Câu 2,3,4 nx thôi ạ. Câu 1 có bạn giúp r ạ
1)\(\sqrt{4x^2+12x+9}=2-x\)
\(\Leftrightarrow\sqrt{\left(2x+3\right)^2}=2-x\)
\(\Leftrightarrow\left|2x+3\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2-x\\2x+3=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
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2)\(\sqrt{x^4+2x^2+1}=x^2+5x+4\) ĐK:\(x\ge-1\)
\(\Leftrightarrow\sqrt{\left(x^2+1\right)^2}=x^2+5x+4\)
\(\Leftrightarrow\left|x^2+1\right|=x^2+5x+4\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2+5x+4\\x^2+1=-x^2-5x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\2x^2+5x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\2\left(x+\dfrac{5}{4}\right)^2+\dfrac{15}{8}=0\left(voli\right)\end{matrix}\right.\)
3)\(\sqrt{5x+1}=4\)
\(\Leftrightarrow5x+1=16\Leftrightarrow5x=15\Leftrightarrow x=3\)
4)\(\sqrt{3-x}=7\Leftrightarrow3-x=49\Leftrightarrow x=-46\)
2) \(\sqrt{x^4+2x^2+1}=x^2+5x+4\)
\(\Leftrightarrow x^4+2x^2+1=x^4+2x^2\left(5x+4\right)+\left(5x+4\right)^2\)
\(\Leftrightarrow-10x^3-31x^2-40x-15=0\)
\(\Leftrightarrow x+\dfrac{3}{5}=0\)
\(\Leftrightarrow x=-\dfrac{3}{5}\)
Vậy \(S=\left\{-\dfrac{3}{5}\right\}\)
3) \(\sqrt{5x+1}=4\)
\(\Leftrightarrow5x+1=16\)
\(\Leftrightarrow5x=15\) \(\Leftrightarrow x=3\)
Vậy \(S=\left\{3\right\}\)
4) \(\sqrt{3-x}=7\)
\(\Leftrightarrow3-x=49\)
\(\Leftrightarrow x=-46\)
Vậy \(S=\left\{-46\right\}\)
3) Ta có: \(\sqrt{5x+1}=4\)
\(\Leftrightarrow5x+1=16\)
\(\Leftrightarrow5x=15\)
hay x=3
4) Ta có: \(\sqrt{3-x}=7\)
nên 3-x=49
hay x=-46
