...................=66

...................=110

=66

=220

1+2+........+1+2 = 66

5+5+........+5+5 = 220

k mk nha

Cả 4 ý đều sai bạn nhé! 

Cả bốn câu sai đúng như mình suy rằng cả bốn phép tính đều sai còn các bạn khác có như đáp án của mình và khánh lưa ko nhớ nhắn cho mình nhé hi hi😀😂

cả 4 ý đều sai nhé

a: \(P=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{1}-\sqrt{3}-\sqrt{2}\)

\(=2+\sqrt{3}+2-\sqrt{2}-\sqrt{3}-\sqrt{2}\)

\(=4-2\sqrt{2}\)

b: \(N=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\sqrt{5}\right)\left(-\sqrt{5}-1\right)\)

\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)

a) \(P=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)\)

\(P=\dfrac{\sqrt{3}\cdot\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\cdot\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\sqrt{2}-\sqrt{3}\)

\(P=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-\sqrt{3}\)

\(P=2\)

b) \(N=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

\(N=\left[1-\dfrac{\sqrt{5}\left(1+\sqrt{5}\right)}{1+\sqrt{5}}\right]\left[1+\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\)

\(N=\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)\)

\(N=1^2-\left(\sqrt{5}\right)^2\)

\(N=-4\)

c) \(Q=\left(\dfrac{5+2\sqrt{5}}{2-\sqrt{5}}-2\right)\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\)

\(Q=\left[\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+2\right]\left[\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}-2\right]\)

\(Q=\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)\)

\(Q=\left(\sqrt{5}\right)^2-2^2\)

\(Q=1\)

Lời giải chi tiết:

5-2<4              5-4<2             4+1=5

5-2=3              5-3>1            5-1<5 

5-2>2              5-1=4            5-4>0

câu cuối là 999,7650928....

rồi mà, câu 3 là 100 

khó quá tui chưa học

1) \(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}=9+4\sqrt{5}\)

2) \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{2}-\sqrt{5}}=\dfrac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{-\left(\sqrt{5}-\sqrt{2}\right)}=-\sqrt{10}\)

3) \(\dfrac{\sqrt{20}-3\sqrt{10}}{3-\sqrt{5}}=\dfrac{\sqrt{10}\left(\sqrt{5}-3\right)}{-\left(\sqrt{5}-3\right)}=-\sqrt{10}\)

4) \(\dfrac{6-2\sqrt{5}}{3+\sqrt{5}}=\dfrac{\left(6-2\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}=\dfrac{18-6\sqrt{5}-6\sqrt{5}+10}{4}=\dfrac{28-12\sqrt{5}}{4}=7-3\sqrt{5}\)

5)\(\dfrac{9+4\sqrt{5}}{\sqrt{5}+2}=\sqrt{5}+2\)

2/5x2+4x2/5+2/5x5-2/5

=2/5x(2+4+5-1)

=2/5x10

=4

\(\dfrac{2}{5}\times2+4\times\dfrac{2}{5}+\dfrac{2}{5}\times5-\dfrac{2}{5}\\ =\left(2+4+5-1\right)\times\dfrac{2}{5}\\ =10\times\dfrac{2}{5}\\ =4\)

2/5 x (2+4+5+1)

=2/5 x 12

=62/5

\(C=5+5^3+5^5+...+5^{101}\)

\(5^2\cdot C=5^2\cdot\left(5+5^3+...+5^{101}\right)\)

\(25C=5^3+5^5+...+5^{103}\)

\(25C-C=\left(5^3+5^5+....+5^{103}\right)-\left(5+5^3+5^5+...+5^{101}\right)\)

\(24C=\left(5^3-5^3\right)+\left(5^5-5^5\right)+...+\left(5^{103}-5\right)\)

\(24C=5^{103}-5\)

\(C=\dfrac{5^{103}-5}{24}\) 

_____________

\(D=2^{100}-2^{99}+2^{98}-...+2^2-2+1\)

\(2D=2\cdot\left(2^{100}-2^{99}+2^{98}-...-2+1\right)\)

\(2D=2^{101}-2^{100}+2^{99}-...-2^2+2\)

\(2D+D=2^{101}-2^{100}+...-2^2+2+2^{100}-2^{99}+...-2+1\)

\(D=2^{101}+1\) 

wow gioir thaatj

\(A=2^{2010}\cdot5^{2010}\)

\(A=10^{2010}\)

suy ra có 2011 chữ số

Nhớ bấm   L I K E   cho mk nhá :))))

đấy tóan lớp 5 á :)))