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a) ĐKXD:...
\(pt\Leftrightarrow\left(\sqrt{x+2}+\sqrt{x-2}\right)^2=6-2x\)
\(\Leftrightarrow\sqrt{x+2}+\sqrt{x-2}=\sqrt{6-2x}\)
Đến đây dễ rồi
c1 cậu đặt cái trong căn =a
=>pt<=> a^2-2x=2xa-a
c2 cậu đưa về dang a^2=b^2
bài 2 nhé
đặt \(a=\sqrt{x+2}\)
ta có pt<=>
\(2a^3=3x\left(x+2\right)-x^3\Leftrightarrow2a^3=3xa^2-x^3\)
\(\Leftrightarrow2a^3-3xa^2+x^3=0\Leftrightarrow2a^3-2a^2x+x^2-xa^2=0\)
\(\Leftrightarrow\left(a-x\right)\left(2a^2-ax-x^2\right)\)
ĐKXĐ: \(-4\le x\le1\)
Đặt \(\sqrt{x+4}-\sqrt{1-x}=t\)
\(\Rightarrow t^2=5-2\sqrt{\left(x+4\right)\left(1-x\right)}\Rightarrow\sqrt{\left(x+4\right)\left(1-x\right)}=\frac{5-t^2}{2}\)
Pt trở thành:
\(t\left(1+\frac{5-t^2}{2}\right)=3\Leftrightarrow t\left(7-t^2\right)=6\)
\(\Leftrightarrow t^3-7t+6=0\Leftrightarrow\left(t+3\right)\left(t-1\right)\left(t-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=-3\\t=1\\t=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x+4}-\sqrt{1-x}=-3\\\sqrt{x+4}-\sqrt{1-x}=1\\\sqrt{x+4}-\sqrt{1-x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+4}+3=\sqrt{1-x}\left(vn\right)\\\sqrt{x+4}=1+\sqrt{1-x}\\\sqrt{x+4}=2+\sqrt{1-x}\end{matrix}\right.\) (1 vô nghiệm do \(VT\ge3;VP\le\sqrt{5}< 3\))
\(\Leftrightarrow\left[{}\begin{matrix}x+4=2-x+2\sqrt{1-x}\\x+4=5-x+4\sqrt{1-x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{1-x}\left(x\ge-1\right)\\2x-1=4\sqrt{1-x}\left(x\ge\frac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1=1-x\\4x^2-4x+1=16-16x\end{matrix}\right.\) \(\Leftrightarrow...\)
a/ ĐKXĐ: \(x\ge\frac{3}{4}\)
\(\Leftrightarrow6x+1+2\sqrt{5x^2+5x}=6x+1+2\sqrt{8x^2+10x-12}\)
\(\Leftrightarrow\sqrt{5x^2+5x}=\sqrt{8x^2+10x-12}\)
\(\Leftrightarrow5x^2+5x=8x^2+10x-12\)
\(\Leftrightarrow3x^2+5x-12=0\Rightarrow\left[{}\begin{matrix}x=-3< \frac{3}{4}\left(l\right)\\x=\frac{4}{3}\end{matrix}\right.\)
b/ \(\Leftrightarrow x^2+x+1+2\sqrt{x^2+x+1}-3=0\)
Đặt \(\sqrt{x^2+x+1}=t>0\)
\(\Rightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+x+1}=1\)
\(\Leftrightarrow x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)