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
Tính : Gợi ý của thầy mình là nhân 2 vế cho \(\sqrt{2}\) đó các bạn .
a) A = \(\sqrt{3+\sqrt{5}-\sqrt{3-\sqrt{5}}}-\sqrt{2}\)
b) B = \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)
#)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
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
a)
\(\sqrt{3-\sqrt{5}}=\frac{\sqrt{6-2\sqrt{5}}}{2}=\sqrt{\frac{5-2\sqrt{5}.1+1}{2}}\)
\(=\sqrt{\frac{\left(\sqrt{5}-1\right)^2}{2}}=\frac{\sqrt{10}-\sqrt{2}}{2}\)
Tương tự ta cũng có: \(\sqrt{3+\sqrt{5}}=\frac{\sqrt{10}+\sqrt{2}}{2}\)
Suy ra \(A=\sqrt{\frac{\sqrt{10}+\sqrt{2}}{2}-\frac{\sqrt{10}-\sqrt{2}}{2}}-\sqrt{2}\)
\(=\sqrt[4]{2}-\sqrt{2}\)
Có gì đó sai sai ạ!
1. A=\(\sqrt{4+\sqrt{ }7}\)+ \(\sqrt{4-\sqrt{ }7}\) 2. B=\(\dfrac{\sqrt{\sqrt{7-\sqrt{3}}}-\sqrt{\sqrt{7+\sqrt{3}}}}{\sqrt{\sqrt{7}-\sqrt{2}}}\) 3. C=\(\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{4+2\sqrt{3}}}}}\) 4. D=\(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\) 5. so sánh Cho A=\(\sqrt{11+\sqrt{96}}\) B=\(\dfrac{2\sqrt{2}}{\sqrt{1+\sqrt{2-\sqrt{3}}}}\) so sánh A và B
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: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
1. Tìm các số nguyên dương a; b sao cho:
\(\dfrac{4}{a}\) \(+\) 3\(\sqrt{4-b}\) \(=\) 3\(\sqrt{4+4\sqrt{b}+b}\) \(+\) 3\(\sqrt{4-4\sqrt{b}+b}\)
2. Giải phương trình nghiệm nguyên
\(x^3-y^3-6x^2+12x=27\)
đăng câu hỏi kiểu j mà đặng đc lên như thế này đấy
1.
Đặt \(\sqrt[3]{2+\sqrt{b}}=x;\sqrt[3]{2-\sqrt{b}}=y\)
Do \(x>0\Rightarrow x^2+y^2-xy=\dfrac{3}{4}x^2+\left(\dfrac{1}{2}x-y\right)^2>0\)
\(PT\Leftrightarrow\dfrac{x^3+y^3}{a}+xy=x^2+y^2\Leftrightarrow\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{a}=x^2-xy+y^2\\ \Leftrightarrow\left(x^2-xy+y^2\right)\left(\dfrac{x+y}{a}-1\right)=0\\ \Leftrightarrow\dfrac{x+y}{a}=1\\ \Leftrightarrow\sqrt[3]{2+\sqrt{b}}+\sqrt[3]{2-\sqrt{b}}=a\left(1\right)\\ \Leftrightarrow\left(\sqrt[3]{2+\sqrt{b}}+\sqrt[3]{2-\sqrt{b}}\right)^3=a^3\\ \Leftrightarrow4+3a\sqrt[3]{4-b}=a^3\left(2\right)\\ \Rightarrow4-b=\left(\dfrac{a^3-4}{3a}\right)^3\)
Mặt khác \(b\in \mathbb{Z^+}\)
\(\Rightarrow\left(a^3-4\right)⋮3a\Rightarrow\left(a^3-4\right)⋮a\\ \Rightarrow4⋮a\Rightarrow a\in\left\{1;2;4\right\}\)
Với \(a=1\Rightarrow4-b=1\Rightarrow b=5\)
Với \(a=2;a=4\Rightarrow b\notin \mathbb{Z}\)
Vậy \(\left(a;b\right)=\left(1;5\right)\)
Cho a=\(\sqrt[3]{7+4\sqrt{3}}\);b=\(\sqrt[3]{7-4\sqrt{3}}\)
Đặt S=a+b;Tính P=S3-3S
\(S^3=\left(\sqrt[3]{7+4\sqrt{3}+}\sqrt[3]{7-4\sqrt{3}}\right)^3\)
= \(7+4\sqrt{3}+7-4\sqrt{3}+3.\sqrt{7+4\sqrt{3}}.\sqrt{7-4\sqrt{3}}.\left(a+b\right)\)
= 14+\(3.\sqrt{49-48}.S\)
= 14+3S
