(5/2-x)(3x+1)=0

=>\(\left[{}\begin{matrix}\dfrac{5}{2}-x=0\\3x+1=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

Có 4 giao điểm 

1: \(D=3^0+3^1+...+3^{302}\)

=>\(3D=3+3^2+...+3^{303}\)

=>\(3D-D=3+3^2+...+3^{303}-3^0-3^1-...-3^{202}\)

=>\(2D=3^{303}-1\)

=>\(2D+1=3^{303}\)

=>\(27n=3^{303}\)

=>\(n=3^{300}\)

1)\(D=3^0+3^1+...+3^{302}\)

\(\Rightarrow3D=3\left(1+3+3^2+...+3^{302}\right)\)

\(\Rightarrow3D=3+3^2+3^3+...+3^{302}+3^{303}\)

\(\Rightarrow3D-D=\left(3+3^2+3^3+...+3^{303}\right)-\left(3^0+3^1+...+3^{302}\right)\)

\(\Rightarrow2D=3^{303}-3^0\)

\(\Rightarrow2D=3^{303}-1\)

\(\Rightarrow2D-1=3^{303}\)

\(Do3^{303}=\left(3^3\right)^{101}=27^{101}\)

\(\Rightarrow2D+1=27^{101}=27^n\)

\(\Rightarrow n=101\)

được giảm số phần trăm là :

1200000:8000000=0,15=15%

Bác An đã được giảm giá số % là : 

1,200,000 : 8,000,000 = 0,15 = 15% 

           Đ/S : .....................

Đặt A = 1/101 + 1/102 + 1/103 + ... + 1/199 + 1/200

Số số hạng của A:

200 - 101 + 1 = 100 (số hạng)

Ta có:

1/101 < 1/100

1/102 < 1/100

1/103 < 1/100

...

1/200 < 1/100

Cộng vế với vế, ta có:

1/101 + 1/102 + 1/103 + ... + 1/199 + 1/200 < 1/100 + 1/100 + 1/100 + ... + 1/100

⇒ A < 100/100 = 1

Vậy A < 1

\(\dfrac{1}{101}\)+\(\)....+\(\dfrac{1}{200}\)<\(\dfrac{1}{101}\).(200-101+1)

                      =\(\dfrac{100}{101}\)<1

\(\dfrac{1}{101}\)

Đặt A = 1/4 + 1/(4.7) + 1/(7.10) + ... + 1/(94.97) + 1/(97.100)

= 1/3 . (1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/94 - 1/97 + 1/97 - 1/100)

= 1/3 . (1 - 1/100)

= 1/3 - 1/300 < 1/3

Vậy A < 1/3

(2/3+2/7-1/14) / (-1-3/7+3/28)

=(2/3+2/7-2/28) / (-3/3-3/7+3/28)

=[2.(1/3+1/7-1/28)]/[(-3).(1/3+1/7-1/28)]

=2/-3=-2/3

a;   \(\dfrac{122}{13}\) - (\(\dfrac{12}{5}\) + \(\dfrac{57}{13}\))

A = \(\dfrac{122}{13}\) - \(\dfrac{12}{5}\) - \(\dfrac{57}{13}\)

A =  \(\dfrac{122}{13}\) - \(\dfrac{57}{13}\) - \(\dfrac{12}{5}\)

A = 5 - \(\dfrac{12}{5}\)

A = \(\dfrac{13}{5}\)

b; (\(\dfrac{4}{17}\) - \(\dfrac{4}{49}\) - \(\dfrac{4}{131}\)) : (\(\dfrac{3}{17}\) - \(\dfrac{3}{49}\) - \(\dfrac{3}{131}\))

 = 4.(\(\dfrac{1}{17}-\dfrac{1}{49}-\dfrac{1}{131}\)) : [3.(\(\dfrac{1}{17}-\dfrac{1}{49}-\dfrac{1}{131}\))]

= \(\dfrac{4}{3}\)

a: \(\left(\dfrac{1}{17}-\dfrac{1}{57}+\dfrac{1}{23}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

\(=\left(\dfrac{1}{17}-\dfrac{1}{57}+\dfrac{1}{23}\right)\cdot\dfrac{3-2-1}{6}\)

=0

b: \(\left(\dfrac{17}{3}-\dfrac{29}{6}\right)+\dfrac{7}{6}=\dfrac{17}{3}-\dfrac{29}{6}+\dfrac{7}{6}\)

\(=\dfrac{17}{3}-\dfrac{22}{6}=\dfrac{17}{3}-\dfrac{11}{3}=\dfrac{6}{3}=2\)

Các phép suy luận sau đây là tương đương.

Do 8, 3 và 23 nguyên tố cùng nhau nên:

(3a+4b) chia hết 23 <==> 8(3a+4b) chia hết 23

<==> 8(3a+4b) = 24a + 32b = (24a + 9b) +23b  chia hết 23

<==> 24a+ 9b chia hết 23 <==> 3(8a+3b) chia hết 23

<==> 8a+3b chia hết 23 (đpcm).