CHU HOANG TRUNG

$\frac{x}{y+z+t}+\frac{y}{z+t+x}+\frac{z}{t+x+y}+\frac{t}{x+y+z}+\frac{y+z+t}{x}+\frac{z+t+x}{y}+\frac{t+x+y}{z}+\frac{x+y+z}{t}$

$=(\sum \frac{x}{y+z+t}+\sum \frac{y+z+t}{9x})+\frac{8}{9}(\sum \frac{y+z+t}{x})$

$=(\sum \frac{x}{y+z+t}+\sum \frac{y+z+t}{9x})+\frac{8}{9}(\frac{y}{x}+\frac{z}{x}+\frac{t}{x}+\frac{z}{y}+\frac{t}{y}+\frac{x}{y}+\frac{t}{z}+\frac{x}{z}+\frac{y}{z}+\frac{x}{t}+\frac{y}{t}+\frac{z}{t})\geq \frac{8}{3}+\frac{8}{9}.12=13\frac{1}{3}$

 

P/s: Đề sai thì phải 

Đề chuyên Lí vào 10 1995-1996 của Lam Sơn Thanh Hóa

1/Cho 3 số a,b,c đôi một khác nhau .CMR 

 $\frac{(a+b)^2}{a-b}+\frac{(b+c)^2}{b-c}+\frac{(c+a)^2}{c-a}\geq 2$

2/$\frac{a^3+b^3}{2ab}+\frac{b^3+c^3}{2bc}+\frac{c^3+a^3}{2ca}\geq a+b+c$

Với a,b,c>0

1.Pt tương đương (x-2)(2x-1)(2x^2+11x+2)=0

2.pt có nghiệm là :$x_1=2+\sqrt{5};x_2=2-\sqrt{5};x_3=\frac{\sqrt{5}-1}{2};x_4=\frac{-\sqrt{5}-1}{2}$

3.pt <=>$(x-2)(x+2)(x^2-x-1)=0$

187/$ \left\{\begin{matrix} \sqrt{x}+\sqrt{1998-y}=\sqrt{1998} & & \\ \sqrt{1998-x}+\sqrt{y}=\sqrt{1998} & & \end{matrix}\right.$

188/ $ \left\{\begin{matrix} x^{2}+xy +y^{2}=7 & & \\ x^{4}+x^{2}y^{2}+y^{4}=21 & & \end{matrix}\right.$

189/$\left\{\begin{matrix} x^{2}+\frac{1}{y^{2}}+\frac{x}{y}=3 & & \\ x+\frac{1}{y}+\frac{x}{y}=3 & & \end{matrix}\right.$

190/ $ \left\{\begin{matrix} 2x^{2}-y^{2}=1 & & \\ xy + x^{2}=2 & & \end{matrix}\right.$

191/ $ \left\{\begin{matrix} x+xy+y=1 & & \\ x+yz+z=3 & & \\ z+zx+x=7 & & \end{matrix}\right.$

192/ $  \left\{\begin{matrix} x+y=3 & & \\ xz+yt=4 & & \\ xz^{2}+yt^{2}=6 & & \\ xz^{3}+yt^{3}=10 & & \end{matrix}\right.$

193/$\left\{\begin{matrix} x^{3}(y^{2}+3y+3)=3y^{2} & & \\ y^{3}(z^{2}+3z+3)=3z^{2} & & \\ z^{3}(x^{2}+3x+3)=3x^{2} \end{matrix}\right.$

194/$\left\{\begin{matrix} x+y=z^{2} & & \\ x=2(y+z) & & \\ xy=2(z+1)\end{matrix}\right.$

195/ $\left\{\begin{matrix} xy=x+3y & & \\ yz=2(2y+z) & & \\ zx=3(3z+2x) \end{matrix}\right.$

196/$\left\{\begin{matrix} (x+y+z)^2=12t & & \\ (y+z+t)^2=12x & & \\ (z+t+x)^3=12y & & \\ (t+x+y)^2=12z & & \end{matrix}\right.$

197/$\left\{\begin{matrix} x^3=2y-x & & \\ y^3=2x-y & & \end{matrix}\right.$

198/ $\left\{\begin{matrix} x-y=(\sqrt{y}-\sqrt{x})(1+xy) & & \\ x^3+y^3=54 & & \end{matrix}\right.$

199/$\left\{\begin{matrix} (x+y)-\sqrt{\frac{x+y}{x-y}}=\frac{12}{x-y} & & \\ xy=15 & & \end{matrix}\right.$

61/

pttt: $\sqrt{x}(\sqrt{3x+1}-\sqrt{x-1}-2\sqrt{x})=0$

<=> $\sqrt{x}=0$ hoặc $\sqrt{3x+1}-\sqrt{x-1}-2\sqrt{x}=0$

<=> x=0 (tm) hoặc $\sqrt{3x+1}-\sqrt{x-1}-2\sqrt{x}=0$(1)

 ĐK :$x\geq 1$

(1)<=> 

 $\sqrt{3x+1}=\sqrt{x-1}+2\sqrt{x}$

<=>$ 3x+1=x-1+4\sqrt{(x-1)x}+4x$

<=> $x-1+2\sqrt{(x-1)x}$=0

<=>$(\sqrt{x-1}+\sqrt{x})^{2}=x$

<=>$(\sqrt{x-1})(\sqrt{x-1}+2\sqrt{x})=0$

<=>$\sqrt{x-1}=0$hoặc $\sqrt{x-1}=-2\sqrt{x}$(vô lí)

           Vậy x=0 hoặc x=1