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19 tháng 8 2020
a) \(-0.8\sqrt{\left(-0.125\right)^2}=-0.8\left|-0.125\right|=-0.8\times0.125=0,1\)
b) \(\sqrt{\left(-2\right)^6}=\sqrt{2^6}=\sqrt{\left(2^3\right)^2}=\left|8\right|=8\)
QP
20 tháng 7 2017
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
20 tháng 7 2017
hình như ở câu a) chỗ sau dấu bằng đầu tiên bạn bị sai dấu trừ cuối cùng
17 tháng 7 2018
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
3 tháng 9 2019
a) \(\sqrt{25}+\sqrt{9}-\sqrt{16}\) = \(\sqrt{5^2}+\sqrt{3^2}-\sqrt{4^2}\) = 5 + 3 - 4 = 4
b) \(\sqrt{0,16}+\sqrt{0,01}+\sqrt{0,25}\) = 0,4 + 0,1 + 0,5 = 1
c) \(\left(\sqrt{3^2}\right)-\left(\sqrt{2^2}\right)+\left(\sqrt{5^2}\right)\)
= 3 - 2 + 5 = 6
d) \(\sqrt{4}-\left(-\sqrt{3}\right)^2+\sqrt{49}\) = 2 - 3 + 7 = 6
e) \(\left(2\sqrt{2}\right)^2-\left(3\sqrt{3}\right)^2\)
= \(\left(\sqrt{8}\right)^2-\left(\sqrt{27}\right)^2\) = 8 - 27 = -19
f) \(\left(-2\sqrt{2}\right)^2+\left(3\sqrt{3}\right)^2\) = 8 + 27 = 35
KN
27 tháng 6 2019
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) \(\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3}\)\(=2\sqrt{5}\)
KN
27 tháng 6 2019
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
TN
13 tháng 6 2019
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
24 tháng 8 2020
a) Ta có: \(\left(\sqrt{6}+\sqrt{2}\right)\cdot\left(\sqrt{3}-2\right)\cdot\left(\sqrt{2+\sqrt{3}}\right)\)
\(=\sqrt{2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{4+2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\left|\sqrt{3}+1\right|\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)(Vì \(\sqrt{3}>1>0\))
\(=\left(4+2\sqrt{3}\right)\cdot\left(\sqrt{3}-2\right)\)
\(=2\cdot\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)\)
\(=2\cdot\left(3-4\right)\)
\(=-2\)
b) Ta có: \(\sqrt{2}\cdot\left(\sqrt{2-\sqrt{3}}\right)\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}+1\right)\)
\(=\left|\sqrt{3}-1\right|\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)(Vì \(\sqrt{3}>1\))
\(=3-1=2\)
c) Ta có: \(\left(\sqrt{10}-\sqrt{6}\right)\cdot\left(\sqrt{4-\sqrt{15}}\right)\)
\(=\sqrt{2}\cdot\sqrt{4-\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=8-2\sqrt{15}\)
d) Ta có: \(\left(\sqrt{3}-\sqrt{12}\right)\cdot\left(\sqrt{5+2\sqrt{6}}\right)\)
\(=\sqrt{3}\cdot\left(1-2\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)
\(=-\sqrt{3}\cdot\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=-\sqrt{3}\cdot\left|\sqrt{3}+\sqrt{2}\right|\)
\(=-\sqrt{3}\cdot\left(\sqrt{3}+\sqrt{2}\right)\)(Vì \(\sqrt{3}>\sqrt{2}>0\))
\(=-3-\sqrt{6}\)
e) Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\cdot\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\left|\sqrt{3}-1\right|\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)\left(\sqrt{3}+2\right)\)(Vì \(\sqrt{3}>1\))
\(=\frac{\left(4-2\sqrt{3}\right)\left(4+2\sqrt{3}\right)}{2}\)
\(=\frac{16-12}{2}=\frac{4}{2}=2\)
f) Ta có: \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+2\cdot2\cdot\sqrt{3}+3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left|2+\sqrt{3}\right|}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)(Vì \(2>\sqrt{3}>0\))
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left|5-\sqrt{3}\right|}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)(Vì \(5>\sqrt{3}\))
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+\sqrt{25}}\)
\(=\sqrt{4+5}=\sqrt{9}=3\)
TT
1 tháng 7 2018
\(\left(\sqrt{3}-\sqrt{5}\right)^2=3-2\sqrt{15}+5=8-2\sqrt{15}\)
\(\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)=7-2=5\)
\(\left(\sqrt{3}+\sqrt{5}\right)^2=3+2\sqrt{15}+5=8+2\sqrt{15}\)
By Thành Trương
\(-0,8\sqrt{\left(-0,125\right)^2}\)
\(=-0,8\cdot\left|-0,125\right|\)
\(=-0,8\cdot0,125\)
\(=-0,1\)
nhanh đới