Tính:
a.\(\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}\)
b.\(\left(2+\sqrt{5}\right)\sqrt{9-4\sqrt{5}}\)
Những câu hỏi liên quan
Tính:Asqrt{4+2sqrt{3}}-sqrt{4-2sqrt{3}}Bsqrt{9-4sqrt{5}}+sqrt{9+4sqrt{5}}Csqrt{4-sqrt{7}}-sqrt{4+sqrt{7}}Dsqrt{5sqrt{3+5sqrt{48-10sqrt{7+4sqrt{3}}}}}Eleft(4+sqrt{15}right)left(sqrt{10}-sqrt{6}right)sqrt{4-sqrt{15}}(2 cách)Fdfrac{sqrt{17-12sqrt{2}}}{sqrt{3-2sqrt{2}}}-dfrac{sqrt{17}+12sqrt{2}}{sqrt{3+2sqrt{2}}}
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Tính:
A=\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
B=\(\sqrt{9-4\sqrt{5}}+\sqrt{9+4\sqrt{5}}\)
C=\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
D=\(\sqrt{5\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
E=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)(2 cách)
F=\(\dfrac{\sqrt{17-12\sqrt{2}}}{\sqrt{3-2\sqrt{2}}}-\dfrac{\sqrt{17}+12\sqrt{2}}{\sqrt{3+2\sqrt{2}}}\)
\(A=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}-\sqrt{3+1-2\sqrt{3.1}}\)
\(=\sqrt{(\sqrt{3}+1)^2}-\sqrt{(\sqrt{3}-1)^2}=|\sqrt{3}+1|-|\sqrt{3}-1|=2\)
\(B=\sqrt{4+5-2\sqrt{4.5}}+\sqrt{4+5+2\sqrt{4.5}}=\sqrt{(\sqrt{4}-\sqrt{5})^2}+\sqrt{(\sqrt{4}+\sqrt{5})^2}\)
\(=|\sqrt{4}-\sqrt{5}|+|\sqrt{4}+\sqrt{5}|=2\sqrt{5}\)
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\(C\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7+1-2\sqrt{7.1}}-\sqrt{7+1+2\sqrt{7.1}}\)
\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}\)
\(=|\sqrt{7}-1|-|\sqrt{7}+1|=-2\Rightarrow C=-\sqrt{2}\)
----------------------------
\(7+4\sqrt{3}=(2+\sqrt{3})^2\Rightarrow 10\sqrt{7+4\sqrt{3}}=10(2+\sqrt{3})\)
\(\Rightarrow \sqrt{48-10\sqrt{7+4\sqrt{3}}}=\sqrt{28-10\sqrt{3}}=\sqrt{(5-\sqrt{3})^2}=5-\sqrt{3}\)
\(\Rightarrow 3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}=3+5(5-\sqrt{3})=28-5\sqrt{3}\)
\(\Rightarrow D=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
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Cách 1:
\(E=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})=(4+\sqrt{15})(8-2\sqrt{15})\)
\(=2(4+\sqrt{15})(4-\sqrt{15})=2(16-15)=2\)
Cách 2:
\(E^2=(4+\sqrt{15})^2(\sqrt{10}-\sqrt{6})^2(4-\sqrt{15})=(4+\sqrt{15})(4-\sqrt{15})(4+\sqrt{15}).(16-4\sqrt{15})\)
\(=(16-15)(4+\sqrt{15})(4-\sqrt{15}).4=(16-15)(16-15).4=4\)
Vì $E>0$ nên $E=2$
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Xem thêm câu trả lời
a) \(\left(2+\sqrt{3}\right)\cdot\sqrt{7-4\sqrt{3}}\)
b) \(\sqrt{4+2\sqrt{3}}-\sqrt{5+2\sqrt{6}}+\sqrt{2}+\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
c) \(\sqrt{3-\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(\left(2+\sqrt{3}\right)\left(\sqrt{7-4\sqrt{3}}\right)=\left(2+\sqrt{3}\right)\sqrt{4-4\sqrt{3}+3}\)
\(=\left(2+\sqrt{3}\right).\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left(2+\sqrt{3}\right)\left|2-\sqrt{3}\right|\)
\(=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)\)( Vì \(2-\sqrt{3}>0\))
\(=4-2=1\)
mk
