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1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)
\(\Leftrightarrow5-2x=36\)
\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)
2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)
\(\Leftrightarrow2-x=x+1\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)
\(\Leftrightarrow\left|x-5\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
\(1)\) ĐKXĐ : \(x\ge3\)
\(\sqrt{x^2-4x+3}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x^2-4x+4\right)-1}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x-2\right)^2-1}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x-2-1\right)\left(x-2+1\right)}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x-3\right)\left(x-1\right)}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{x-1}\left(\sqrt{x-3}+1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\sqrt{x-1}=0\\\sqrt{x-3}+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x\in\left\{\varnothing\right\}\end{cases}}}\)
Vậy \(x=1\)
\(2)\)\(\sqrt{x^2-2x+1}-\sqrt{x^2-6x+9}=10\)
\(\Leftrightarrow\)\(\sqrt{\left(x-1\right)^2}-\sqrt{\left(x-3\right)^2}=10\)
\(\Leftrightarrow\)\(\left|x-1\right|-\left|x-3\right|=10\)
+) Với \(\hept{\begin{cases}x-1\ge0\\x-3\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\ge3\end{cases}\Leftrightarrow}x\ge3}\) ta có :
\(x-1-x+3=10\)
\(\Leftrightarrow\)\(0=8\) ( loại )
+) Với \(\hept{\begin{cases}x-1< 0\\x-3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 1\\x< 3\end{cases}\Leftrightarrow}x< 1}\) ta có :
\(1-x+x-3=10\)
\(\Leftrightarrow\)\(0=12\) ( loại )
Vậy không có x thỏa mãn đề bài
Chúc bạn học tốt ~
PS : mới lp 8 sai đừng chửi nhé :v
Giải PT
a) \(3\sqrt{9x}+\sqrt{25x}-\sqrt{4x} = 3\)
\(\Leftrightarrow\) \(3.3\sqrt{x} +5\sqrt{x} - 2\sqrt{x} = 3 \)
\(\Leftrightarrow\) \(9\sqrt{x}+5\sqrt{x}-2\sqrt{x} = 3 \)
\(\Leftrightarrow\) \(12\sqrt{x} = 3\)
\(\Leftrightarrow\) \(\sqrt{x} = 4 \)
\(\Leftrightarrow\) \(\sqrt{x^2} = 4^2\)
\(\Leftrightarrow\) \(x=16\)
b) \(\sqrt{x^2-2x-1} - 3 =0\)
\(\Leftrightarrow\) \(\sqrt{(x-1)^2} -3=0\)
\(\Leftrightarrow\) \(|x-1|=3\)
* \(x-1=3\)
\(\Leftrightarrow\) \(x=4\)
* \(-x-1=3\)
\(\Leftrightarrow\) \(-x=4\)
\(\Leftrightarrow\) \(x=-4\)
c) \(\sqrt{4x^2+4x+1} - x = 3\)
<=> \(\sqrt{(2x+1)^2} = 3+x\)
<=> \(|2x+1|=3+x\)
* \(2x+1=3+x\)
<=> \(2x-x=3-1\)
<=> \(x=2\)
* \(-2x+1=3+x\)
<=> \(-2x-x = 3-1\)
<=> \(-3x=2\)
<=> \(x=\dfrac{-2}{3}\)
d) \(\sqrt{x-1} = x-3\)
<=> \(\sqrt{(x-1)^2} = (x-3)^2\)
<=> \(|x-1| = x^2-2.x.3+3^2\)
<=> \(|x-1| = x-6x+9\)
<=> \(|x-1| = -5x+9\)
* \(x-1= -5x+9\)
<=> \(x+5x = 9+1\)
<=> \(6x=10\)
<=> \(x= \dfrac{10}{6} =\dfrac{5}{3}\)
* \(-x-1 = -5x+9\)
<=> \(-x+5x = 9+1\)
<=> \(4x = 10\)
<=> \(x= \dfrac{10}{4} = \dfrac{5}{2}\)
Vì \(\sqrt{x^2+1}\)\(\ge\) 0
nên x2+1 \(\ge\)0
mà x2+1 > 0
nên \(\sqrt{4x^2-4x+5}=0\)
\(\Rightarrow\)4x2-4x+5 =0
mà 4x2-4x+1+4
=(2x-1)2+4>0
\(\Rightarrow\)Phương trình vô nghiệm
a/ ĐKXĐ: \(x\ge\frac{3}{4}\)
\(\Leftrightarrow6x+1+2\sqrt{5x^2+5x}=6x+1+2\sqrt{8x^2+10x-12}\)
\(\Leftrightarrow\sqrt{5x^2+5x}=\sqrt{8x^2+10x-12}\)
\(\Leftrightarrow5x^2+5x=8x^2+10x-12\)
\(\Leftrightarrow3x^2+5x-12=0\Rightarrow\left[{}\begin{matrix}x=-3< \frac{3}{4}\left(l\right)\\x=\frac{4}{3}\end{matrix}\right.\)
b/ \(\Leftrightarrow x^2+x+1+2\sqrt{x^2+x+1}-3=0\)
Đặt \(\sqrt{x^2+x+1}=t>0\)
\(\Rightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+x+1}=1\)
\(\Leftrightarrow x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
\(5x^2+4x+7-4x\sqrt{x^2+x+2}-4\sqrt{3x+1}=0\)
ĐK: \(x\ge-\frac{1}{3}\)
