Câu hỏi của Nguyễn.T.Huệ

Question

giải phương trình : căn bậc 4(1-x)+căn bậc 4(x+15)=2

Nguyen · Accepted Answer

$\sqrt[4]{1-x}+\sqrt[4]{x+15}=2$(ĐK: $-15\le x\le1$)

$\Leftrightarrow\dfrac{\sqrt{1-x^2}}{\sqrt[4]{1+x}}+\dfrac{\sqrt{x+15}-4}{\sqrt[4]{x+15}+2}=0$

$\Leftrightarrow\dfrac{1-x^2}{\sqrt{1+x}}=\dfrac{x-1}{\left(\sqrt[4]{x+15}+2\right)^2}$

$\Leftrightarrow\left(x-1\right)\left[\dfrac{1}{\left(\sqrt[4]{x+15}+2\right)^2}-\dfrac{\left(1+x\right)}{\sqrt{1-x}}\right]=0$

$\Leftrightarrow x=1$(Vì $\dfrac{1}{\left(\sqrt[4]{x+15}+2\right)^2}-\dfrac{1+x}{\sqrt{1-x}}$$\ne0$

Vậy pt có nghiệm là x=1.

Câu trả lời:

$\sqrt[4]{1-x}+\sqrt[4]{x+15}=2$(ĐK: $-15\le x\le1$)

$\Leftrightarrow\dfrac{\sqrt{1-x^2}}{\sqrt[4]{1+x}}+\dfrac{\sqrt{x+15}-4}{\sqrt[4]{x+15}+2}=0$

$\Leftrightarrow\dfrac{1-x^2}{\sqrt{1+x}}=\dfrac{x-1}{\left(\sqrt[4]{x+15}+2\right)^2}$

$\Leftrightarrow\left(x-1\right)\left[\dfrac{1}{\left(\sqrt[4]{x+15}+2\right)^2}-\dfrac{\left(1+x\right)}{\sqrt{1-x}}\right]=0$

$\Leftrightarrow x=1$(Vì $\dfrac{1}{\left(\sqrt[4]{x+15}+2\right)^2}-\dfrac{1+x}{\sqrt{1-x}}$$\ne0$

Vậy pt có nghiệm là x=1.