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Tính:Asqrt{27}-2sqrt{48}+3sqrt{75}Bsqrt{left(sqrt{5}-2right)^2}-sqrt{left(sqrt{5}-3right)^2}Csqrt{left(2sqrt{3}+1right)^2}+sqrt{left(2sqrt{3}-5right)^2}Dsqrt{9-4sqrt{5}}-sqrt{14+6sqrt{5}}Edfrac{4}{sqrt{5}-2}-dfrac{32}{sqrt{5}+1}Mdfrac{10}{3sqrt{2}-4}+dfrac{28}{3sqrt{2}+2} please help ;-;
Đọc tiếp
Tính:
\(A=\sqrt{27}-2\sqrt{48}+3\sqrt{75}\)
\(B=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(C=\sqrt{\left(2\sqrt{3}+1\right)^2}+\sqrt{\left(2\sqrt{3}-5\right)^2}\)
\(D=\sqrt{9-4\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)
\(E=\dfrac{4}{\sqrt{5}-2}-\dfrac{32}{\sqrt{5}+1}\)
\(M=\dfrac{10}{3\sqrt{2}-4}+\dfrac{28}{3\sqrt{2}+2}\)
please help ;-;
\(a,\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{9-4\sqrt{5}}\)
\(b,\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}\)
\(c,\frac{\sqrt{7}-5}{2}-\frac{6}{\sqrt{7}-2}+\frac{1}{3+\sqrt{7}}+\frac{2}{5+2\sqrt{7}}\)
a) \(=\left|\sqrt{5}-3\right|+\sqrt{5-2\sqrt{20}+4}\)
\(=3-\sqrt{5}+\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=3-\sqrt{5}+\left|\sqrt{5}-2\right|\)
\(=3-\sqrt{5}+\sqrt{5}-2\)
\(=1\)
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b)\(=\left(\sqrt{5}+1\right)\cdot\sqrt{2}\cdot\left(6-2\sqrt{5}\right)\cdot\sqrt{3+\sqrt{5}}\)
\(=\left(\sqrt{5}+1\right)\left(6-2\sqrt{5}\right)\sqrt{6+2\sqrt{5}}\)
\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)^2\)
\(=16\)
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c)\(=\frac{\sqrt{7}-5}{2}-\frac{6\left(\sqrt{7}+2\right)}{7-4}+\frac{\left(3-\sqrt{7}\right)}{9-7}+\frac{2\left(5-2\sqrt{7}\right)}{25-28}\)
\(=\frac{\sqrt{7}-5}{2}-\frac{6\left(\sqrt{7}+2\right)}{3}+\frac{3-\sqrt{7}}{2}-\frac{2\left(5-2\sqrt{7}\right)}{3}\)
\(=\frac{\sqrt{7}-\sqrt{7}-5+3}{2}-\frac{6\sqrt{7}+12+10-4\sqrt{7}}{3}\)
\(=-1-\frac{2\sqrt{7}+22}{3}\)
\(=\frac{-3-2\sqrt{7}-22}{3}\)
\(=-\frac{25+2\sqrt{7}}{3}\)
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Tính:Asqrt{20}-10sqrt{dfrac{1}{5}}+sqrt{left(sqrt{5}-1right)^2}B2sqrt{32}+5sqrt{8}-4sqrt{32}Csqrt{left(3-sqrt{2}^2right)}-sqrt{left(1-sqrt{2}right)^2}Dsqrt{left(5-1right)^2}+sqrt{left(sqrt{5}-3right)^2}Eleft(3+dfrac{5-sqrt{5}}{sqrt{5}-1}right)left(3-dfrac{5+sqrt{5}}{sqrt{5}-1}right)Fsqrt{6+2sqrt{5}}-sqrt{9-4sqrt{5}}Gdfrac{3sqrt{2}-2sqrt{3}}{sqrt{3}-sqrt{2}}+dfrac{sqrt{15}-sqrt{5}}{sqrt{3}-1}Hdfrac{10}{sqrt{3}-1}-dfrac{55}{2sqrt{3}+1} help
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Tính:
\(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(C=\sqrt{\left(3-\sqrt{2}^2\right)}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(D=\sqrt{\left(5-1\right)^2}+\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(E=\left(3+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3-\dfrac{5+\sqrt{5}}{\sqrt{5}-1}\right)\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(G=\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\)
