\(x+3\sqrt{x}-4=0\)
1) x-\(7\sqrt{x-3}\) -9=0 2) \(\sqrt{x+3}\) =5-\(\sqrt{x-2}\) 3) \(\sqrt{x-4\sqrt{x+4}}\) =3 4) \(\sqrt{8-\dfrac{2}{3}x}-5\sqrt{2}\) =0 5) \(\sqrt{x^2-4x+4}\) =2-x
1, \(\sqrt{x-1}+\sqrt{x-4}=5\)
2, \(2x-7\sqrt{x}+5=0\)
3, \(\sqrt{2x+1}+\sqrt{x-3}=2\sqrt{x}\)
4, \(x-4\sqrt{x}+2021\sqrt{x-4}+4=0\)
5, \(\sqrt{2x-3}-\sqrt{x+1}=7\left(4-x\right)\)
1. ĐKXĐ: $x\geq 4$
PT $\Leftrightarrow \sqrt{x-1}=5-\sqrt{x-4}$
$\Rightarrow x-1=25+x-4-10\sqrt{x-4}$
$\Leftrightarrow 22=10\sqrt{x-4}$
$\Leftrightarrow 2,2=\sqrt{x-4}$
$\Leftrightarrow 4,84=x-4\Leftrightarrow x=8,84$
(thỏa mãn)
2. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow (2x-2\sqrt{x})-(5\sqrt{x}-5)=0$
$\Leftrightarrow 2\sqrt{x}(\sqrt{x}-1)-5(\sqrt{x}-1)=0$
$\Leftrightarrow (\sqrt{x}-1)(2\sqrt{x}-5)=0$
$\Leftrightarrow \sqrt{x}-1=0$ hoặc $2\sqrt{x}-5=0$
$\Leftrightarrow x=1$ hoặc $x=\frac{25}{4}$ (tm)
3. ĐKXĐ: $x\geq 3$
Bình phương 2 vế thu được:
$3x-2+2\sqrt{(2x+1)(x-3)}=4x$
$\Leftrightarrow 2\sqrt{(2x+1)(x-3)}=x+2$
$\Leftrightarrow 4(2x+1)(x-3)=(x+2)^2$
$\Leftrightarrow 4(2x^2-5x-3)=x^2+4x+4$
$\Leftrightarrow 7x^2-24x-16=0$
$\Leftrightarrow (x-4)(7x+4)=0$
Do $x\geq 3$ nên $x=4$
Thử lại thấy thỏa mãn
Vậy $x=4$
4. ĐKXĐ: $x\geq 4$
PT $\Leftrightarrow (x-4\sqrt{x}+4)+2021\sqrt{x-4}=0$
$\Leftrightarrow (\sqrt{x}-2)^2+2021\sqrt{x-4}=0$
Ta thấy, với mọi $x\geq 4$ thì:
$(\sqrt{x}-2)^2\ge 0$
$2021\sqrt{x-4}\geq 0$
Do đó để tổng của chúng bằng $0$ thì:
$\sqrt{x}-2=\sqrt{x-4}=0$
$\Leftrightarrow x=4$ (tm)
Giải phương trình :
a.\(x^2+5x^2-3=0\)
b.\(x^2-\left(2\sqrt{3}-1\right)x+4\sqrt{3}-6=0\)
c.\(x^2-6x+9=0\)
d.\(x^2-4\sqrt{3}x-4=0\)
c: \(\Leftrightarrow x-3=0\)
hay x=3
1, \(K=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
2, \(\sqrt{x-3}-2.\sqrt{x^2-3x}=0\)
3, \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
4, \(x-5\sqrt{x}+4=0\)
1,\(K=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{x}}\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{5}-1\right|+\sqrt{5}+1\right)\)\(=\dfrac{1}{\sqrt{2}}\left|\sqrt{5}-1+\sqrt{5}+1\right|=\dfrac{1}{\sqrt{2}}.2\sqrt{5}\)\(=\sqrt{10}\)
2, \(\sqrt{x-3}-2\sqrt{x^2-3x}=0\left(đk:x\ge3\right)\)
\(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1-2\sqrt{x}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\left(ktm\right)\end{matrix}\right.\)
Vậy pt có nghiệm x=3
3, \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\left(đk:x>-\dfrac{5}{7}\right)\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow x=6\left(tm\right)\)
4, \(x-5\sqrt{x}+4=0\)(đk: \(x\ge0\))
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=16\end{matrix}\right.\) (tm)
Vậy...
