Cho S=\(\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}=a+b\sqrt{3}\) trong đó a, b là các số nguyên . Tính tổng
T=a+b
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T
4 tháng 6 2019
#)Giải :
a) A = √(3+√5)-√(3-√5)-√2
<=>A√2=√(6+2√5)-√(6-2√5)-2
<=>A√2=√(√5+1)^2-√(√5-1)-2
<=>A√2=√5+1-√5+1-2
<=>A√2=0
<=>A=0
=>√(3+√5)-√(3-√5)-√2 =0
b) B=√(4-√7)-√ (4+√7)+√7
<=>B√2=√(8-2√7)-√(8+2√7)+2√7
<=>B√2=√(√7-1)^2-√(√7+1)^2+2√7
<=>B√2=√7-1-√7-1+2√7
<=>B√2=2√7-2
<=>B=(2√7-2)/√2
=√14-√2
#~Will~be~Pens~3
4 tháng 6 2019
Câu a) hình như sai đề đúng không bạn ?
b) \(B=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)
Xét \(\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)^2\)
\(=4-\sqrt{7}-2\sqrt{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}+4+\sqrt{7}\)
\(=8-2\sqrt{16-7}\)
\(=8-2\cdot3\)
\(=2\)
\(\Rightarrow\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=-\sqrt{2}\)( vì \(\sqrt{4-\sqrt{7}}< \sqrt{4+\sqrt{7}}\))
Khi đó : \(B=-\sqrt{2}+\sqrt{7}\)
Góp ý nhẹ với bạn ๖²⁴ʱŤ.Ƥεɳɠʉїɳş༉ ( Team TST 14 ) là không biết thì đừng làm nhé
8 tháng 8 2022
1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
3: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
HA
2
TA
7 tháng 11 2018
\(S^3=\left(\sqrt[3]{7+4\sqrt{3}+}\sqrt[3]{7-4\sqrt{3}}\right)^3\)
= \(7+4\sqrt{3}+7-4\sqrt{3}+3.\sqrt{7+4\sqrt{3}}.\sqrt{7-4\sqrt{3}}.\left(a+b\right)\)
= 14+\(3.\sqrt{49-48}.S\)
= 14+3S
=> S3-3S=14+3S-3S=14
7 tháng 11 2018
\(P=S^3-3S\)
\(P=\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)^3-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=7+4\sqrt{3}+3\left(\sqrt[3]{7+4\sqrt{3}}\right)^2.\sqrt[3]{7-4\sqrt{3}}+3.\sqrt[3]{7+4\sqrt{3}}\left(\sqrt[3]{7-4\sqrt{3}}\right)^2+7-4\sqrt{3}\text{}\text{}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=14+3\sqrt[3]{7+4\sqrt{3}}.\sqrt[3]{7-4\sqrt{3}}\text{}\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\text{}\text{}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=14+3\sqrt[3]{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\text{}\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\text{}\text{}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=14+3\sqrt[3]{49-48}\text{}\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\text{}\text{}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=14+3\text{}\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\text{}\text{}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=14\)
8 tháng 8 2022
1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
3: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
10 tháng 10 2021
\(\sqrt{53-20\sqrt{7}}=a+b\sqrt{7}\)
\(\Leftrightarrow a+b\sqrt{7}=-5+2\sqrt{7}\)
=> a=-5; b=2
HN
0
6 tháng 8 2019
a, A= \(\frac{\sqrt{48-12\sqrt{7}}}{2}-\frac{\sqrt{48+12\sqrt{7}}}{2}\)
= \(\frac{\sqrt{\left(\sqrt{42}-\sqrt{6}\right)^2}}{2}-\frac{\sqrt{\left(\sqrt{42}+\sqrt{6}\right)^2}}{2}\)
= \(\frac{-2\sqrt{6}}{2}\)
= \(-\sqrt{6}\)
26 tháng 7 2023
a: \(A=\dfrac{\left(\sqrt{7}+\sqrt{3}\right)^2+\left(\sqrt{7}-\sqrt{3}\right)^2}{4}\)
\(=\dfrac{10+2\sqrt{21}+10-2\sqrt{21}}{4}=\dfrac{20}{4}=5\)
b: \(B=6\sqrt{3}+\sqrt{3}-1-2\sqrt{2}\)
\(=7\sqrt{3}-2\sqrt{2}-1\)
Ta có: \(a+b\sqrt{3}=\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(\Leftrightarrow a+b\sqrt{3}=2-\sqrt{3}-2-\sqrt{3}\)
\(\Leftrightarrow a+b\sqrt{3}=-2\sqrt{3}\)
\(\Leftrightarrow a=0;b=-2\)
T=a+b=0+(-2)=-2
\(S=\sqrt{\left(\sqrt{3}\right)^2-2\cdot2\sqrt{3}+2^2}-\sqrt{\left(\sqrt{3}\right)^2+2\cdot2\cdot\sqrt{3}+2^2}\)
\(S=\sqrt{\left(\sqrt{3}-2\right)^2}-\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(S=\left|\sqrt{3}-2\right|-\left|\sqrt{3}+2\right|=-\sqrt{3}+2-\sqrt{3}-2=0+\left(-2\right)\sqrt{3}\)
\(a=0,b=-2\)
\(T=0+-2=-2\)