\(\sqrt{x^2+9}+\sqrt{x^2-2x+17}=x^2+6x-3\left(x>=0\right)\)
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30 tháng 8 2021
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
26 tháng 10 2023
a: ĐKXĐ: x>=-3/2
\(\sqrt{x^2+4}=\sqrt{2x+3}\)
=>\(x^2+4=2x+3\)
=>\(x^2-2x+1=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1(nhận)
b: \(\sqrt{x^2-6x+9}=2x-1\)(ĐKXĐ: \(x\in R\))
=>\(\sqrt{\left(x-3\right)^2}=2x-1\)
=>\(\left\{{}\begin{matrix}\left(2x-1\right)^2=\left(x-3\right)^2\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(2x-1-x+3\right)\left(2x-1+x-3\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+2\right)\left(3x-4\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>x=4/3(nhận) hoặc x=-2(loại)
c:
Sửa đề: \(\sqrt{4x+12}=\sqrt{9x+27}-5\)
ĐKXĐ: \(x>=-3\)
\(\sqrt{4x+12}=\sqrt{9x+27}-5\)
=>\(2\sqrt{x+3}=3\sqrt{x+3}-5\)
=>\(-\sqrt{x+3}=-5\)
=>x+3=25
=>x=22(nhận)
d: ĐKXĐ: \(\left[{}\begin{matrix}x< =\dfrac{3-\sqrt{5}}{4}\\x>=\dfrac{3+\sqrt{5}}{4}\end{matrix}\right.\)
\(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)
=>\(\sqrt{\left(4x^2-6x+1\right)}=\sqrt{4x^2-20x+25}\)
=>\(4x^2-6x+1=4x^2-20x+25\)
=>\(-6x+20x=25-1\)
=>\(14x=24\)
=>x=12/7(nhận)
18 tháng 1
a: ĐKXĐ: \(\left[{}\begin{matrix}x>=2\\x< =-3\end{matrix}\right.\)
\(\sqrt{\left(x-2\right)\left(x+3\right)}=5\)
=>\(\sqrt{x^2+x-6}=5\)
=>\(x^2+x-6=25\)
=>\(x^2+x-31=0\)
=>\(\left[{}\begin{matrix}x=\dfrac{-1+5\sqrt{5}}{2}\left(nhận\right)\\x=\dfrac{-1-5\sqrt{5}}{2}\left(nhận\right)\end{matrix}\right.\)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(2x+3\right)^2}=x-5\)
=>\(\left|2x+3\right|=x-5\)
=>\(\left\{{}\begin{matrix}x>=5\\\left(2x+3\right)^2=\left(x-5\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=5\\\left(2x+3-x+5\right)\left(2x+3+x-5\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=5\\\left(x+8\right)\left(3x-2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=5\\\left[{}\begin{matrix}x=-8\left(loại\right)\\x=\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
=>\(x\in\varnothing\)
c: ĐKXĐ: \(x\in R\)
\(\sqrt{x^2-6x+9}=x+7\)
=>\(\sqrt{\left(x-3\right)^2}=x+7\)
=>\(\left|x-3\right|=x+7\)
=>\(\left\{{}\begin{matrix}x+7>=0\\\left(x-3\right)^2=\left(x+7\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-7\\\left(x-3-x-7\right)\left(x-3+x+7\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-7\\-10\left(2x+4\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-7\\x+2=0\end{matrix}\right.\)
=>x=-2
d: ĐKXĐ: x>=3/2
\(\sqrt{2x-3}=x-1\)
=>\(\left\{{}\begin{matrix}2x-3=\left(x-1\right)^2\\x>=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-2x+1=2x-3\\x>=\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-4x+4=0\\x>=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\x>=\dfrac{3}{2}\end{matrix}\right.\)
=>x=2
PT
1
H9
HT.Phong (9A5)
CTVHS
12 tháng 8 2023
a) \(\sqrt{x}-x-0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\1-\sqrt{x}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
b) \(x-\sqrt{2x-9}=6\)
\(\Leftrightarrow\sqrt{2x-9}=x-6\) (ĐK: \(x\ge\dfrac{9}{2}\))
\(\Leftrightarrow2x-9=\left(x-6\right)^2\)
\(\Leftrightarrow2x-9=x^2-12x+36\)
\(\Leftrightarrow x^2-14x+45=0\)
\(\Leftrightarrow x^2-5x-9x+45=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\)
c) \(3x-\sqrt{6x-\left(3-2\right)}=0\) (ĐK: \(x\ge\dfrac{1}{6}\))
\(\Leftrightarrow3x-\sqrt{6x-1}=0\)
\(\Leftrightarrow\sqrt{6x-1}=3x\)
\(\Leftrightarrow6x-1=9x^2\)
\(\Leftrightarrow9x^2-6x+1=0\)
\(\Leftrightarrow\left(3x-1\right)^2=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\left(tm\right)\)
HN
5 tháng 8 2016
\(\Leftrightarrow-\left(x^2-2x\right)+\sqrt{6\left(x^2-2x\right)+7}=0\) ĐK \(\sqrt{6x^2-12x+7}\ge0\)
Đặt \(t=x^2-2x\left(t\ge0\right)\Leftrightarrow pt:-t+\sqrt{6t+7}=0\Leftrightarrow\sqrt{6t+7}=t\\ 6t+7-t^2=0\Leftrightarrow\left[\begin{array}{nghiempt}t=7\left(tm\right)\\t=-1\left(ktm\right)\end{array}\right.\)
Với \(t=7\Leftrightarrow x^2-2x-7=0\Leftrightarrow x=1\pm2\sqrt{2}\left(tm\right)\)
Vậy S={\(1\pm2\sqrt{2}\)}
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 2 2020
a/ ĐKXĐ: ...