1) Bạn tự làm
2) ĐK: \(x\ge3\)
PT \(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\2\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\left(loại\right)\end{matrix}\right.\)
Vậy ...
3) ĐK: \(x>-\dfrac{5}{7}\)
PT \(\Rightarrow9x-7=7x+5\) \(\Leftrightarrow x=6\)
Vậy ...
4) ĐK: \(x\ge0\)
PT \(\Leftrightarrow x-4\sqrt{x}-\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=1\end{matrix}\right.\)
Vậy ...
1.\(\sqrt{-4x^2+25}=x\)
2.\(\sqrt{3x^2-4x+3}=1-2x\)
3. \(\sqrt{4\left(1-x\right)^2}-\sqrt{3}=0\)
4.\(\dfrac{3\sqrt{x+5}}{\sqrt{ }x-1}< 0\)
5. \(\dfrac{3\sqrt{x-5}}{\sqrt{x+1}}\ge0\)
1) x-7\(\sqrt{x-3}\) -9=0 2) \(\sqrt{x+3}\) =5-\(\sqrt{x-2}\) 3) \(\sqrt{x-4\sqrt{x+4}}\) =3
1.
ĐKXĐ: \(x\ge3\)
Đặt \(\sqrt{x-3}=t\ge0\Rightarrow x=t^2+3\)
Pt trở thành:
\(t^2+3-7t-9=0\)
\(\Leftrightarrow t^2-7t-6=0\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{7-\sqrt{73}}{2}< 0\left(loại\right)\\t=\dfrac{7+\sqrt{73}}{2}\end{matrix}\right.\)
\(\Rightarrow\sqrt{x-3}=\dfrac{7+\sqrt{73}}{2}\)
\(\Rightarrow x=\dfrac{67+7\sqrt{73}}{2}\)
Nghiệm xấu quá, em nói giáo viên ra đề kiểm tra lại đề là \(x-7\sqrt{x-3}-9=0\) hay \(x-7\sqrt{x-3}+9=0\) nhé
2.
ĐKXĐ: \(x\ge2\)
\(\sqrt{x+3}+\sqrt{x-2}=5\)
\(\Leftrightarrow2x+1+2\sqrt{\left(x+3\right)\left(x-2\right)}=25\)
\(\Leftrightarrow\sqrt{x^2+x-6}=12-x\) (\(x\le12\))
\(\Rightarrow x^2+x-6=\left(12-x\right)^2\)
\(\Leftrightarrow x^2+x-6=144-24x+x^2\)
\(\Rightarrow x=6\)
Cách 2:
\(\Leftrightarrow\sqrt{x+3}-3+\sqrt{x-2}-2=0\)
\(\Leftrightarrow\dfrac{x-6}{\sqrt{x+3}+3}+\dfrac{x-6}{\sqrt{x-2}+2}=0\)
\(\Leftrightarrow\left(x-6\right)\left(\dfrac{1}{\sqrt{x+3}+3}+\dfrac{1}{\sqrt{x-2}+2}\right)=0\)
\(\Leftrightarrow x=6\)
3.