\(H=\dfrac{10}{\sqrt{3}-1}-\dfrac{55}{2\sqrt{3}+1}\)
help
a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)
\(=\sqrt{5}-1\)
b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
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Tính:
a) \(\sqrt{\left(-5\right)^2}+\sqrt{5^2}-\sqrt{\left(-3\right)^2}-\sqrt{3^2}-\left(\sqrt{7}\right)^2\)
b) \(\left[\sqrt{4^2}\right]+\sqrt{\left(-4\right)^2}.\left(\sqrt{5}\right)^2-\sqrt{5^{-2}}\)
c) \(\sqrt{\left(-10\right)^2}+10.\left(-\sqrt{5}\right)^2\)
HỘ MK BÀI NÀY NHA MỌI NGƯỜI
a) \(\sqrt{\left(-5\right)^2}+\sqrt{5^2}-\sqrt{\left(-3\right)^2}-\sqrt{3^2}-\left(\sqrt{7}\right)^2=\sqrt{25}+\sqrt{25}-\sqrt{9}-\sqrt{9}\)
\(=5+5-3-3\)
\(=4\)
c) \(\sqrt{\left(-10\right)^2}+10.\left(-\sqrt{5}\right)^2=\sqrt{100}+10.5\)
\(=10+10.5\)
\(=10+50\)
\(=60\)
Học tốt nha^^
a, \(\sqrt{\left(-5\right)^2}+\sqrt{5^2}-\sqrt{\left(-3\right)^2}-\sqrt{3^2}-\left(\sqrt{7}\right)^2=5+5-3-3-7=-4\)
b, \(\left[\sqrt{4^2}\right]+\sqrt{\left(-4\right)^2}.\left(\sqrt{5}\right)^2-\sqrt{5^{-2}}\)
\(=4+4.5-\sqrt{\frac{1}{25}}=4+20-\frac{1}{5}=\frac{119}{5}\)
c, \(\sqrt{\left(-10\right)^2}+10.\left(-\sqrt{5}\right)^2=10+10.5=10+50=60\)
Xem thêm câu trả lời
rút gọna) left(-7sqrt{7}right)left(-2sqrt{8}right)b) -sqrt{33}.3sqrt{3}c) left(3sqrt{5}right).left(-10sqrt{3}right)d) dfrac{1}{2}sqrt{5}.left(-6sqrt{2}right)e) dfrac{2}{3}sqrt{7}.left(-dfrac{9}{16}sqrt{3}right)f) 15sqrt{6}:5sqrt{3}g) -25sqrt{12}:left(-5sqrt{6}right)h) 36sqrt{8}:12sqrt{2}i) 4sqrt{27}:left(-2sqrt{3}right)
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rút gọn
a) \(\left(-7\sqrt{7}\right)\left(-2\sqrt{8}\right)\)
b) \(-\sqrt{33}.3\sqrt{3}\)
c) \(\left(3\sqrt{5}\right).\left(-10\sqrt{3}\right)\)
d) \(\dfrac{1}{2}\sqrt{5}.\left(-6\sqrt{2}\right)\)
e) \(\dfrac{2}{3}\sqrt{7}.\left(-\dfrac{9}{16}\sqrt{3}\right)\)
f) \(15\sqrt{6}:5\sqrt{3}\)
g) \(-25\sqrt{12}:\left(-5\sqrt{6}\right)\)
h) \(36\sqrt{8}:12\sqrt{2}\)
i) \(4\sqrt{27}:\left(-2\sqrt{3}\right)\)
i: =-12*căn 3/2căn 3=-6
h: =72căn 2/12căn 2=6
g: =25căn 12/5căn 6=5căn 2
f: =(15:5)*căn 6:3=3căn 2
d: =-1/2*6*căn 10=-3căn 10
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Tính:1) ( 2sqrt{5}-sqrt{7} ) left(2sqrt{5}+sqrt{7}right)2) left(5sqrt{2}+2sqrt{3}right)left(2sqrt{3}-5sqrt{2}right)3) sqrt{left(sqrt{7}-2right)^2}+sqrt{left(sqrt{7}+2right)^2}4) sqrt{left(sqrt{3}+sqrt{2}right)^2}+sqrt{left(sqrt{3}-sqrt{2}right)^2}5) left(sqrt{5}-sqrt{6}right)^26) left(sqrt{3}-sqrt{5}right)^27) left(2sqrt{2}+sqrt{3}right)^2
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Tính:
1) ( \(2\sqrt{5}-\sqrt{7}\) ) \(\left(2\sqrt{5}+\sqrt{7}\right)\)
2) \(\left(5\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{3}-5\sqrt{2}\right)\)
3) \(\sqrt{\left(\sqrt{7}-2\right)^2}+\sqrt{\left(\sqrt{7}+2\right)^2}\)
4) \(\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
5) \(\left(\sqrt{5}-\sqrt{6}\right)^2\)
6) \(\left(\sqrt{3}-\sqrt{5}\right)^2\)
