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a,( √6+2)(√3-√2)
<=> ( √2√3+2)(√3-√2)
<=> √2(√3+√2)(√3-√2)
<=> √2( (√3)2-(√2)2) = √2
b, (√3+1)2-2√3+4
<=> (√3)2 +2√3 +1 -2√3+4 =8
c, (1+√2-√3)(√2+√3)
<=>√2+√3+(√2)2+√6-√6-(√3)2
<=> √2+√3-1
d, √3(√2-√3)2-(√3+√2)
<=> √3( 2-2√6+3)-√3-√2
<=> 5√3-2√18-√3-√2
<=> 4√3-√2(√36-1)
<=> 4√3 - 3√2
e, (1+2√3-√2)(1+2√3+√2)
<=> (1+2√3)2-(√2)2
<=> (1+4√3+(2√3)2)-2
<=> 1+4√3+12-2= 11+4√3
g, (1-√3)2(1+2√3)2
<=>(1-2√3+3)(1+4√3+12)
<=>( 4-2√3)(13+4√3)
<=> 52+16√3-26√3-24
<=> -10√3+28
a) \(\sqrt{(3-2\sqrt{2})^2}+\sqrt{(3+2\sqrt{2})^2}=3-2\sqrt{2}+3-2\sqrt{2}=6\)
b\(\sqrt{(5-2\sqrt{6})^2}+\sqrt{(5+2\sqrt{6})^2}=5-2\sqrt{6}+5+2\sqrt{6}=10 \)
các ý còn lại làm tương tự
hình như ở câu a) chỗ sau dấu bằng đầu tiên bạn bị sai dấu trừ cuối cùng
a) \(\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}=\sqrt{2}+\sqrt{3}\)
b) \(\sqrt{\left(\sqrt{3}-2\right)^2}=\sqrt{3}-2\)
c) \(\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3}\)\(=2\sqrt{5}\)
d) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}=\sqrt{2}-\sqrt{3}-1-\sqrt{3}\)
\(=\sqrt{12}-\sqrt{2}-1\)
e) \(\sqrt{\left(\sqrt{3-1}^2\right)-\sqrt{3}}=\sqrt{\sqrt{2}^2-\sqrt{3}}=\sqrt{2-\sqrt{3}}\)
P/S: Ko chắc
B1 :
a) \(\sqrt{1,2.270}=\sqrt{0,4.3.90.3}=3\sqrt{36}=3.6=18\)
\(\sqrt{55.77.35}=\sqrt{5.11.7.11.7.5}=\sqrt{25.49.212}=\sqrt{25}.\sqrt{49}.\sqrt{121}=5.7.11=385\)
b) \(\left(\sqrt{3}-\sqrt{2}\right)^2=3-2.\sqrt{3}.\sqrt{2}+2=5-2\sqrt{6}\)
\(\left(3\sqrt{2}-1\right)\left(3\sqrt{2}+1\right)=3\sqrt{2}.3\sqrt{2}+3\sqrt{2}-3\sqrt{2}-1=18-1\)
\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-2\right)=\sqrt{6}.\sqrt{3}-2\sqrt{6}+2\sqrt{3}-4=\sqrt{18}-2\sqrt{6}+2\sqrt{3}-4\)\(=3\sqrt{2}-2\sqrt{6}+2\sqrt{3}-4\)
\(c,\left(\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{2}{3}}\right)=\dfrac{\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{2}}{\sqrt{3}}=\dfrac{3-2}{\sqrt{2}\sqrt{3}}\) = \(\dfrac{1}{\sqrt{6}}\)
\(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}=\sqrt{\dfrac{8}{3}}.\sqrt{6}-\sqrt{24}.\sqrt{6}+\sqrt{\dfrac{50}{3}}.\sqrt{6}\) = \(\dfrac{\sqrt{8}.\sqrt{6}}{\sqrt{3}}-\sqrt{144}+\dfrac{\sqrt{50}.\sqrt{6}}{\sqrt{3}}=\dfrac{\sqrt{48}}{\sqrt{3}}-12+\dfrac{\sqrt{300}}{\sqrt{3}}=\sqrt{\dfrac{48}{3}}-12+\sqrt{\dfrac{300}{3}}=4-12+10=2\)
B2 :
a) \(\sqrt{\dfrac{1}{8}}.\sqrt{2}.\sqrt{125}.\sqrt{\dfrac{1}{5}}=\sqrt{\dfrac{1}{8}.2.125.\dfrac{1}{5}}=\sqrt{\dfrac{25}{4}}=\dfrac{5}{2}\)
\(\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2+\sqrt{2}-\sqrt{2}-1}=1\)
b) \(\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}=\left|\sqrt{2}-3\right|.\sqrt{2+6\sqrt{2}+9}=\left(\sqrt{2}-3\right).\sqrt{\left(\sqrt{2}+3\right)^2}=\left(\sqrt{2}-3\right)\)\(\left(\sqrt{2}+3\right)=2+3\sqrt{2}-3\sqrt{2}-9=-7\)