ĐKXĐ: \(x\ge8+8\sqrt{2}\)
Đặt \(\sqrt{x+4}=t>0\) \(\Rightarrow x=t^2-4\)
Pt trở thành:
\(\sqrt{t^2-4-4t}=3\)
\(\Leftrightarrow t^2-4t-4=9\)
\(\Leftrightarrow t^2-4t-13=0\)
\(\Rightarrow\left[{}\begin{matrix}t=2+\sqrt{17}\\t=2-\sqrt{17}< 0\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+4}=2+\sqrt{17}\)
\(\Leftrightarrow x=17+4\sqrt{17}\)
Như câu 1, em nhờ giáo viên ra đề kiểm tra lại là \(\sqrt{x-4\sqrt{x+4}}=3\) hay \(\sqrt{x-4\sqrt{x-4}}=3\)
1) x-7\(\sqrt{x-3}\) -9=0 2) \(\sqrt{x+3}\)=5-\(\sqrt{x-2}\) 3) \(\sqrt{x-4\sqrt{x+4}}\) =3
1) x-7\(\sqrt{x-3}\) -9=0 2) \(\sqrt{x+3}\)=5-\(\sqrt{x-2}\) 3) \(\sqrt{x-4\sqrt{x+4}}\) =3
2. ĐKXĐ: $x\geq 2$
PT \(\Rightarrow x+3=(5-\sqrt{x-2})^2\)
\(\Leftrightarrow x+3=25+x-2-10\sqrt{x-2}\)
\(\Leftrightarrow 20=10\sqrt{x-2}\Leftrightarrow x-2=4\Leftrightarrow x=6\)
Thử lại thấy thỏa mãn
Vậy $x=6$
3. ĐKXĐ: $x\geq -4$
PT $\Leftrightarrow \sqrt{(x+4)-4\sqrt{x+4}+4}=3$
$\Leftrightarrow \sqrt{(\sqrt{x+4}-2)^2}=3$
$\Leftrightarrow |\sqrt{x+4}-2|=3$
$\Leftrightarrow \sqrt{x+4}-2=\pm 3$. TH $\sqrt{x+4}-2=-3$ loại vì $\sqrt{x+4}-2\geq -2> -3$
Do đó: $\sqrt{x+4}-2=3$
$\Leftrightarrow \sqrt{x+4}=5$
$\Leftrightarrow x+4=25$
$\Leftrightarrow x=21$ (thỏa mãn)
Vậy $x=21$
** Lần sau bạn chú ý ghi đầy đủ yêu cầu của đề.
Lời giải:
1. ĐKXĐ: $x\geq 3$
PT $\Leftrightarrow x-9=7\sqrt{x-3}$
\(\Leftrightarrow \left\{\begin{matrix} x\geq 9\\ (x-9)^2=49(x-3)\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 9\\ x^2-67x+228=0\end{matrix}\right.\Rightarrow x=\frac{67+7\sqrt{73}}{2}\)
Giải các phương trình sau:
1) \(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
2) \(x^2-2x-12+4\sqrt{\left(4-x\right)\left(2+x\right)}=0\)
3) \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}=2x+\dfrac{1}{2x}-7\)
4) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
5)\(\left(x-7\right)\sqrt{\dfrac{x+3}{x-7}}=x+4\)
6) \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)
7) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
Giúp mình với ajk, mink đang cần gấp
a) 1/2 * sqrt(x - 1) - sqrt(4x - 4) + 3 = 0 c) sqrt(7 - x + 1) = x b) sqrt(x ^ 2 - 4x + 4) + x - 2 = 0
a: ĐKXĐ: x>=1
\(\dfrac{1}{2}\sqrt{x-1}-\sqrt{4x-4}+3=0\)
=>\(3+\dfrac{1}{2}\sqrt{x-1}-2\sqrt{x-1}=0\)
=>\(3-\dfrac{3}{2}\sqrt{x-1}=0\)
=>\(\dfrac{3}{2}\sqrt{x-1}=3\)
=>\(\sqrt{x-1}=2\)
=>x-1=4
=>x=5(nhận)
b: \(\sqrt{x^2-4x+4}+x-2=0\)
=>\(\sqrt{\left(x-2\right)^2}=-x+2\)
=>|x-2|=-(x-2)
=>x-2<=0
=>x<=2
c:
ĐKXĐ: 7-x>=0
=>x<=7
\(\sqrt{7-x}+1=x\)
=>\(\sqrt{7-x}=x-1\)
=>\(\left\{{}\begin{matrix}x-1>=0\\7-x=x^2-2x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1< =x< =7\\x^2-2x+1-7+x=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}1< =x< =7\\x^2-x-6=0\end{matrix}\right.\Leftrightarrow x=3\)