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Bài rút gọn
\(\sqrt{\left(x-1\right)^2}-x=\left|x-1\right|-x\)
\(=\left(x-1\right)-x=x-1-x=-1\left(x>1\right)\)
Bài gpt:
\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)
Đk:\(-1\le x\le3\)
\(pt\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}+\sqrt{x-3}\right)=0\)
Dễ thấy:\(\sqrt{x-2}+\sqrt{x-3}=0\) vô nghiệm
Nên \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)
-1; -6
b) ĐK: \(x^2+7x+7\ge0\) (đk xấu quá em ko giải đc;v)
PT \(\Leftrightarrow3x^2+21x+18+2\left(\sqrt{x^2+7x+7}-1\right)=0\)
\(\Leftrightarrow3\left(x+1\right)\left(x+6\right)+2\left(\frac{x^2+7x+6}{\sqrt{x^2+7x+7}+1}\right)=0\)
\(\Leftrightarrow3\left(x+1\right)\left(x+6\right)+\frac{2\left(x+1\right)\left(x+6\right)}{\sqrt{x^2+7x+7}+1}=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)\left[3+\frac{1}{\sqrt{x^2+7x+7}+1}\right]=0\)
Hiển nhiên cái ngoặc vuông > 0 nên vô nghiệm suy ra x = -1 (TM) hoặc x = -6 (TM)
Vậy....
P/s: Cũng may nghiệm đẹp chứ chứ nghiệm xấu thì tiêu rồi:(
a/ Giải rồi
b/ ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)
\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\) (1)
Pt trở thành:
\(t=t^2-6\Leftrightarrow t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=3\)
\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=9\)
\(\Leftrightarrow2\sqrt{2x^2+5x+3}=5-3x\left(x\le\frac{5}{3}\right)\)
\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(5-3x\right)^2\)
\(\Leftrightarrow...\)
e/ ĐKXD: \(x>0\)
\(5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=t\ge\sqrt{2}\)
\(\Rightarrow t^2=x+\frac{1}{4x}+1\)
Pt trở thành:
\(5t=2\left(t^2-1\right)+4\)
\(\Leftrightarrow2t^2-5t+2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=2\)
\(\Leftrightarrow2x-4\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{2\pm\sqrt{2}}{2}\)
\(\Rightarrow x=\frac{3\pm2\sqrt{2}}{2}\)
1) Đk: x \(\ge\)1
Ta có: \(\sqrt{3x-2}+\sqrt{x-1}=3\)
<=> \(3x-2+x-1+2\sqrt{\left(3x-2\right)\left(x-1\right)}=9\)
<=> \(2\sqrt{\left(3x-2\right)\left(x-1\right)}=12-4x\)
<=> \(\sqrt{3x^2-5x+2}=6-2x\)(\(1\le x\le3\))
<=> \(3x^2-5x+2=4x^2-24x+36\)
<=> \(x^2-19x+34=0\)
<=> \(x^2-17x-2x+34=0\)
<=> \(\left(x-17\right)\left(x-2\right)=0\)
<=> \(\orbr{\begin{cases}x=17\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)
Vậy S = {2}
3.Đk: x \(\ge\)0
\(3x-7\sqrt{x}+4=0\)
<=> \(3x-3\sqrt{x}-4\sqrt{x}+4=0\)
<=> \(\left(3\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\)
<=> \(\orbr{\begin{cases}x=\frac{16}{9}\\x=1\end{cases}}\left(tm\right)\)
4. Đk: x \(\ge\)0; x \(\ne\)16; x \(\ne\)49
Ta có: \(\frac{\sqrt{x}-2}{\sqrt{x}-4}=\frac{6-\sqrt{x}}{7-\sqrt{x}}\)
=> \(\left(\sqrt{x}-2\right)\left(\sqrt{x}-7\right)=\left(\sqrt{x}-4\right)\left(\sqrt{x}-6\right)\)
<=> \(x-9\sqrt{x}+14=x-10\sqrt{x}+24\)
<=> \(\sqrt{x}=10\) <=> x = 100 (tm)
5.Đk: x \(\ge\)1
\(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
<=> \(-\sqrt{x-1}=-17\) <=> \(x-1=17\) <=> x = 18 (tm)