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4,ĐK :\(\forall x\in R\)
Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))
\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)
\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)
\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy ....
a) \(\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=-x^2+6x-5\) (ĐKXĐ : \(1\le x\le5\) )\
Ta có : \(\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=\sqrt{3\left(x^2-6x+9\right)+1}+\sqrt{4\left(x^2-6x+9\right)+9}=\sqrt{3\left(x-3\right)^2+1}+\sqrt{4\left(x-3\right)^2+9}\)
\(\Rightarrow\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}\ge1+3=4\)
Lại có : \(-x^2+6x-5=-\left(x^2-6x+9\right)+4=-\left(x-3\right)^2+4\le4\)
Do đó, phương trình tương đương với : \(\begin{cases}1\le x\le5\\\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=4\\-x^2+6x-5=4\end{cases}\)\(\Rightarrow x=3\left(TM\right)\)
Vậy nghiệm của phương trình là x = 3
b) \(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2+1}+\sqrt{\left(x-2\right)^2+4}+\sqrt{\left(x-2\right)^2+5}=3+\sqrt{5}\)
Mặt khác, ta có : \(\begin{cases}\sqrt{\left(x-2\right)^2+1}\ge1\\\sqrt{\left(x-2\right)^2+4}\ge2\\\sqrt{\left(x-2\right)^2+5}\ge\sqrt{5}\end{cases}\)\(\Rightarrow\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}\ge3+\sqrt{5}\)
Dấu đẳng thức xảy ra <=> x = 2.
Vậy nghiệm của phương trình : x = 2
a,ĐK: x≥4
Ta có: \(2\sqrt{x-4}-\dfrac{1}{3}\sqrt{9x-36}=4-\sqrt{x-4}\)
\(\Leftrightarrow2\sqrt{x-4}-\sqrt{x-4}=4-\sqrt{x-4}\)
\(\Leftrightarrow2\sqrt{x-4}=4\)
\(\Leftrightarrow\sqrt{x-4}=2\Leftrightarrow x-4=4\Leftrightarrow x=8\left(tm\right)\)
b, ĐK: x≥2
Ta có: \(3\sqrt{x-2}-\sqrt{x^2-4}=0\)
\(\Leftrightarrow3\sqrt{x-2}-\sqrt{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(3-\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\3-\sqrt{x+2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x+2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=7\end{matrix}\right.\)
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) \(đk:x\ge\dfrac{8}{3}\)
\(pt\Leftrightarrow3x+4=9x^2-48x+64\)
\(\Leftrightarrow9x^2-51x+60=0\)
\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)
a,
\(\sqrt{9-12x+4x^2}=4\\ \sqrt{\left(3-2x\right)^2}=4\\ \left|3-2x\right|=4\\ \Rightarrow\left[{}\begin{matrix}3-2x=4\\3-2x=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=-1\\2x=7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{7}{2}\end{matrix}\right.\)
đặt S=vế trái
ta có:S=\(\sqrt{3\left(x^2-6x+9\right)+1}+\sqrt{4\left(x^2-6x+9\right)+9}\)
S=\(\sqrt{3\left(x-3\right)^2+1}+\sqrt{4\left(x-3\right)^2+9}\)
ta thấy:\(\sqrt{3\left(x-3\right)^2+1}\ge\sqrt{1}=1\);\(\sqrt{4\left(x-3\right)^2+9}\ge\sqrt{9}=3\)
→S\(\ge\)4; xét vế phải :\(-5-x^2+6x=4-\left(x-3\right)^2\)\(\le\)4
vậy pt xảy ra khi x-3=0↔x=3
(đề là -5 -x2+6x thì khả nghi hơn)
Lời giải:
a. ĐKXĐ: $x\geq 0$
$2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28$
$\Leftrightarrow 2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28$
$\Leftrightarrow 13\sqrt{2x}=28$
$\Leftrightarrow \sqrt{2x}=\frac{28}{13}$
$\Leftrightarrow 2x=\frac{784}{169}$
$\Leftrightarrow x=\frac{392}{169}$
b. ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$
$\Leftrightarrow x-5=4$
$\Leftrightarrow x=9$ (tm)
c. ĐKXĐ: $x\geq \frac{2}{3}$ hoặc $x< -1$
PT $\Leftrightarrow \frac{3x-2}{x+1}=9$
$\Rightarrow 3x-2=9(x+1)$
$\Leftrightarrow x=\frac{-11}{6}$ (tm)