\(\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\dfrac{1}{3-\sqrt{3}}}=\left|\sqrt{3}-3\right|.\dfrac{1}{3-\sqrt{3}}=-\left(3-\sqrt{3}\right).\left(\dfrac{1}{3-\sqrt{3}}\right)=-1\)
a/ \(\sqrt{\left(\frac{1}{\sqrt{2}}-\frac{1}{2}\right)^2}=\) \(|\frac{1}{\sqrt{2}}-\frac{1}{2}|\)
\(=|\frac{\sqrt{2}-1}{2}|\)
\(=\frac{\sqrt{2}-1}{2}\)
các câu còn lại tương tự nha
chúc bn học tốt
a) \(\sqrt{3^2}-\sqrt{7^2}+\sqrt{\left(-1\right)^2}=|3|-|7|+|-1|=3-7+1=-3\)
b) \(-2\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}+\sqrt{3^2}=-2|2|+|-5|+\left|3\right|=-4+5+3=4\)
c) \(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2+\sqrt{2}\right)^2}=\left|2-\sqrt{2}\right|+\left|2+\sqrt{2}\right|=2-\sqrt{2}+2+\sqrt{2}=4\)
d) \(\sqrt{\left(3\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}=\left|3\sqrt{2}\right|-\left|1-\sqrt{2}\right|=3\sqrt{2}-\sqrt{2}+1=2\sqrt{2}+1\)
e) \(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{2}+1\right)^2}=\left|\sqrt{2}-1\right|+\left|\sqrt{2}+1\right|=\sqrt{2}-1+\sqrt{2}+1=2\sqrt{2}\)
f) \(\sqrt{9-4\sqrt{5}}+\sqrt{9+4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(\sqrt{5}+2\right)^2}=\left|\sqrt{5}-2\right|+\left|\sqrt{5}+2\right|=\sqrt{5}-2+\sqrt{5}+2=2\sqrt{5}\)
g) \(\sqrt{9-4\sqrt{2}}+\sqrt{11-6\sqrt{2}}=\sqrt{9-2\sqrt{8}}+\sqrt{2-2\sqrt{2}.3+9}=\sqrt{\left(\sqrt{8}-1\right)^2}+\sqrt{\left(\sqrt{2}-3\right)^2}=\sqrt{8}-1+3-\sqrt{2}=2-\sqrt{2}+\sqrt{8}\)
h) \(\sqrt{12+8\sqrt{2}}+\sqrt{6-4\sqrt{2}}=\sqrt{12+2\sqrt{4}\sqrt{8}}+\sqrt{6-2\sqrt{2}\sqrt{4}}=\sqrt{\left(\sqrt{4}+\sqrt{8}\right)^2}+\sqrt{\left(\sqrt{4}-\sqrt{2}\right)^2}=\sqrt{4}+\sqrt{8}+\sqrt{4}-\sqrt{2}\)
k) \(\left(2-\sqrt{3}\right)\sqrt{7+4\sqrt{3}}=\left(2-\sqrt{3}\right)\sqrt{\left(\sqrt{3}+2\right)^2}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3=1\)
1/ \(=2+\sqrt{5}-\left|2-\sqrt{5}\right|=2+\sqrt{5}-\sqrt{5}+2=4\)
2/ bạn coi lại đề
3/ \(=\sqrt{2}+1-\left|1-\sqrt{2}\right|=\sqrt{2}+1-\sqrt{2}+1=2\)
4/ \(=\sqrt{3}+2-\left|\sqrt{3}-2\right|=\sqrt{3}+2-2+\sqrt{3}=2\sqrt{3}\)
5/ \(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
6/ \(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}+1-\sqrt{3}+1=2\)
Các bạn giúp mình với, tối nay mình nộp rồi.
Câu 6 sửa lại đề giúp mình như này nhé:
\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
a, \(\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}\)
\(=\left|\sqrt{2}-1\right|+\left|3\sqrt{2}-2\right|\)
\(=\sqrt{2}-1+3\sqrt{2}-2=4\sqrt{2}-3\)
b, \(2\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2\sqrt{3}+1\right)^2}\)
\(=2\left|\sqrt{3}-1\right|-\left|2\sqrt{3}+1\right|\)
\(=2\sqrt{3}-2-2\sqrt{3}-1=-3\)
c, \(4\sqrt{\left(2-\dfrac{\sqrt{3}}{2}\right)^2}-3\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=4\left|2-\dfrac{\sqrt{3}}{2}\right|-3\left|\sqrt{3}-1\right|\)
\(=8-2\sqrt{3}-3\sqrt{3}+3=11-5\sqrt{3}\)