Question

tìm x

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

Answer 1

Gửi em

Phương trình tương đương:

\(\begin{gathered} \frac{{{x^2}}}{{x + 2\sqrt {x + 1} + 2}} = x - 4 \hfill \\ \Leftrightarrow \frac{{{x^2}}}{{x + 2\sqrt {x + 1} + 2}} - x + 4 = 0 \hfill \\ \Leftrightarrow \frac{{{x^2} - \left[ {{x^2} + 6x + 2\sqrt {x + 1} \left( {x + 4} \right) + 8} \right]}}{{x + 2\sqrt {x + 1} + 2}} = 0 \hfill \\ \Leftrightarrow \frac{{{x^2} - \left[ {{x^2} + 6x + 8 + 2\sqrt {x + 1} \left( {x + 4} \right)} \right]}}{{x + 2\sqrt {x + 1} + 2}} = 0 \hfill \\ \Leftrightarrow \frac{{{x^2} - \left[ {\left( {x + 4} \right)\left( {x + 2} \right) + 2\sqrt {x + 1} \left( {x + 4} \right)} \right]}}{{x + 2\sqrt {x + 1} + 2}} = 0 \hfill \\ \Leftrightarrow \frac{{{x^2} - \left[ {\left( {x + 4} \right)\left( {x + 2 + 2\sqrt {x + 1} } \right)} \right]}}{{x + 2\sqrt {x + 1} + 2}} = 0 \hfill \\ \Leftrightarrow {x^2} - x - 4 = 0 \hfill \\ \Leftrightarrow x = \frac{{1 \pm \sqrt {17} }}{2} \hfill \\ \end{gathered} \)

Answer 2

ĐK \(x\ge-1\)

Pt<=> \(\left(\frac{x+1-1}{\sqrt{x+1}+1}\right)^2=x-4\)

<=> \(\left(\frac{\left(\sqrt{x+1}-1\right)\left(\sqrt{x+1}+1\right)}{\sqrt{x+1}+1}\right)^2=x-4\)

<=>\(\left(\sqrt{x+1}-1\right)^2=x-4\)

<=> \(x+2-2\sqrt{x+1}=x-4\)

=> \(\sqrt{x+1}=3\)=> \(x=8\)(tm ĐK)

Vậy x=4

Answer 3

Gửi lại:

ĐKXĐ: \(x\ge-1\)

Phương trình tương đương:

\(\begin{array}{l} {\left( {\dfrac{x}{{\sqrt {x + 1} + 1}}} \right)^2} = x - 4\\ \Leftrightarrow x - 2\sqrt {x + 1} + 2 - x + 4 = 0\\ \Leftrightarrow - 2\sqrt {x + 1} + 6 = 0\\ \Leftrightarrow - 2\sqrt {x + 1} = - 6\\ \Leftrightarrow \sqrt {x + 1} = 3\\ \Leftrightarrow {\left( {\sqrt {x + 1} } \right)^2} = {3^2}\\ \Leftrightarrow x + 1 = 9\\ \Leftrightarrow x = 8\left( {TM} \right) \end{array}\)