7) \(\left(2\sqrt{2}+\sqrt{3}\right)^2\)
\(1,=20-7=13\\ b,=12-50=-38\\ c,=\sqrt{7}-2+\sqrt{7}+2=2\sqrt{7}\\ d,=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\\ e,=11+2\sqrt{30}\\ f,=8-2\sqrt{15}\\ g,=11+2\sqrt{6}\)
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1) \(=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)
2) \(=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
3) \(=\sqrt{7}-2+\sqrt{7}+2=2\sqrt[]{7}\)
4) \(=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
5) \(=5+6-2\sqrt{5.6}=11-2\sqrt{30}\)
6) \(=3+5-2\sqrt{3.5}=8-4\sqrt{2}\)
7) \(=\left(2\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+2\sqrt{2\sqrt{2}.3}=11+2\sqrt{6\sqrt{2}}\)
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Tính
A=\(\left(\frac{15}{\sqrt{7}+2}+\frac{12}{\sqrt{7}-1}-\frac{8}{3-\sqrt{7}}\right)\cdot\left(3\sqrt{7}+20\right)\)
B=\(\left(9+4\sqrt{5}\right):\left(\frac{\sqrt{5}+2}{\sqrt{5}-2}\right)\)
Tính (Rút gọn):
a) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
b)\(\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)
c)\(\left(\sqrt{5+2\sqrt{9\sqrt{5}-19}}-\sqrt{7-\sqrt{5}}\right):2\sqrt{\sqrt{5}-2}\)
d)\(\frac{\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{3-2\sqrt{2}}\)
a)
\((4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4-\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{3+5-2\sqrt{3.5}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})^2=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})\)
\(=2(4^2-15)=2\)
b)
\(\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}=\sqrt{(8+2\sqrt{15})+2+2(\sqrt{6}+\sqrt{10})}\)
\(=\sqrt{(\sqrt{5}+\sqrt{3})^2+2\sqrt{2}(\sqrt{3}+\sqrt{5})+2}\)
\(=\sqrt{(\sqrt{5}+\sqrt{3}+\sqrt{2})^2}=\sqrt{5}+\sqrt{3}+\sqrt{2}\)
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c)
\((\sqrt{5+2\sqrt{9\sqrt{5}-19}}-\sqrt{7-\sqrt{5}}):(2\sqrt{\sqrt{5}-2})\)
\(=(\sqrt{(5+2\sqrt{9\sqrt{5}-19})(\sqrt{5}+2)}-\sqrt{(7-\sqrt{5})(\sqrt{5}+2)}):(2\sqrt{(\sqrt{5}-2)(\sqrt{5}+2)})\)
\(=[\sqrt{10+5\sqrt{5}+2\sqrt{(9\sqrt{5}-19)(9+4\sqrt{5})}}-\sqrt{9+5\sqrt{5}}]:2\)
\(=[\sqrt{10+5\sqrt{5}+2\sqrt{9+5\sqrt{5}}}-\sqrt{9+5\sqrt{5}}]:2\)
\(=[\sqrt{(9+5\sqrt{5})+2\sqrt{9+5\sqrt{5}}+1}-\sqrt{9+5\sqrt{5}}]:2\)
\(=[\sqrt{(\sqrt{9+5\sqrt{5}}+1)^2}-\sqrt{9+5\sqrt{5}}]:2\)
\(=[\sqrt{9+5\sqrt{5}}+1-\sqrt{9+5\sqrt{5}}]:2=\frac{1}{2}\)
d)
\((\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}})^2=18+2\sqrt{(9+\sqrt{5})(9-\sqrt{5})}=18+4\sqrt{19}\)
\(\Rightarrow \sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}=\sqrt{18+4\sqrt{19}}\)
Do đó:
\(\frac{\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{3-2\sqrt{2}}=\frac{\sqrt{18+4\sqrt{19}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{2+1-2\sqrt{2.1}}\)
\(=\frac{\sqrt{2}.\sqrt{9+2\sqrt{19}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{(\sqrt{2}-1)^2}=\sqrt{2}-(\sqrt{2}-1)=1\)
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