\(\frac{7}{5}+\frac{4}{5}=\frac{7+4}{5}=\frac{11}{5}\)
\(\frac{7}{5}+\frac{4}{5}=\frac{7+4}{5}=\frac{11}{5}\)
\(\frac{7}{5}+\frac{4}{5}=\frac{7+4}{5}=\frac{11}{5}\)
\(\frac{7}{5}+\frac{4}{5}=\frac{7+4}{5}=\frac{11}{5}\)
\(\frac{7}{5}+\frac{4}{5}=\frac{7+4}{5}=\frac{11}{5}\)
\(\frac{7}{5}+\frac{4}{5}=\frac{7+4}{5}=\frac{11}{5}\)
1+2+3+4+5+6+7+8+9-1-2-3-4-5-6-7-8-9*1*2*3*4*5*6*7*8*9:1:2:3:4:5:6:7:8:9=???
THỰC HIỆN PHÉP TÍNH
1,\(\sqrt{3+\sqrt{5}}.\sqrt{2}\)
2,\(\sqrt{3-\sqrt{5}.\sqrt{8}}\)
3,\((\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\sqrt{\dfrac{4}{3})}.\sqrt{12}\)
4,\((\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{7}):\sqrt{7}\)
5, \(\sqrt{36-12\sqrt{5}}:\sqrt{6}\)
6,\(\sqrt{3-\sqrt{5}:}\sqrt{2}\)
Tính
a) \(\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
b) \(\sqrt{4+\sqrt{7}} -\sqrt{4-\sqrt{7}}\)
c) \(\sqrt{4-\sqrt{10-2\sqrt{5}}}-\sqrt{4+\sqrt{10-2\sqrt{5}}}\)
a) √(4-√7)-√(4+√7) b)√(3+√5)+√(3-√5)
c)√(2+√3)-√(2-√3)
\(\frac{5}{\sqrt{2}-7}-\frac{4}{3\sqrt{2}+5}-\frac{7}{4-5\sqrt{2}}\)
phân tích thành nhân tử
5+√5
√33 +√32
3+√3
√14+√7
√15-√6
7-√7
10-2√10
4-4√5
5-2√5
Rút gọn bth sau:
a) (√7+√5)/(√7-√5)+(√7-√5)/(√7+√5)
b) (4+√2-√3-√6+√8)/(2+√2-√3)
c) M=(√3+1).√(14-6√3)/(5+√3)
Tính:
\(a,\frac{2}{4-3\sqrt{2}}-\frac{2}{4+3\sqrt{2}}\)
\(b,\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
Cmr nếu a+b+c=0 thì:
a) \(10\left(a^7+b^7+c^7\right)=7\left(a^2+b^2+c^2\right)\left(a^5+b^5+c^5\right)\)
b) \(a^5\left(b^2+c^2\right)+b^5\left(c^2+a^2\right)+c^5\left(a^2+b^2\right)=\dfrac{1}{2}\left(a^3+b^3+c^3\right)\left(a^4+b^4+c^4\right)\)
\(A=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)\(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)