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\(1)\) 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) \(\sqrt{x-1} = x-3\)
<=> \(\sqrt{(x-1)^2} = (x-3)^2\)
<=> \(|x-1| = x^2-2.x.3+3^2\)
<=> \(|x-1| = x-6x+9\)
<=> \(|x-1| = -5x+9\)
* \(x-1= -5x+9\)
<=> \(x+5x = 9+1\)
<=> \(6x=10\)
<=> \(x= \dfrac{10}{6} =\dfrac{5}{3}\)
* \(-x-1 = -5x+9\)
<=> \(-x+5x = 9+1\)
<=> \(4x = 10\)
<=> \(x= \dfrac{10}{4} = \dfrac{5}{2}\)
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 : \(x\ge0\)
\(x+\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=2\)
+) \(\hept{\begin{cases}\sqrt{x}=1\\\sqrt{x}+1=2\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=1\end{cases}\Leftrightarrow}x=1\left(TM\right)}\)
+) \(\hept{\begin{cases}\sqrt{x}=2\\\sqrt{x}+1=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=\sqrt{2}\\x=0\end{cases}}}\left(TM\right)\)
Làm hơi tắt xíu, có gì ko hiểu cmt nha :>
\(a.\sqrt{x-1}=3\left(ĐK:x\ge1\right)\Leftrightarrow x-1=9\Leftrightarrow x=10\)
\(b.\sqrt{x^2-4x+4}=2\\ \Leftrightarrow\sqrt{\left(x-2\right)^2}=2\\ \Leftrightarrow\left|x-2\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2\left(x\ge2\right)\\2-x=2\left(x< 2\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
\(c.\sqrt{25x^2-10x+1}=4x-9\\ \Leftrightarrow\sqrt{\left(5x-1\right)^2}=4x-9\\ \Leftrightarrow\left|5x-1\right|=4x-9\\\Leftrightarrow \left[{}\begin{matrix}5x-1=4x-9\left(x\ge\frac{1}{5}\right)\\1-5x=4x-9\left(x< \frac{1}{5}\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-8\left(ktm\right)\\x=\frac{10}{9}\left(ktm\right)\end{matrix}\right.\)
\(d.\sqrt{x^2+2x+1}=\sqrt{x+1}\left(ĐK:x\ge-1\right)\\ \Leftrightarrow x^2+2x+1=x+1\\ \Leftrightarrow x^2+x=0\Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
e. ĐK: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)
\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\\ \Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\\ \Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\\ \Leftrightarrow\sqrt{x-3}=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\)
Câu cuối chưa nghĩ ra, sorry :<
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 \(\sqrt{x^2+x+1}=t>0\)
\(\Rightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+x+1}=1\)
\(\Leftrightarrow x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Bài 1: Giải phương trình
a) ĐKXĐ: \(x\ge3\)
Ta có: \(\sqrt{100\cdot\left(x-3\right)}=\sqrt{20}\)
\(\Leftrightarrow\left|100\cdot\left(x-3\right)\right|=\left|20\right|\)
\(\Leftrightarrow100\cdot\left|x-3\right|=20\)
\(\Leftrightarrow\left|x-3\right|=\frac{1}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=\frac{1}{5}\\x-3=-\frac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{16}{5}\left(nhận\right)\\x=\frac{14}{5}\left(loại\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{16}{5}\right\}\)
b) Ta có: \(\sqrt{\left(x-3\right)^2}=7\)
\(\Leftrightarrow\left|x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=7\\x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\)
Vậy: S={10;-4}
c) Ta có: \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-7}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{5}{2};\frac{-7}{2}\right\}\)
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)