6) \(\sqrt{x^2+12x+36}=-x-6\)
7) \(\sqrt{9x^2-12x+4}=3x-2\)
8) \(\sqrt{16-24x+9x^2}=2x-10\)
9) \(\sqrt{x^2-6x+9}==2x-3\)
10) \(\sqrt{x^2-3x+\dfrac{9}{4}}=\dfrac{3}{x}x-4\)
K
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25 tháng 11 2023
2: ĐKXĐ: x>=0
\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)
=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)
=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)
=>\(-2\sqrt{3x}=-4\)
=>\(\sqrt{3x}=2\)
=>3x=4
=>\(x=\dfrac{4}{3}\left(nhận\right)\)
3:
ĐKXĐ: x>=0
\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)
=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)
=>\(13\sqrt{2x}=20+3\sqrt{2}\)
=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)
=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)
=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)
4: ĐKXĐ: x>=-1
\(\sqrt{16x+16}-\sqrt{9x+9}=1\)
=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>\(\sqrt{x+1}=1\)
=>x+1=1
=>x=0(nhận)
5: ĐKXĐ: x<=1/3
\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)
=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)
=>\(5\sqrt{1-3x}=10\)
=>\(\sqrt{1-3x}=2\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1(nhận)
6: ĐKXĐ: x>=3
\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)
=>x-3=16
=>x=19(nhận)
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
HP
2 tháng 2 2021
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
HP
2 tháng 2 2021
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
28 tháng 8 2021
3: Ta có: \(\sqrt{4x+1}=x+1\)
\(\Leftrightarrow x^2+2x+1=4x+1\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
4: Ta có: \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)
\(\Leftrightarrow3\sqrt{x-1}=15\)
\(\Leftrightarrow x-1=25\)
hay x=26
5: Ta có: \(\sqrt{4x^2-12x+9}=7\)
\(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
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 \(x\in\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}\)
6) ĐKXĐ: \(x\le-6\)
\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)
\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)
Vậy \(x\le-6\)
7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)
Vậy \(x\ge\dfrac{2}{3}\)
8) ĐKXĐ: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)
\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)
9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)