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\(a,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}=\left|2-\sqrt{3}\right|+\sqrt{\left(\sqrt{3}-1\right)^2}=2-\sqrt{3}+\left|\sqrt{3}-1\right|=2-1=1\)
\(\text{[}3-\sqrt{\left(\sqrt{3}-1\right)^2}\text{]}^2+\sqrt{147}=\left(3-\sqrt{3}+1\right)^2+7\sqrt{3}=\left(4-\sqrt{3}\right)^2+7\sqrt{3}=7-8\sqrt{3}+7\sqrt{3}=7-\sqrt{3}\)
\(\frac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}-\frac{\sqrt{10}-\sqrt{15}}{\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-\frac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{5}}-\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}=\sqrt{3}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}=\sqrt{3}\)
a)\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(=2-\sqrt{3}+\left[\left(\sqrt{3}\right)^2-2\sqrt{3+1^2}\right]\)
=\(2-\sqrt{3}+3-2\sqrt{3}+1\)
=6-3\(\sqrt{3}\)
b)\(\left[3-\sqrt{\left(\sqrt{3-1}\right)^2}\right]-\sqrt{147}\)
=\(\left[3-\sqrt{3}+1\right]^2-7\sqrt{3}\)
=\(\left(2\sqrt{3}\right)^2-7\sqrt{3}\)
=12-7\(\sqrt{3}\)
a) \(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}\)
\(=\sqrt{75}-\sqrt{\frac{16}{3}}+\frac{9}{2}\sqrt{\frac{8}{3}}+2\sqrt{27}\)
\(=5\sqrt{3}-\frac{4}{\sqrt{3}}+3\sqrt{6}+6\sqrt{3}\)
\(=-\frac{4}{\sqrt{3}}+5\sqrt{3}+3\sqrt{6}+6\sqrt{3}\)
\(=-\frac{4}{\sqrt{3}}+11\sqrt{3}+3\sqrt{6}\)
\(=-\frac{4\sqrt{3}}{3}+11\sqrt{3}+3\sqrt{6}\)
b) \(\sqrt{48}-\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}\)
\(=\sqrt{48}-\sqrt{\frac{16}{3}}+2\sqrt{75}-5\sqrt{\frac{4}{3}}\)
\(=4\sqrt{3}-\frac{4}{\sqrt{3}}+10\sqrt{3}-\frac{10}{\sqrt{3}}\)
\(=-\frac{4}{\sqrt{3}}-\frac{10}{\sqrt{3}}+4\sqrt{3}+10\sqrt{3}\)
\(=-\frac{14\sqrt{3}}{3}+4\sqrt{3}+10\sqrt{3}\)
\(=-\frac{14\sqrt{3}}{3}+14\sqrt{3}\)
c)\(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}\)
\(=27+12\sqrt{5}+12\sqrt{5}\)
\(=27+24\sqrt{5}\)
d)\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\sqrt{6}+2-\sqrt{3}-\sqrt{2}\)
e) \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4\)
\(=4+2\sqrt{3}-2\sqrt{3}+4\)
= 8
f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
\(=\frac{7-4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+\frac{7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{14}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
= 14
a) \(2\sqrt{2}.\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)
\(=2\sqrt{2}.\left(\sqrt{3}-2\right)+9+4\sqrt{2}-2\sqrt{6}\)
\(=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}\)
= 9 (đpcm)
b) \(\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2\left(\sqrt{2}-1\right)}\)
\(=\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2^{\frac{1}{2}}\left(\sqrt{2}-1\right)}\)
\(=\sqrt{2\left(\sqrt{2}-1\right)}\) (đpcm)
a)\(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}=5\sqrt{3}-\frac{\sqrt{15}}{3}+3\sqrt{3}+6\sqrt{3}=14\sqrt{3}-\frac{\sqrt{15}}{3}\)
b) \(\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}=4\sqrt{3}+\frac{\sqrt{15}}{3}+10\sqrt{3}-\frac{5\sqrt{3}}{3}=\frac{12\sqrt{3}+30\sqrt{3}-5\sqrt{3}}{3}+\frac{\sqrt{15}}{3}=\frac{37\sqrt{3}+\sqrt{15}}{3}\)
c) \(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}=\left[\left(\sqrt{15}\right)^2+4\sqrt{45}+\left(2\sqrt{3}\right)^2\right]+12\sqrt{5}=15+12\sqrt{5}+12+12\sqrt{5}=27+24\sqrt{5}\)
d) \(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{18}-\sqrt{12}+\sqrt{6}-2\sqrt{2}=3\sqrt{2}-2\sqrt{3}+\sqrt{6}-2\sqrt{2}=\sqrt{2}-2\sqrt{3}+\sqrt{6}\)
e) \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=\left(\sqrt{3}\right)^2+2\sqrt{3}+1-2\sqrt{3}+4=3+2\sqrt{3}+1-2\sqrt{3}+4=8\)
f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)
g) \(\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right)\frac{1}{\left(\sqrt{2}+1\right)^2}=\left(\frac{\sqrt{5}+2-\sqrt{5}+2+5-2}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\right)\frac{1}{3+2\sqrt{2}}=\frac{7}{3}.\frac{1}{3+2\sqrt{2}}=\frac{7}{9+6\sqrt{2}}\)
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 :)
a) \(\left(\frac{\sqrt{9}}{2}+\frac{\sqrt{1}}{2}-\sqrt{2}\right)\sqrt{2}\)
\(=\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-2\)
\(=\frac{4\sqrt{2}}{2}-2=2\sqrt{2}-2\)
b) \(\left(\frac{\sqrt{8}}{3}-\sqrt{24}+\frac{\sqrt{50}}{3}\right)\sqrt{6}\)
\(=\frac{4\sqrt{3}}{3}-12+\frac{10\sqrt{3}}{3}\)
\(=\frac{14\sqrt{3}}{3}-12\)
c) \(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{1}\right)\) (đã sửa đề)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\sqrt{2}\)
\(=\left(3-1\right)\sqrt{2}\)
\(=2\sqrt{2}\)
d) \(\left(3\sqrt{2}+1\right)\left(\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\left(\sqrt{3\sqrt{2}+1}\cdot\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{18-1}\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{17}\)
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