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a) \(\sqrt{\left(\sqrt{7}-4\right)^2}-\sqrt{28}=\left|\sqrt{7}-4\right|-2\sqrt{7}=4-\sqrt{7}-2\sqrt{7}=4-3\sqrt{7}\)
b) \(\sqrt{\left(\sqrt{5}-3\right)^2}-\sqrt{75}=\left|\sqrt{5}-3\right|-5\sqrt{3}=3-\sqrt{5}-5\sqrt{3}\)
Giải bsif toán tổng hợp
Vào Fx gõ cẩn thận ra
a)\(\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\)
b)\(\sqrt{\left(2-\sqrt{11}\right)^2}=2-\sqrt{11}\)
c)\(2\sqrt{a^2}=2a\) vì a≥0
1: =3+căn 2-3+căn 2
=2căn 2
2: =(căn 3-2)(căn 3+2)
=3-4=-1
Lời giải:
a.
\(=\sqrt{5+2.2\sqrt{5}+2^2}-\sqrt{5-2.2\sqrt{5}+2^2}\)
$=\sqrt{(\sqrt{5}+2)^2}-\sqrt{(\sqrt{5}-2)^2}$
$=|\sqrt{5}+2|-|\sqrt{5}-2|=(\sqrt{5}+2)-(\sqrt{5}-2)=4$
b.
$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$
$=\sqrt{(\sqrt{3}-3)^2}+\sqrt{(\sqrt{3}+3)^2}$
$=|\sqrt{3}-3|+|\sqrt{3}+3|$
$=(3-\sqrt{3})+(\sqrt{3}+3)=6$
c.
$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$
$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$
$=(\sqrt{2}+3)-(3-\sqrt{2})=2\sqrt{2}$
9: \(A=\dfrac{\sqrt{8+2\sqrt{15}}-\sqrt{14-6\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
10: \(A=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
11: \(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
12: \(B=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)
=9-3=6
13: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)
\(=\sqrt{5}-2-3+\sqrt{5}=2\sqrt{5}-5\)
Bài 20:
a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)
b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)
\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)
c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=2
d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)
\(=8+4\sqrt{3}-4\sqrt{3}-6\)
=2
Hihi mình cũng học lớp 9, để mình giúp cậu nha!
a) \(\sqrt{3-2\sqrt{2}}+\sqrt{9+4\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(1+\sqrt{8}\right)^2}\)
\(=\left|\sqrt{2}-1\right|+\left|1+\sqrt{8}\right|=\sqrt{2}-1+1+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\)
b) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(1+\sqrt{7}\right)^2}\)
\(=\left|\sqrt{7}-1\right|-\left|1+\sqrt{7}\right|=\sqrt{7}-1-1-\sqrt{7}=-2\)
c) \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{2}+3\right|-\left|3-\sqrt{2}\right|=\sqrt{2}+3-3+\sqrt{2}=2\sqrt{2}\)
(Nhớ click cho mình với nhoa!)