=> S3-3S=14+3S-3S=14
\(P=S^3-3S\)
\(P=\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)^3-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=7+4\sqrt{3}+3\left(\sqrt[3]{7+4\sqrt{3}}\right)^2.\sqrt[3]{7-4\sqrt{3}}+3.\sqrt[3]{7+4\sqrt{3}}\left(\sqrt[3]{7-4\sqrt{3}}\right)^2+7-4\sqrt{3}\text{}\text{}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=14+3\sqrt[3]{7+4\sqrt{3}}.\sqrt[3]{7-4\sqrt{3}}\text{}\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\text{}\text{}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=14+3\sqrt[3]{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\text{}\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\text{}\text{}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=14+3\sqrt[3]{49-48}\text{}\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\text{}\text{}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=14+3\text{}\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\text{}\text{}-3\left(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right)\)
\(P=14\)
1. A=\(\sqrt{4+\sqrt{7}}\) +\(\sqrt{4-\sqrt{7}}\)
2. B= \(\dfrac{\sqrt{\sqrt{7-\sqrt{3}}-\sqrt{7+\sqrt{3}}}}{\sqrt{7-\sqrt{2}}}\)
3. C=\(\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{4+2\sqrt{3}}}}}\)
4. D=\(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
5 E=\(\dfrac{1+\sqrt{5}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\) +\(\dfrac{1-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
6. so sánh Cho A=\(\sqrt{11+\sqrt{96}}\)
B= \(\dfrac{2\sqrt{2}}{1+\sqrt{2-\sqrt{3}}}\) so sánh A và b
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: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
cho \(\sqrt{53-20\sqrt{7}}=a+b\sqrt{7}\) (với a, b là các số nguyên). Khi đó a-b bằng bao nhiêu
\(\sqrt{53-20\sqrt{7}}=a+b\sqrt{7}\)
\(\Leftrightarrow a+b\sqrt{7}=-5+2\sqrt{7}\)
=> a=-5; b=2
so sánh A\(=\sqrt[3]{4+\sqrt{7}}-\sqrt[3]{4-\sqrt{7}}\)
B \(=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
Thực hiện các phép tính sau:
a, A=\(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
b, B=\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
c, C=\(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
a, A= \(\frac{\sqrt{48-12\sqrt{7}}}{2}-\frac{\sqrt{48+12\sqrt{7}}}{2}\)
= \(\frac{\sqrt{\left(\sqrt{42}-\sqrt{6}\right)^2}}{2}-\frac{\sqrt{\left(\sqrt{42}+\sqrt{6}\right)^2}}{2}\)
= \(\frac{-2\sqrt{6}}{2}\)
= \(-\sqrt{6}\)
a) A = (sqrt(7) + sqrt(3))/(sqrt(7) - sqrt(3)) + (sqrt(7) - sqrt(3))/(sqrt(7) + sqrt(3)) b) B = 2sqrt(27) + sqrt((1 - sqrt(3)) ^ 2) - 4/(sqrt(2))
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: \(B=6\sqrt{3}+\sqrt{3}-1-2\sqrt{2}\)
\(=7\sqrt{3}-2\sqrt{2}-1\)