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1) \(\sqrt{\text{x^2− 20x + 100 }}=10\)
<=> \(\sqrt{\left(x-10\right)^2}=10\)
<=> \(\left|x-10\right|=10\)
=> \(\left[{}\begin{matrix}x-10=10\\x-10=-10\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=10+10\\x=\left(-10\right)+10\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=20\\x=0\end{matrix}\right.\)
Vậy S = \(\left\{20;0\right\}\)
2) \(\sqrt{x +2\sqrt{x}+1}=6\)
<=> \(\sqrt{\left(\sqrt{x^2}+2.\sqrt{x}.1+1^2\right)}=6\)
<=> \(\sqrt{\left(\sqrt{x}+1\right)^2}=6\)
<=> \(\left|\sqrt{x}+1\right|=6\)
=> \(\left[{}\begin{matrix}\sqrt{x}+1=6\\\sqrt{x}+1=-6\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{x}=6-1=5\\\sqrt{x}=\left(-6\right)-1=-7\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=25\\x=-49\left(loai\right)\end{matrix}\right.\)
Vậy S = \(\left\{25\right\}\)
3) \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)
<=> \(\sqrt{\left(x-3\right)^2}=\sqrt{\sqrt{3^2}+2.\sqrt{3}.1+1^2}\)
<=> \(\left|x-3\right|=\sqrt{\left(\sqrt{3}+1\right)^2}\)
<=> \(\left|x-3\right|=\sqrt{3}+1\)
=> \(\left[{}\begin{matrix}x-3=\sqrt{3}+1\\x-3=-\left(\sqrt{3}+1\right)\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=\sqrt{3}+4\\x=-\sqrt{3}+2\end{matrix}\right.\)
Vậy S = \(\left\{\sqrt{3}+4;-\sqrt{3}+2\right\}\)
4) \(\sqrt{3x+2\sqrt{3x}+1}=5\)
<=> \(\sqrt{\sqrt{3x}^2+2.\sqrt{3x}.1+1^2}=5\)
<=> \(\sqrt{\left(\sqrt{3x}+1\right)^2}=5\)
<=> \(\left|\sqrt{3x}+1\right|=5\)
=> \(\left[{}\begin{matrix}\sqrt{3x}+1=5\\\sqrt{3x}+1=-5\end{matrix}\right.\)=> \(\left[{}\begin{matrix}\sqrt{3x}=5-1=4\\\sqrt{3x}=\left(-5\right)-1=-6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3x=16\\3x=-6\left(loai\right)\end{matrix}\right.\)=> x = \(\dfrac{16}{3}\) Vậy S = \(\left\{\dfrac{16}{3}\right\}\)
5) \(\sqrt{x^2+2x\sqrt{3}+3}=\sqrt{4-2\sqrt{3}}\)
<=> \(\sqrt{\left(x-\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{3}-1\right)^2}\)
<=> \(\left|x-\sqrt{3}\right|=\sqrt{3}-1\)
<=> \(\left[{}\begin{matrix}x-\sqrt{3}=\sqrt{3}-1\\x-\sqrt{3}=-\left(\sqrt{3}-1\right)\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=-1\\x=-2\sqrt{3}+1\end{matrix}\right.\)
Vậy S = \(\left\{-1;-2\sqrt{3}+1\right\}\)
6) \(\sqrt{6x+4\sqrt{6x}+4}=7\)
<=> \(\sqrt{\sqrt{6x}^2+2.\sqrt{6x}.2+2^2}=7\)
<=> \(\sqrt{\left(\sqrt{6}+2\right)^2}=7\)
<=> \(\left|\sqrt{6x}+2\right|=7\)
=> \(\left[{}\begin{matrix}\sqrt{6x}+2=7\\\sqrt{6x}+2=-7\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{6x}=7-2=5\\\sqrt{6x}=\left(-7\right)-2=-9\left(loai\right)\end{matrix}\right.\)
=> \(\sqrt{6x}=5=>6x=25=>x=\dfrac{25}{6}\)
Câu 6:
ĐK: $x\geq 1$
PT $\Leftrightarrow \sqrt{(x-1)-2\sqrt{x-1}+1}-\sqrt{x-1}=1$
$\Leftrightarrow \sqrt{(\sqrt{x-1}-1)^2}=\sqrt{x-1}+1$
