\(\sqrt{x}\)-\(\sqrt{x+1}\)-\(\sqrt{x+4}\)+\(\sqrt{x+9}\)=0
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28 tháng 7 2023
a: =>2*căn x+5+căn x+5-1/3*3*căn x+5=4
=>2*căn(x+5)=4
=>căn (x+5)=2
=>x+5=4
=>x=-1
b: =>\(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)
=>2*căn x-1=16
=>x-1=64
=>x=65
HQ
Hà Quang Minh
Giáo viên
28 tháng 7 2023
c, \(\sqrt{\left(x-3\right)^2}-2\sqrt{\left(x-1\right)^2}+\sqrt{x^2}=0\\ \Leftrightarrow\left|x-3\right|-2\left|x-1\right|+\left|x\right|=0\left(1\right)\)
TH1: \(x\ge3\)
\(\left(1\right)\Rightarrow x-3-2x+2+x=0\\ \Leftrightarrow-1=0\left(loại\right)\)
TH2: \(2\le x< 3\)
\(\left(1\right)\Rightarrow3-x-2x+2+x=0\\ \Leftrightarrow-2x=-5\\ \Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)
TH3: \(0\le x< 2\)
\(\left(1\right)\Rightarrow3-x+2x-2+x=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
TH4: \(x< 0\)
\(\left(1\right)\Rightarrow3-x+2x-2-x-=0\\ \Leftrightarrow1=0\left(loại\right)\)
Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}\)
7 tháng 10 2021
c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=4\)
hay x=5
e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)
\(\Leftrightarrow\left|2x-7\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=5\\2x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
8 tháng 10 2021
a. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{(x-2)^2}=2-x$
$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$
$\Leftrightarrow x\leq 2$
b. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 2\sqrt{x-2}-\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 1=2\sqrt{x-2}$
$\Leftrightarrow \frac{1}{2}=\sqrt{x-2}$
$\Leftrightarrow \frac{1}{4}=x-2$
$\Leftrightarrow x=\frac{9}{4}$ (tm)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 7 2021
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
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 7 2021
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:
\(\Leftrightarrow\sqrt{x+3}-3+\sqrt{x-2}-2=0\)
\(\Leftrightarrow\dfrac{x-6}{\sqrt{x+3}+3}+\dfrac{x-6}{\sqrt{x-2}+2}=0\)
\(\Leftrightarrow\left(x-6\right)\left(\dfrac{1}{\sqrt{x+3}+3}+\dfrac{1}{\sqrt{x-2}+2}\right)=0\)
\(\Leftrightarrow x=6\)
AH
Akai Haruma
Giáo viên
14 tháng 7 2021
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 $x=6$
AH
Akai Haruma
Giáo viên
14 tháng 7 2021
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 $x=21$
H9
HT.Phong (9A5)
CTVHS
5 tháng 11 2023
6) \(\sqrt{x^2-4x+1}=x\left(x\ge0\right)\)
\(\Leftrightarrow x^2-4x+1=x^2\)
\(\Leftrightarrow x^2-x^2=4x-1\)
\(\Leftrightarrow4x=1\)
\(\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)
8) \(\sqrt{x^2-x-6}=\sqrt{x-3}\left(x\ge3\right)\)
\(\Leftrightarrow x^2-x-6=x-3\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
9) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\left(x\ge1\right)\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=1+1\)
\(\Leftrightarrow x=2\left(tm\right)\)
15 tháng 8 2021
1: Ta có: \(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-9-\left(x-9\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
Để \(A=-\dfrac{1}{\sqrt{x}}\) thì \(x+\sqrt{x}=-\sqrt{x}+3\)
\(\Leftrightarrow x+2\sqrt{x}-3=0\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow x=1\left(nhận\right)\)
2: Để A nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{-1;1;2;-2;4;-4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{2;4;5;1;7\right\}\)
\(\Leftrightarrow x\in\left\{16;25;1;49\right\}\)
DT
2
29 tháng 10 2021
\(a,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+6\sqrt{x-1}=14\\ \Leftrightarrow7\sqrt{x-1}=14\\ \Leftrightarrow\sqrt{x-1}=2\Leftrightarrow x-1=4\\ \Leftrightarrow x=5\left(tm\right)\\ b,ĐK:-2\le x\le2\\ PT\Leftrightarrow\sqrt{2-x}\left(1-\sqrt{2+x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2-x=0\\2+x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
29 tháng 10 2021
a) ĐKXĐ: \(x\ge1\)
\(pt\Leftrightarrow\sqrt{x-1}+6\sqrt{x-1}=14\)
\(\Leftrightarrow7\sqrt{x-1}=14\Leftrightarrow\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\Leftrightarrow x=5\left(tm\right)\)
b) ĐKXĐ: \(-2\le x\le2\)
\(pt\Leftrightarrow\sqrt{2-x}-\sqrt{\left(2-x\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\sqrt{2-x}\left(1-\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+2=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
16 tháng 5 2021
`A=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)(3-sqrtx)(x>=0,x ne 4, x ne 9)`
`=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)(sqrtx-3)`
`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`
16 tháng 5 2021
`A=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4, x ne 9)`
`=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`
`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`
Điều kiện: \(x\ge0\)
\(\sqrt{x}-\sqrt{x+1}-\sqrt{x+4}+\sqrt{x+9}=0\)
\(\Leftrightarrow\sqrt{x}+\sqrt{x+9}=\sqrt{x+1}+\sqrt{x+4}\)
\(\Leftrightarrow2x+9+2\sqrt{x}.\sqrt{x+9}=2x+5+2\sqrt{x+1}.\sqrt{x+4}\)
\(\Leftrightarrow2+\sqrt{x^2+9x}=\sqrt{x^2+5x+4}\)
\(\Leftrightarrow4+4\sqrt{x^2+9x}+x^2+9x=x^2+5x+4\)
\(\Leftrightarrow\sqrt{x^2+9x}+x=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x+9}+\sqrt{x}\right)=0\)
\(\Leftrightarrow x=0\)