nếu \(\sqrt{14-6\sqrt{5}}=a+b\sqrt{5}\)
thì a+b=?
Giải chi tiết giùm mình luôn nha.
bài1: Nếu \(\sqrt{14-6\sqrt{5}}\)= \(a+b\sqrt{5}\)Với a,b \(\in\)Z, thì a+b=?
bài2: Nếu \(\sqrt{55-6\sqrt{6}}=a+b\sqrt{6}\)Với a,b \(\in\)Z, thì a-b=?
1/ Ta có √(14 - 6√5) = √(9 - 6√5 +5) = 3 - √5
Từ đó a + b = 2
2/ Đề sai sửa lại là
√(15 - 6√6) = √(9 - 6√6 + 6) = (3 - √6)
Vậy a = 3; b = -1
=> a + b = 2
Nếu , với , thì ....
\(\sqrt{14-6\sqrt{5}}=\sqrt{14-2\sqrt{9.5}}=\sqrt{\left(3-\sqrt{5}\right)^2}\)
=\(3-\sqrt{5}\)
=> a=3 và b=-1
=> a+b=3-1=2
Nếu \(\frac{23\sqrt{2}}{\sqrt{2}+\sqrt{14+5\sqrt{3}}}=a+b\sqrt{3}\)với a;b là các số hữu tỉ thì ab=....
Tính:
\(a.\) \(A=\sqrt{12}-2\sqrt{48}+\dfrac{7}{5}\sqrt{75}\)
\(b.\) \(B=\sqrt{14-6\sqrt{5}}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(c.\) \(C=\left(\sqrt{6}-\sqrt{2}\right)\sqrt{2+\sqrt{3}}\)
\(d.\) \(D=\dfrac{5+\sqrt{5}}{\sqrt{5}+2}+\dfrac{\sqrt{5}-5}{\sqrt{5}}-\dfrac{11}{2\sqrt{5}+3}\)
a)A=\(2\sqrt{3}-8\sqrt{3}+7\sqrt{3}=\sqrt{3}\)
b)B\(=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}=3-\sqrt{5}+\sqrt{5}-2=1\)
d)\(=\dfrac{\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)}{1}+1-\sqrt{5}-\dfrac{11\left(2\sqrt{5}-3\right)}{11}=5\sqrt{5}+5-10-2\sqrt{5}+1-\sqrt{5}-2\sqrt{5}+3=-1\)
Nếu \(\sqrt{14-6\sqrt{5}}=a+b\sqrt{5}\) với \(a,b\in Z\) thì \(a+b=\) ?
\(\sqrt{14-6\sqrt{5}}=\left(3-\sqrt{5}\right)^{ }\)
suy ra a=3 ; b=-1
suy ra a+b=3+(-1)=2
Rút gọn:
A=\(\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
B=\(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
C=\(\sqrt{14-6\sqrt{5}-\sqrt{14+6\sqrt{5}}}\)
\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(A=\sqrt{5}-1-\sqrt{5}-1\)
\(A=-2\)
\(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(B=\sqrt{5}+2-\sqrt{5}+2\)
\(B=4\)
Sửa đề :
\(C=\sqrt{14-6\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)
\(C=\sqrt{\left(3-\sqrt{5}\right)^2}-\sqrt{\left(3+\sqrt{5}\right)^2}\)
\(C=3-\sqrt{5}-3-\sqrt{5}\)
\(C=-2\sqrt{5}\)
* Tính
a. A=\(\left(\dfrac{6+\sqrt{20}}{3+\sqrt{5}}+\dfrac{\sqrt{14}-\sqrt{2}}{\sqrt{7}-1}\right):\left(2+\sqrt{2}\right)\)
b. B=\(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\dfrac{11}{2\sqrt{3}+1}\)
a: Ta có: \(A=\left(\dfrac{6+\sqrt{20}}{3+\sqrt{5}}+\dfrac{\sqrt{14}-\sqrt{2}}{\sqrt{7}-1}\right):\left(2+\sqrt{2}\right)\)
\(=\left(2+\sqrt{2}\right):\left(2+\sqrt{2}\right)\)
=1
b: Ta có: \(B=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\dfrac{11}{2\sqrt{3}+1}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+1\)
=1
a) \(\sqrt{28-2\sqrt{3}}+\sqrt{7}.\sqrt{7}+\sqrt{84}\)
b) \(\sqrt{14-6\sqrt{5}+\sqrt{14+6\sqrt{5}}}\)
Rút gọn
a) A=\(\sqrt{14+6\sqrt{5}}+\sqrt{14-6\sqrt{5}}\)
b)B=\(\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
Giúp e với ạ
\(A=\sqrt{14+6\sqrt{5}}+\sqrt{14-6\sqrt{5}}\)
\(A=\sqrt{9+6\sqrt{5}+5}+\sqrt{9-6\sqrt{5}+5}\)
\(A=\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(A=3+\sqrt{5}+3-\sqrt{5}=6\)
b) \(B=\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(B=\sqrt{3-4\sqrt{3}+4}-\sqrt{3+4\sqrt{3}+4}\)
\(B=\sqrt{\left(\sqrt{3}-2\right)^2}-\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(B=2-\sqrt{3}-\sqrt{3}-2=-2\sqrt{3}\)
Câu a tách 14 thành 5+9 . Có hằng đẳng thức
Câu b tương tự tách 7 thành 4+ 3 nhé