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![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\Leftrightarrow\sqrt{3\left(x-1\right)^2+16}+\sqrt{\left(x-1\right)^2+25}=9-\left(x-1\right)^2\)
Ta có: \(\left(x-1\right)^2\ge0\Rightarrow\sqrt{3\left(x-1\right)^2+16}\ge\sqrt{16}=4\)
\(\sqrt{\left(x-1\right)^2+25}\ge\sqrt{25}=5\)
\(\Rightarrow VT\ge4+5=9\)
\(VP=9-\left(x-1\right)^2\le9\le VT\)
Dấu "=" xảy ra khi và chỉ khi \(\left(x-1\right)^2=0\Leftrightarrow x=1\)
Vậy pt có nghiệm duy nhất \(x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(VT=\sqrt{3x^2-6x+19}+\sqrt{x^2-2x+26}\)
\(=\sqrt{3\left(x-1\right)^2+16}+\sqrt{\left(x-1\right)^2+25}\ge4+5=9\)
\(VP=8-x^2+2x=9-\left(x-1\right)^2\le9\)
Dấu = xảy ra khi \(x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời :
Con a giai pt vế trái rồi nhân căn bình phương cả 2 vế
Con b cũng giải pt vế phải chuyển vế rồi bình phương cả 2 vế
Chắc vậy
k bt
![](https://rs.olm.vn/images/avt/0.png?1311)
c) \(\sqrt{\left(x-2\right)^2}=10\)
\(x-2=10\)
\(x=12\)
d) \(\sqrt{9x^2-6x+1}=15\)
\(\sqrt{\left(3x\right)^2-2.3x.1+1^2}=15\)
\(\sqrt{\left(3x-1\right)^2}=15\)
\(3x-1=15\)
\(3x=16\)
\(x=\dfrac{16}{3}\)
a) \(đk:x\ge0\)
\(pt\Leftrightarrow3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)
\(\Leftrightarrow4\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=3\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)
b) \(đk:x\ge-2\)
\(pt\Leftrightarrow3\sqrt{x+2}+12\sqrt{x+2}-2\sqrt{x+2}=26\)
\(\Leftrightarrow13\sqrt{x+2}=26\)
\(\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)
c) \(pt\Leftrightarrow\left|x-2\right|=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)
d) \(pt\Leftrightarrow\sqrt{\left(3x-1\right)^2}=15\)
\(\Leftrightarrow\left|3x-1\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=15\\3x-1=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{3}\\x=-\dfrac{14}{3}\end{matrix}\right.\)
e) \(đk:x\ge\dfrac{8}{3}\)
\(pt\Leftrightarrow3x+4=9x^2-48x+64\)
\(\Leftrightarrow9x^2-51x+60=0\)
\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)