2 replies to this topic

[Kira Tatsuya]

1./ $\begin{cases}(y+1)^2-\sqrt{(3x+2)^3}=1+y\sqrt{-3x-2}-3xy\\x^3+3x^2+12x-(3x-1)y+6=0\end{cases}$

2./ $8x^2-13x+7=\left(1+\frac{1}{x}\right).\sqrt[3]{3x^2-2}$

3./ $\sqrt{1+\sqrt{2x-x^2}}+\sqrt{1-\sqrt{2x-x^2}}=2(x-1)^4(2x^2-4x+1)$

[leminhnghiatt]

2./ $8x^2-13x+7=\left(1+\frac{1}{x}\right).\sqrt[3]{3x^2-2}$

 

$\iff 8x^3-13x^2+7x=(x+1)\sqrt[3]{3x^2-2}$

 

$\iff [(8x^3-7x^2-x)-(8x^2-7x-1)]+[(x+1)(2x-1)-(x+1)\sqrt[3]{3x^2-2}]=0$

 

$\iff (x-1)^2(8x+1)+(x+1)(2x-1-\sqrt[3]{3x^2-2})=0$

 

$\iff (x-1)^2(8x+1)+(x+1)\dfrac{8x^3-15x^2+6x+1}{(2x-1)^2+(2x-1)\sqrt[3]{3x^2-2}+\sqrt[3]{3x^2-2}^2}=0$

 

Đặt $(2x-1)^2+(2x-1)\sqrt[3]{3x^2-2}+\sqrt[3]{3x^2-2}^2=t > 0$

 

$\iff (x-1)^2(8x+1)+\dfrac{(x+1)(8x+1)(x-1)^2}{t}=0$

 

$\iff (x-1)^2(8x+1)(1+\dfrac{x+1}{t})=0$

 

$\iff x=1$  v   $x=\dfrac{-1}{8}$   v   $1+\dfrac{x+1}{t}=0$

 

Xét $1+\dfrac{x+1}{(2x-1)^2+(2x-1)\sqrt[3]{3x^2-2}+\sqrt[3]{3x^2-2}^2}=0$

 

$\iff (2x-1)^2+(2x-1)\sqrt[3]{3x^2-2}+\sqrt[3]{3x^2-2}^2+x+1=0$

 

Đặt $2x-1=a ; \sqrt[3]{3x^2-2}=b$

 

$\iff a^2+ab+b^2+x+1=0$

 

$\iff (\dfrac{a^2}{2}+ab+b^2)+(\dfrac{(2x-1)^2}{2}+x+1)=0$

 

$\iff (\dfrac{a^2}{2}+ab+b^2)+(2x^2-x+\dfrac{1}{2})=0 \ (*)$

 

Ta thấy $\dfrac{a^2}{2}+ab+b^2 >0$ ; $2x^2-x+\dfrac{1}{2} > 0$ nên $(\dfrac{a^2}{2}+ab+b^2)+(2x^2-x+\dfrac{1}{2}) >0$ 

 

Vậy (*) vô nghiệm.

 

PT có nghiệm S={$1;\dfrac{-1}{8}$}

Edited by leminhnghiatt, 11-01-2016 - 15:40.

[Kira Tatsuya]

bạn làm hay thật :3 , giờ mình post lời giải luôn, ngắn hơn đôi chút, chứ mình 1 mình giải không nổi :

$8x^2-13x+7=\left ( 1+\frac{1}{x} \right )\sqrt[3]{3x^2-2}\\\Leftrightarrow 8x^3-13x^2+7x=\left ( 1+x \right )\sqrt[3]{3x^2-2}\\\Leftrightarrow (2x-1)^3-(x^2-x-1)=\left ( x+1 \right )\sqrt[3]{(x+1)(2x-1)+x^2-x-1}$

Đặt $u=2x-1;v=\sqrt[3]{3x^2-2}$, ta có hệ : 

$\begin{cases}u^3-(x^2-x-1)=(x+1).v\\v^3-(x^2-x-1)=(x+1).u\end{cases}\\\Rightarrow (u-v)(u^2+uv+v^2+x+1)=0$

.Đến dây thì dễ rồi 

Edited by Kira Tatsuya, 11-01-2016 - 15:53.