chiu

giáo viên dạy lý 

làm nhà bác học đi

TL :

   4 + 5 - 6 =

=   9     - 6 

= 3 

_HT_

= 1+2 nhé !

4+5-6=9-6

         =3

HT

.-.

:)

mik thik cậu ả

1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9

= (1 - 2) + (3 - 4) + (5 - 6) + (7 - 8) + 9

= -1 - 1 - 1 - 1 + 9

= -4 + 9

= 5

Câu hỏi

1-2+3-4+5-6+7-8+9

=(1-2)+(3-4)+(5-6)+(7-8)+9

=(-1)+(-1)+(-1)+(-1)+9

=(-4)+9

=5.

Hok tốt ''Đặng Trà My''

@^.^@

1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9 = 5 

Học tốt

Nhớ t.i.c.k

#Vii

\(P=4^{25}\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{7}\right)\cdot...\cdot\left(\dfrac{1}{2}-\dfrac{1}{101}\right)\)

\(P=4^{25}\cdot\left(\dfrac{1}{2}\cdot\left(\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}-...+\dfrac{1}{101}\right)\right)\)

\(P=4^{25}\cdot\left(\dfrac{1}{2}\cdot\left(\dfrac{1}{101}-\dfrac{1}{3}\right)\div2+1\right)\)

\(P=4^{25}\cdot\left(\dfrac{1}{2}\cdot\dfrac{1}{50}\right)\)

\(P=4^{25}\cdot\dfrac{1}{100}\)

\(C=1+3+3^2+3^3+...+3^{19}+3^{20}\)

\(3.C=3+3^2+3^3+3^4+...+3^{20}+3^{21}\)

\(3.C-C=3^{21}-1\)

\(\Rightarrow C=\frac{3^{21}-1}{2}\)

\(D=\)\(5+5^2+5^3+...+5^{99}+5^{100}\)

\(5.D=5^2+5^3+5^4+...+5^{100}+5^{101}\)

\(5.D-D=5^{101}-5\)

\(\Rightarrow D=\frac{5^{101}-5}{4}\)

\(C=1+3+3^2+...+3^{20}\)

\(3C=3+3^2+3^3+...+3^{21}\)

\(3C-C=\left(3+3^2+3^3+...+3^{21}\right)-\left(1+3+3^2+...+3^{20}\right)\)

\(2C=\left(3-3\right)+\left(3^2-3^2\right)+\left(3^3-3^3\right)+...+\left(3^{20}-3^{20}\right)+3^{21}-1\)

\(2C=3^{21}-1\)

\(C=\frac{3^{21}-1}{2}\)

\(D=5+5^2+5^3+...+5^{100}\)

\(5D=5^2+5^3+5^4+...+5^{101}\)

\(5D-D=\left(5^2+5^3+5^4+...+5^{101}\right)-\left(5+5^2+5^3+...+5^{100}\right)\)

\(4D=\left(5^2-5^2\right)+\left(5^3-5^3\right)+\left(5^4-5^4\right)+...+\left(5^{100}-5^{100}\right)+5^{101}-5\)

\(4D=5^{101}-5\)

\(D=\frac{5\left(5^{100}-1\right)}{4}=\frac{5}{4}\left(5^{100}-1\right)\)