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T sợ chỉ dám liên hợp thôi, nhường cách bình phương cho 1 ng` chăm chỉ :(
\(pt\Leftrightarrow6x+3x\sqrt{9x^2+3}+4x+2+\left(4x+2\right)\sqrt{x^2+x+1}=0\)
\(\Leftrightarrow2\left(5x+1\right)+\left(3x\sqrt{9x^2+3}+\dfrac{6\sqrt{21}}{25}\right)+\left(\left(4x+2\right)\sqrt{x^2+x+1}-\dfrac{6\sqrt{21}}{25}\right)=0\)
\(\Leftrightarrow2\left(5x+1\right)+\dfrac{\dfrac{27}{625}\left(5x-1\right)\left(5x+1\right)\left(75x^2+28\right)}{3x\sqrt{9x^2+3}-\dfrac{6\sqrt{21}}{25}}+\dfrac{\dfrac{4}{625}\left(5x+1\right)\left(5x+4\right)\left(100x^2+100x+109\right)}{\left(4x+2\right)\sqrt{x^2+x+1}+\dfrac{6\sqrt{21}}{25}}=0\)
\(\Leftrightarrow\left(5x+1\right)\left(2+\dfrac{\dfrac{27}{625}\left(5x-1\right)\left(75x^2+28\right)}{3x\sqrt{9x^2+3}-\dfrac{6\sqrt{21}}{25}}+\dfrac{\dfrac{4}{625}\left(5x+4\right)\left(100x^2+100x+109\right)}{\left(4x+2\right)\sqrt{x^2+x+1}+\dfrac{6\sqrt{21}}{25}}\right)=0\)
\(\Rightarrow5x+1=0\Rightarrow x=-\dfrac{1}{5}\)
Lời giải:
ĐKXĐ: $x\geq -1$
Đặt $\sqrt{x+1}=a(a\geq 0)$ thì PT trở thành:
$x^3-3x(x+1)+2\sqrt{(x+1)^3}=0$
$\Leftrightarrow x^3-3xa^2+2a^3=0$
$\Leftrightarrow (x^3-xa^2)-(2xa^2-2a^3)=0$
$\Leftrightarrow x(x-a)(x+a)-2a^2(x-a)=0$
$\Leftrightarrow (x-a)(x^2+ax-2a^2)=0$
$\Leftrightarrow (x-a)[(x+a)(x-a)+a(x-a)]=0$
$\Leftrightarrow (x-a)^2(x+2a)=0$
Nếu $x-a=0$
$\Rightarrow x^2=a^2\Leftrightarrow x^2=x+1$
$\Rightarrow x=\frac{1\pm \sqrt{5}}{2}$. Vì $x=a\geq 0$ nên $x=\frac{1+\sqrt{5}}{2}$
Nếu $x+2a=0$
$\Rightarrow x^2=4a^2\Leftrightarrow x^2=4(x+1)$
$\Rightarrow x=2\pm 2\sqrt{2}$. Mà $x=-2a\leq 0$ nên $x=2-2\sqrt{2}$
Vậy..........
ĐKXĐ: ...
\(\Leftrightarrow x^3-3x\left(x+1\right)+2\sqrt{\left(x+1\right)^3}=0\)
Đặt \(\left\{{}\begin{matrix}x=a\\\sqrt{x+1}=b\end{matrix}\right.\)
\(\Rightarrow a^3-3ab^2+2b^3=0\)
\(\Leftrightarrow\left(a+2b\right)\left(a-b\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}2b=-a\\a=b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=-x\left(x\le0\right)\\x=\sqrt{x+1}\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-4=0\\x^2-x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2-2\sqrt{2}\\x=\frac{1+\sqrt{5}}{2}\end{matrix}\right.\)
ví dụ x âm thì sao căn x2 bằng x được em?
\(\sqrt{3x^2}-\left(1-\sqrt{3}\right)x-1=0\)
\(\Leftrightarrow\sqrt{3}x-x-\sqrt{3}x-1=0\)
\(\Leftrightarrow-x-1=0\)
\(\Leftrightarrow-x=1\)
\(\Leftrightarrow x=-1\)