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a) \(A=\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-3\right)^2}\)
\(=\left|x-1\right|+\left|x-3\right|\ge\left|\left(x-1\right)+\left(3-x\right)\right|=2\)
Vậy\(A_{min}=2\Leftrightarrow\left(x-1\right)\left(3-x\right)\ge0\)
\(TH1:\hept{\begin{cases}x-1\ge0\\3-x\ge0\end{cases}}\Leftrightarrow1\le x\le3\)
\(TH1:\hept{\begin{cases}x-1\le0\\3-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge3\end{cases}}\left(L\right)\)
Vậy \(A_{min}=2\Leftrightarrow1\le x\le3\)
b) Cách làm cũng giống như thế :v
ĐKXĐ: \(x\ge\frac{1}{2}\)
\(PT\Leftrightarrow\left(x-1\right)\left(\frac{4x+6}{\sqrt{2x-1}+1}+\frac{x}{\sqrt{x+3}+2}+x\right)=0\)
\(\Leftrightarrow x=1\) (TMĐK)
a) ĐKXĐ: \(x\ge1\).
\(PT\Leftrightarrow x\left(\sqrt{x-1}-1\right)+\left(2x+1\right)\left(\sqrt{x+2}-2\right)+\left(x^3-4x^2+6x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{x}{\sqrt{x-1}+1}+\frac{2x+1}{\sqrt{x+2}+2}+x^2-2x+2\right)=0\)
\(\Leftrightarrow x=2\left(TMĐK\right)\)
Ta có \(a,\sqrt{9(x-1)}=21 \)
<=> \(3\sqrt{x-1}=21 \)
<=> \(\sqrt{x-1}=7 \)
<=>\(x-1=49\)
<=>x=50
b, \(\sqrt{4(x-1)^2}-6=0 \)
<=>\(2|x-1|-6=0\)
<=>\(|x-1|=3\)
<=>x=4 hoặc x=-2
c,\(\sqrt{(x-5)^2}=8 \)
<=>|x-5|=8
<=>x=-3 hoặc x=13
d,\(\sqrt{(2x-1)^2}=3 \)
<=>|2x-1|=3
=> x=2 hoặc x=-1
e, \(\sqrt{(2x+3)^2}=3 \)
<=>|2x+3|=3
=>x=0 hoặc x=-3
f, \(\sqrt{x^2-4x+4}=2x-3 \)
<=>\(\sqrt{(x-2)^2}=2x-3 \)
<=>|x-2|=2x-3
Với x-2=2x-3
=>x-1=0
<=>x=1
Với 2-x=2x-3
=>x=\(\frac{5}{3}\)
Bài 3: \(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)
\(\Leftrightarrow\left(3-8x\right)\sqrt{2x^2+1}=3x^2+x+3\)
\(\Rightarrow\left(3-8x\right)^2\left(2x^2+1\right)=\left(3x^2+x+3\right)^2\)
\(\Leftrightarrow119x^4-102x^3+63x^2-54x=0\)
\(\Leftrightarrow x\left(7x-6\right)\left(17x^2+9\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{6}{7}\end{cases}}\)
Thử lại, ta nhận được \(x=0\)là nghiệm duy nhất của phương trình
\(x^2+4x=23-10\sqrt{2}\Leftrightarrow x^2+4x+4=\left(x+2\right)^2=27-10\sqrt{2}=25-10\sqrt{2}+2=5^2-5.2\sqrt{2}+\left(\sqrt{2}\right)^2=\left(5-\sqrt{2}\right)^2=\left(\sqrt{2}-5\right)^2\Leftrightarrow\left[{}\begin{matrix}x+2=5-\sqrt{2}\\x+2=\sqrt{2}-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3-\sqrt{2}\\x=\sqrt{2}-7\end{matrix}\right.\)
\(b,+,x>\frac{1}{2}\Rightarrow2x>1\Rightarrow1-2x< 0\Rightarrow\left|1-2x\right|=-\left(1-2x\right)=2x-1\Rightarrow\left(2x-1\right)^2=\left(2x-1\right)\Leftrightarrow\left(2x-1\right)\left(2x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(loai\right)\\x=1\left(thoaman\right)\end{matrix}\right.\)\(+,x\le\frac{1}{2}\Rightarrow2x\le1\Rightarrow1-2x\ge0\Rightarrow\left|1-2x\right|=1-2x\Rightarrow\left(2x-1\right)^2=1-2x\Leftrightarrow2x\left(2x-1\right)=0\Leftrightarrow x\left(2x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(thoaman\right)\\x=\frac{1}{2}\left(thoaman\right)\end{matrix}\right..\)
\(c,Taco:\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(2x+1\right)^2\ge0\end{matrix}\right.mà:\left(x-2\right)^2+\left(2x+1\right)^2=0nên:\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(2x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-\frac{1}{2}\end{matrix}\right.\left(voli\right).Nên:x\in\varnothing\)
cảm ơn rất nhìu