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![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 1:
a:\(\sqrt{\left(\sqrt{3}-2\right)^2}\)+\(\sqrt{\left(1+\sqrt{3}\right)^2}\)
=\(\sqrt{3}-2+1+\sqrt{3}\)
=\(2\sqrt{3}-1\)
b; dài quá mink lười làm thông cảm
bài 2:
\(\sqrt{x^2-2x+1}=7\)
=>\(\sqrt{\left(x-1\right)^2}=7
\)
=>\(\orbr{\begin{cases}x-1=7\\x-1=-7\end{cases}}\)
=>\(\orbr{\begin{cases}x=8\\x=-6\end{cases}}\)
b: \(\sqrt{4x-20}-3\sqrt{\frac{x-5}{9}}=\sqrt{1-x}\)
=>\(\sqrt{4\left(x-5\right)}-9\sqrt{x-5}=\sqrt{1-x}\)
\(=2\sqrt{x-5}-9\sqrt{x-5}=\sqrt{1-x}\)
=>\(-7\sqrt{x-5}=\sqrt{1-x}\)
=\(-7.\left(x-5\right)=1-x\)
=>\(-7x+35=1-x\)
=>\(-7x+x=1-35\)
=>\(-6x=-34\)
=>\(x\approx5.667\)
mink sợ câu b bài 2 sai đó bạn
1 a)\(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
= \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
= \(|2-\sqrt{3}|+|1+\sqrt{3}|\)
= \(2-\sqrt{3}+1+\sqrt{3}\)
= \(2+1\)= \(3\)
b) \(\left(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}\right)\cdot\left(3\sqrt{\frac{2}{3}}-\sqrt{12}-\sqrt{6}\right)\)
= \(\left(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{6}{3^2}}-4\sqrt{\frac{6}{2^2}}\right)\cdot\left(3\sqrt{\frac{6}{3^2}}-\sqrt{6}\sqrt{2}-\sqrt{6}\right)\)
= \(\left(\frac{3}{2}\sqrt{6}+\frac{2}{3}\sqrt{6}-\frac{4}{2}\sqrt{6}\right)\cdot\left(\frac{3}{3}\sqrt{6}-\sqrt{6}\cdot\sqrt{2}-\sqrt{6}\right)\)
= \(\left(\frac{3}{2}\sqrt{6}+\frac{2}{3}\sqrt{6}-2\sqrt{6}\right)\cdot\left(\sqrt{6}-\sqrt{6}\cdot\sqrt{2}-\sqrt{6}\right)\)
= \(\left(\sqrt{6}\left(\frac{3}{2}+\frac{2}{3}-2\right)\right)\cdot\left(\sqrt{6}\left(1-\sqrt{2}-1\right)\right)\)
= \(\sqrt{6}\frac{1}{6}\cdot\sqrt{6}\left(-\sqrt{2}\right)\)
= \(\sqrt{6}^2\left(\frac{-\sqrt{2}}{6}\right)\)
= \(6\frac{-\sqrt{2}}{6}\)=\(-\sqrt{2}\)
2 a) \(\sqrt{x^2-2x+1}=7\)
<=> \(\sqrt{x^2-2x\cdot1+1^2}=7\)
<=> \(\sqrt{\left(x-1\right)^2}=7\)
<=> \(|x-1|=7\)
Nếu \(x-1>=0\)=>\(x>=1\)
=> \(|x-1|=x-1\)
\(x-1=7\)<=>\(x=8\)(thỏa)
Nếu \(x-1< 0\)=>\(x< 1\)
=> \(|x-1|=-\left(x-1\right)=1-x\)
\(1-x=7\)<=>\(-x=6\)<=> \(x=-6\)(thỏa)
Vậy x=8 hoặc x=-6
b) \(\sqrt{4x-20}-3\sqrt{\frac{x-5}{9}}=\sqrt{1-x}\)
<=> \(\sqrt{4\left(x-5\right)}-3\frac{\sqrt{x-5}}{3}=\sqrt{1-x}\)
<=> \(2\sqrt{x-5}-\sqrt{x-5}=\sqrt{1-x}\)
<=> \(\sqrt{x-5}=\sqrt{1-x}\)
ĐK \(x-5>=0\)<=> \(x=5\)
\(1-x\)<=> \(-x=-1\)<=> \(x=1\)
Ta có \(\sqrt{x-5}=\sqrt{1-x}\)
<=> \(\left(\sqrt{x-5}\right)^2=\left(\sqrt{1-x}\right)^2\)
<=> \(x-5=1-x\)
<=> \(x-x=1+5\)
<=> \(0x=6\)(vô nghiệm)
Vậy phương trình vô nghiệm
Kết bạn với mình nha :)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+2\sqrt{12}}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-2\sqrt{75}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)
\(C=\sqrt{4+5}\)
\(C=3\)
2. a) \(ĐKXĐ:x\ge\frac{1}{3}\)
\(\sqrt{3x-1}=4\)\(\Rightarrow\left(\sqrt{3x-1}\right)^2=4^2\)
\(\Leftrightarrow3x-1=16\)\(\Leftrightarrow3x=17\)\(\Leftrightarrow x=\frac{17}{3}\)( thỏa mãn ĐKXĐ )
Vậy \(x=\frac{17}{3}\)
b) \(ĐKXĐ:x\ge1\)
\(\sqrt{x-1}=x-1\)\(\Rightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow x-1=x^2-2x+1\)\(\Leftrightarrow x^2-2x+1-x+1=0\)
\(\Leftrightarrow x^2-3x+2=0\)\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)( thỏa mãn ĐKXĐ )
Vậy \(x=1\)hoặc \(x=2\)
3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)
Vì \(6>1\)\(\Leftrightarrow\sqrt{6}>\sqrt{1}=1\)\(\Rightarrow\sqrt{6}-1>0\)
\(6>4\)\(\Rightarrow\sqrt{6}>\sqrt{4}=2\)\(\Rightarrow\sqrt{6}-2>0\)
\(\Rightarrow\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|=\left(\sqrt{6}-1\right)-\left(\sqrt{6}-2\right)\)
\(=\sqrt{6}-1-\sqrt{6}+2=1\)
hay \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=1\)
2a) \(\sqrt{3x-1}=4\)( ĐKXĐ : \(x\ge\frac{1}{3}\))
Bình phương hai vế
\(\Leftrightarrow\left(\sqrt{3x-1}\right)^2=4^2\)
\(\Leftrightarrow3x-1=16\)
\(\Leftrightarrow3x=17\)
\(\Leftrightarrow x=\frac{17}{3}\)( tmđk )
Vậy phương trình có nghiệm duy nhất là x = 17/3
b) \(\sqrt{x-1}=x-1\)( ĐKXĐ : \(x\ge1\))
Bình phương hai vế
\(\Leftrightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow x-1=x^2-2x+1\)
\(\Leftrightarrow x^2-2x+1-x+1=0\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}\left(tmđk\right)}\)
Vậy phương trình có hai nghiệm là x = 1 hoặc x = 2
3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)
\(=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}-\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot2+2^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}\)
\(=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)
\(=\sqrt{6}-1-\left(\sqrt{6}-2\right)\)
\(=\sqrt{6}-1-\sqrt{6}+2\)
\(=1\)