}\)}   ( a, đkxđ: \(x\ne1;\pm2\))

b, \(A=0\)<=> \(\frac{2\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}=0\)<=> \(x=-1\)( TM) . Vậy \(A=0\Leftrightarrow x=-1\)

3.    \(B=\frac{3x^2-12}{\left(x-3\right)\left(x^2+4x+4\right)}\)\(=\frac{3\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x+2\right)^2}\) ( a,  đkxđ: \(x\ne3,-2\))

\(b,B=0\Leftrightarrow\frac{3\left(x-2\right)}{\left(x-3\right)\left(x+2\right)}=0\Leftrightarrow x=2\left(tm\right)\). Vậy \(B=0\Leftrightarrow x=2\)

giải pt , \(\sqrt{x^4+4x^2}+\sqrt{x+x^2}=\sqrt{\left(x^2+\sqrt{x}\right)^2+9x^2}.\)

\(x=0\)

\(x^3=0\)

\(x^3=2.0.\sqrt{0}\)

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

\(x^3=2x\sqrt{x}\)     

\(4\left(x^3-2x\sqrt{x}\right)^2=0\)

\(4\left(x^6-4x^4\sqrt{x}+4x^2x\right)=0\)

\(4x^6-16x^4\sqrt{x}+16x^2x=0\)

\(4x^6+16x^3=16x^4\sqrt{x}\)

\(16x^4+4x^5+4x^6+16x^3=16x^4+4x^5+16x^4\sqrt{x}\)

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

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

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

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

\(\left(\sqrt{x^4+4x^2}+\sqrt{x^2+x}\right)^2=\left(x^4+2x^2\sqrt{x}+x\right)+9x^2\)

\(\sqrt{x^4+4x^2}+\sqrt{x^2+x}=\sqrt{\left(x^2+\sqrt{x}\right)^2+9x^2}\)

vậy x=0 là nghiệm của pt  =)) 

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

Đặt t = \(\frac{1}{x}\)=> B= 1-4t+t²= t²-4t+1 = t²-2.t.2+4-3= (t-2)²-3 \(\ge\)-3 

Min của A= -3 khi t-2= 0 => t=2 => \(\frac{1}{x}\)= 2 => x = \(\frac{1}{2}\)

Cho a,b,c>0 và a+b+c=3. CMR: \(a^2+b^2+c^2\ge3\)

Ta cần tìm m, n để bđt sau luôn đúng \(a^2\ge ma+n\) (1) 

tương tự: \(b^2\ge mb+n;c^2\ge mc+n\)

cộng 3 bđt lại ta đc: \(a^2+b^2+c^2\ge m\left(a+b+c\right)+3n=3m+3n\)

dự đoán cực trị xảy ra tại a=b=c=1 nên \(3m+3n=\left(a^2+b^2+c^2\right)_{min}=3\)\(\Rightarrow\)\(n=1-m\)

thay n=1-m vào (1) : \(a^2\ge ma-m+1\)(2)\(\Leftrightarrow\)\(\left(a-1\right)\left(a+1\right)\ge m\left(a-1\right)\)

đồng nhất hệ số : \(a+1=m\)\(\Leftrightarrow\)\(m=a+1=1+1=2\) (dấu "=" xảy ra tại a=1)

thay m=2 vào (2) ta có bđt cần CM: \(a^2\ge2a-1\) ( với \(0< a< 3\) ) 

bđt \(\Leftrightarrow\)\(\left(a-1\right)^2\ge0\) luôn đúng 

do đó: \(a^2+b^2+c^2\ge2a-1+2b-1+2c-1=2\left(a+b+c\right)-3=2.3-3=3\)

dấu "=" xảy ra khi a=b=c=1

A =5+ 5²+5³+..................................................+5¹⁰⁰

B= 5+5³ +5⁵ +...................................................+5⁹⁹

C= 1+ \(\frac{1}{3}\) + \(\frac{1}{3^2}\) +....................................+ \(\frac{1}{3^{100}}\)

D= 1-\(\frac{1}{3}\)+ \(\frac{1}{3^2}\) - ..........................................- \(\frac{1}{3^{99}}\)+\(\frac{1}{3^{100}}\)

E=\(\frac{1}{2}\)+\(\frac{2}{2^2}\)+\(\frac{3}{2^3}\) +................................... +\(\frac{100}{2^{100}}\)

\(\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}\)

\(VP=\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\le\frac{a+b}{2}\sqrt{2\left(a+b\right)}\)\(\Rightarrow\)\(VP^2\le\frac{\left(a+b\right)^3}{2}\) (1) 

chứng minh bổ đề: \(VT^2=\left(\frac{\left(a+b\right)^2}{2}+\frac{a+b}{4}\right)^2\ge\frac{\left(a+b\right)^3}{2}\)

\(\Leftrightarrow\)\(\frac{\left(a+b\right)^4}{4}+\frac{\left(a+b\right)^2}{16}+\frac{\left(a+b\right)^3}{4}\ge\frac{\left(a+b\right)^3}{2}\)

\(\Leftrightarrow\)\(\left(a+b\right)^4+\frac{\left(a+b\right)^2}{4}\ge\left(a+b\right)^3\)

Có: \(\left(a+b\right)^4+\frac{\left(a+b\right)^2}{4}\ge2\sqrt{\frac{\left(a+b\right)^6}{4}}=\left(a+b\right)^3\)\(\Rightarrow\)\(VT^2\ge\frac{\left(a+b\right)^3}{2}\) (2) 

(1) và (2) => \(VT^2\ge VP^2\) => \(VT\ge VP\) ( đpcm )