\(\Leftrightarrow5x^2+4x-9-\left(4x\sqrt{x^2+x+2}-8\right)-\left(4\sqrt{3x+1}-8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+9\right)-4\frac{x^2\left(x^2+x+2\right)-4}{x\sqrt{x^2+x+2}+2}-4\frac{3x+1-4}{\sqrt{3x+1}+2}=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+9\right)-4\frac{\left(x-1\right)\left(x^3+2x^2+4x+4\right)}{x\sqrt{x^2+x+2}+2}-4\frac{3\left(x-1\right)}{\sqrt{3x+1}+2}=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+9-4\frac{\left(x^3+2x^2+4x+4\right)}{x\sqrt{x^2+x+2}+2}-4\frac{3}{\sqrt{3x+1}+2}\right)=0\)
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
\(ĐKXĐ:x\ge\frac{-1}{3}\)
\(5x^2+4x+7-4x\sqrt{x^2+x+2}-4\sqrt{3x+1}=0\)
\(\Leftrightarrow\left(x^2+x+2-4x\sqrt{x^2+x+2}+4x\right)\)\(+\left(3x+1-4\sqrt{3x+1}+4\right)=0\)
\(\Leftrightarrow\left(\sqrt{x^2+x+2}-2x\right)^2+\left(\sqrt{3x+1}-2\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x^2+x+2}=2x\\\sqrt{3x+1}=2\end{cases}}\Leftrightarrow\hept{\begin{cases}x>0\\x^2+x+2=4x\\3x+1=4\end{cases}}\Leftrightarrow x=1\)
Vậy nghiệm duy nhất của phương trình là x = 1
ĐKXĐ: \(\left[{}\begin{matrix}0\le x\le2-\sqrt{3}\\x\ge2+\sqrt{3}\end{matrix}\right.\)
\(2x+2+2\sqrt{x^2-4x+1}=6\sqrt{x}\)
\(\Leftrightarrow\left(2x+2-5\sqrt{x}\right)+\left(\sqrt{4x^2-16x+4}-\sqrt{x}\right)=0\)
\(\Leftrightarrow\dfrac{4x^2-17x+4}{2x+2+5\sqrt{x}}+\dfrac{4x^2-17x+4}{\sqrt{4x^2-16x+4}+\sqrt{x}}=0\)
\(\Leftrightarrow\left(4x^2-17x+4\right)\left(\dfrac{1}{2x+2+5\sqrt{x}}+\dfrac{1}{\sqrt{4x^2-16x+4}+\sqrt{x}}\right)=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
a) \(\sqrt{1-x}=\sqrt[3]{8}\) ( ĐK: \(x\le1\) )
\(\Leftrightarrow\sqrt{1-x}=2\)
\(\Leftrightarrow1-x=4\)
\(\Leftrightarrow x=-3\) ( Thỏa mãn )
b) \(\sqrt{4x^2-12x+9}=x+1\) ( ĐK : \(x\ge-1\) )
\(\Leftrightarrow\sqrt{\left(2x\right)^2-2.2x.3+3^2}=x+1\)
\(\Leftrightarrow\sqrt{\left(2x-3\right)^2}=x+1\)
\(\Leftrightarrow\left|2x-3\right|=x+1\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=x+1\\3-2x=x+1\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{2}{3}\end{cases}}\) ( Thỏa mãn )
c) \(x+\sqrt{x}-2=0\) ( ĐK : \(x\ge0\) )
\(\Leftrightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\sqrt{x}-1=0\)
\(\Leftrightarrow x=1\) ( Thỏa mãn )
+) ĐKXĐ : \(x\le1\)
\(\sqrt{1-x}=\sqrt[3]{8}\)
\(\Leftrightarrow\sqrt{1-x}=2\)
\(\Leftrightarrow1-x=4\)
\(\Leftrightarrow x=-3\left(TM\right)\)
+) \(\sqrt{4x^2-12x+9}=x+1\)
\(\Leftrightarrow\sqrt{\left(2x-3\right)^2}=x+1\)
\(\Leftrightarrow\left|2x-3\right|=x+1\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=x+1\left(x\ge\frac{3}{2}\right)\\2x-3=-x-1\left(x< \frac{3}{2}\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-x=3+1\\2x+x=3-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\3x=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{2}{3}\end{cases}\left(TM\right)}}\)
+) ĐKXĐ : \(x\ge0\)
\(x+\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=2\)
+) \(\hept{\begin{cases}\sqrt{x}=1\\\sqrt{x}+1=2\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=1\end{cases}\Leftrightarrow}x=1\left(TM\right)}\)
+) \(\hept{\begin{cases}\sqrt{x}=2\\\sqrt{x}+1=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=\sqrt{2}\\x=0\end{cases}}}\left(TM\right)\)
a) Đk: \(\hept{\begin{cases}x^2-4x+1\ge0\\x+1\ge0\end{cases}}\)
\(\sqrt{x^2-4x+1}=\sqrt{x+1}\)
\(\Leftrightarrow x^2-4x+1=x+1\)
\(\Leftrightarrow x^2-4x-x=0\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)thỏa mãn điều kiện
Vậy x=0 hoặc x=5
2)\(\sqrt{\left(x-1\right)\left(x-3\right)}+\sqrt{x-1}=0\)(1)
Đk: x>=3 hoặc x=1
pt (1)<=> \(\sqrt{x-1}\left(\sqrt{x-3}+1\right)=0\)
<=> \(\sqrt{x-1}=0\)(vì\(\sqrt{x-3}+1>0\)mọi x )
<=> x-1=0
<=> x=1 ( thỏa mãn điều kiện)