\(\Leftrightarrow\left(x^2-6x\right)\left(\sqrt{17-x^2}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x=0\\\sqrt{17-x^2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x-6\right)=0\\x^2=16\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\left(l\right)\\x=4\\x=-4\end{matrix}\right.\)
b/ĐKXĐ: \(x\ge-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+4=0\\\sqrt{x+3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4\left(l\right)\\x=-3\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 2 2020
c/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ge1\\x\le1\end{matrix}\right.\) \(\Rightarrow x=1\)
Thay \(x=1\) vào pt thấy ko thỏa mãn
Vậy pt vô nghiệm
d/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\\\sqrt{x-2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\left(l\right)\\x=2\end{matrix}\right.\)
2 tháng 7 2019
a) \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
<=> \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9\left(x-1\right)}+24\frac{\sqrt{x-1}}{\sqrt{64}}=-17\)
<=>\(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
<=>\(\sqrt{x-1}\left(\frac{1}{2}-\frac{9}{2}+\frac{6}{2}\right)=-17\)
<=>\(\sqrt{x-1}=-17\)
<=>x-1=17
<=>x=18
Vậy pt có nghiệm là x=18
KH
2 tháng 7 2019
\(a.ĐK:x-1\ge0\Leftrightarrow x\ge1\)
\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
\(\Leftrightarrow\frac{1}{2}\sqrt{x-1}-\frac{27}{2}\sqrt{x-1}+24\sqrt{\frac{x-1}{64}}=-17\)
\(\Leftrightarrow\sqrt{x-1}\left(\frac{1}{2}-\frac{27}{2}+24\sqrt{\frac{1}{64}}\right)=-17\)
\(\Leftrightarrow\sqrt{x-1}.\left(-10\right)=-17\)
\(\Leftrightarrow\sqrt{x-1}=\frac{-17}{-10}=\frac{17}{10}\)
\(\Leftrightarrow x-1=\left(\frac{17}{10}\right)^2\)
\(\Leftrightarrow x=\frac{289}{100}+1=3,89\left(TM\right)\)
Vậy \(S=\left\{3,89\right\}\)
\(b.ĐK:x^2+2\ge0\)
\(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)
\(\Leftrightarrow9\sqrt{x^2+2}+2\sqrt{x^2+2}-25\sqrt{x^2+2}=-3\)
\(\Leftrightarrow\sqrt{x^2+2}\left(9+2-25\right)=-3\)
\(\Leftrightarrow\sqrt{x^2+2}=\frac{-3}{-14}=\frac{3}{14}\)
\(\Leftrightarrow x^2+2=\left(\frac{3}{14}\right)^2\)
\(\Leftrightarrow x=\sqrt{\frac{9}{196}-2}=\sqrt{-\frac{383}{196}}\left(vl\right)\)
Vậy \(S=\varnothing\)
Mấy câu kia làm tương tự
DT
Đỗ Thanh Hải
CTVVIP
29 tháng 6 2021
a) \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\dfrac{1}{\sqrt{x+3}}\)(\(x\ge0,x\ne9\))
b) \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-3}=\sqrt{x}-2\left(x\ge0,x\ne9\right)\)
AT
29 tháng 6 2021
a) \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\dfrac{1}{\sqrt{x}+3}\)
b) \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)
c) \(6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(3-x\right)^2}=6-2x-\left|3-x\right|\)
mà \(x< 3\Rightarrow3-x>0\Rightarrow6-2x-\left|3-x\right|=6-2x-3+x=3-x\)