$\Leftrightarrow |\sqrt{x-1}-1|=\sqrt{x-1}+1$
Nếu $\sqrt{x-1}-1\geq 0$ thì PT trở thành:
$\sqrt{x-1}-1=\sqrt{x-1}+1\Leftrightarrow 2=0$ (vô lý)
Nếu $\sqrt{x-1}-1< 0$ (tương đương với $1\leq x< 2$ thì PT trở thành:
$1-\sqrt{x-1}=\sqrt{x-1}+1$
$\Leftrightarrow \sqrt{x-1}=0\Rightarrow x=1$ (thỏa mãn)
Vậy PT có nghiệm $x=1$
Câu 5:
ĐK: $x\geq 1$
PT $\Leftrightarrow \sqrt{(x-1)-4\sqrt{x-1}+4}+\sqrt{(x-1)-6\sqrt{x-1}+9}=1$
$\Leftrightarrow \sqrt{(\sqrt{x-1}-2)^2}+\sqrt{(\sqrt{x-1}-3)^2}=1$
$\Leftrightarrow |\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1$
Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ ta có:
$|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=|\sqrt{x-1}-2|+|3-\sqrt{x-1}|\geq |\sqrt{x-1}-2+3-\sqrt{x-1}|=1$
Dấu "=" xảy ra khi $(\sqrt{x-1}-2)(3-\sqrt{x-1})\geq 0$
$\Leftrightarrow 3\geq \sqrt{x-1}\geq 2$
$\Leftrightarrow 10\geq x\geq 5$. Kết hợp ĐKXĐ ta thấy những giá trị $x$ thỏa mãn $10\geq x\geq 5$ là nghiệm của pt.
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\) ( SỬA ĐỀ)
\(\sqrt{x-1-2.2.\sqrt{x-1}+4}+\sqrt{x-1-2.3.\sqrt{x-1}+9}=1\)
\(|x-1-2|+|x-1-3|=1\)
\(|x-3|+|x-4|=1\)
Với \(x\le3\)thì PT thành \(3-x+4-x=1\) \(\Rightarrow-2x=-6\Rightarrow x=3\)(thõa mãn)
Với \(3\le x< 4\)thì PT thành \(x-3+4-x=1\Leftrightarrow0x=0\Rightarrow\)Đúng với mọi x từ \(3\le x< 4\)
Với \(x\ge4\)thì PT thành \(x-3+x-4=1\Leftrightarrow2x=8\Leftrightarrow x=4\)(thõa mãn)
Vậy \(3\le x\le4\)
1)
ĐK: \(x\geq 5\)
PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)
2)
ĐK: \(x\geq -1\)
\(\sqrt{x+1}+\sqrt{x+6}=5\)
\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)
\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)
Vì \(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$
\(\Rightarrow x=3\) (thỏa mãn)
Vậy .............
mầy câu 1;3;;4;5 cách làm nhu nhau(nhân liên hop hoac bình phuong lên)
1.
\(DK:x\in\left[-4;5\right]\)
\(\Leftrightarrow\sqrt{x-5}+\left(\sqrt{x+4}-3\right)=0\)
\(\Leftrightarrow\sqrt{x-5}+\frac{x-5}{\sqrt{x+4}+3}=0\)
\(\Leftrightarrow\sqrt{x-5}\left(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}\right)=0\)
Vi \(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}>0\)
\(\Rightarrow\sqrt{x-5}=0\)
\(x=5\left(n\right)\)
Vay nghiem cua PT la \(x=5\)
2.
\(DK:x\ge0\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}-3\right)^2}=1\)
\(\Leftrightarrow|\sqrt{x}-2|+|\sqrt{x}-3|=1\)
Ta co:
\(|\sqrt{x}-2|+|\sqrt{x}-3|=|\sqrt{x}-2|+|3-\sqrt{x}|\ge|\sqrt{x}-2+3-\sqrt{x}|=1\)
Dau '=' xay ra khi \(\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)\ge0\)
TH1:
\(\hept{\begin{cases}\sqrt{x}-2\ge0\\3-\sqrt{x}\ge0\end{cases}\Leftrightarrow4\le x\le9\left(n\right)}\)
TH2:(loai)
Vay nghiem cua PT la \(x\in\left[4;9\right]\)