Nhà hán do Lưu Bang lập ra tồn tại từ 206 TCN đến 220 CN

Lưu Bang ( Hán Cao Tổ )202 TCN-220 SCN

    [\(\dfrac{16}{3}\) - 2².\(\dfrac{5}{3}\)].[\(\dfrac{17}{3}\) - 2².\(\dfrac{5}{3}\)]...[\(\dfrac{30}{3}\) - 2².\(\dfrac{5}{3}\)]

= [\(\dfrac{16}{3}\) - \(\dfrac{20}{3}\)].[\(\dfrac{17}{3}\) - \(\dfrac{20}{3}\)][\(\dfrac{18}{3}\)-\(\dfrac{20}{3}\)][\(\dfrac{19}{3}\)-\(\dfrac{20}{3}\)].[\(\dfrac{20}{3}\)-\(\dfrac{20}{3}\)]...[\(\dfrac{30}{3}\) - \(\dfrac{20}{3}\)]

=[\(\dfrac{16}{3}\) - \(\dfrac{20}{3}\)].[\(\dfrac{17}{3}\)- \(\dfrac{20}{3}\)].[\(\dfrac{18}{3}\) - \(\dfrac{20}{3}\)].[\(\dfrac{19}{3}\) - \(\dfrac{20}{3}\)].0...[\(\dfrac{30}{3}\) - \(\dfrac{20}{3}\)]
= 0

\(a,\left|x-\dfrac{1}{2}\right|+\dfrac{1}{3}=\dfrac{2}{3}\\ \Rightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{1}{3}\\x-\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\\x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\end{matrix}\right.\\ b,\dfrac{x}{-2}=\dfrac{y}{5}=\dfrac{x-y}{-2-5}=\dfrac{14}{-7}=-2\\ \Rightarrow x=-2.\left(-2\right)=4;y=-2.5=-10\)

x-y=-14 nhé

\(\dfrac{4^2+2^3+2^2}{7}\)

\(=\dfrac{16+8+4}{7}\)

\(=\dfrac{24+4}{7}\)

\(=\dfrac{28}{7}\)

\(=4\)

 mới lớp 5

a) x = 1

⇒ 3a - 1 = 5

⇒ 3a = 6

⇒ a = 2

b) x = 5

⇒ 3a - 1 = 1

⇒ 3a = 2

⇒ a = 2/3

\(a,\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{13}\\ =\left(31-36\right)+\left(\dfrac{6}{13}-\dfrac{6}{13}\right)+5\dfrac{9}{41}\\ =-5+0+5\dfrac{9}{41}\\ =\left(-5+5\right)+\dfrac{9}{41}=\dfrac{9}{41}\)

\(b,\dfrac{5}{3}+\left(-\dfrac{2}{7}\right)-\left(-1,2\right)\\ =\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}\\ =\dfrac{5.35-2.15+6.21}{105}=\dfrac{271}{105}\\ c,0,25+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-1\dfrac{1}{4}\\ =\left(-1\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)-\dfrac{1}{8}=-1+1-\dfrac{1}{8}=-\dfrac{1}{8}\)

a) (31 6/13 + 5 9/41) - 36 6/13

= 409/13 + 214/41 - 474/13

= (409/13 - 474/13) + 214/41

= -5 + 214/41

= 9/41

b) 5/3 + (-2/7) - (-1,2)

= 5/3 - 2/7 + 6/5

= 29/21 + 6/5

= 271/105

c) 0,25 + 3/5 - (1/8 - 2/5 + 1 1/4)

= 1/4 + 3/5 - 1/8 + 2/5 - 5/4

= (1/4 - 5/4) + (3/5 + 2/5) - 1/8

= -1 + 1 - 1/8

= -1/8

Lời giải:
Ta có:
$\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}=\frac{-(x-y)}{xy}=\frac{-xy}{xy}=-1$

Gọi 120 số 1 hoặc -1 đó lần lượt là a₁; a₂; a₃; ...; a₁₂₀. Theo đề ta có:

a₁.a₂.a₃ = -1; a₂.a₃.a₄ = -1; a₃.a₄.a₅ = -1; ...;

a₁₁₈.a₁₁₉.a₁₂₀ = -1; a₁₁₉.a₁₂₀.a₁ = -1; a₁₂₀.a₁.a₂ = -1.

\(a_1=a_4=\dfrac{1}{a_2\cdot a_3}\); \(a_2=a_5=\dfrac{1}{a_3\cdot a_4}\); \(a_3=a_6=\dfrac{1}{a_4\cdot a_5}\); ...;

\(a_{118}=a_1=\dfrac{1}{a_{119}\cdot a_{120}}\); \(a_{119}=a_2=\dfrac{1}{a_{120}\cdot a_1}\); \(a_{120}=a_3=\dfrac{1}{a_1\cdot a_2}\).

Từ đây ta suy ra \(a_1=a_4=a_7=...=a_{118}\); \(a_2=a_5=a_8=...=a_{119}\); \(a_3=a_6=a_9=...=a_{120}\). (1)

Do đó \(a_1=\dfrac{1}{a_2\cdot a_3}\); \(a_2=\dfrac{1}{a_3\cdot a_1}\); \(a_3=\dfrac{1}{a_1\cdot a_2}\). Mà a₁.a₂.a₃ = -1 và các số a₁; a₂; a₃; ...; a₁₂₀ chỉ có thể là 1 hoặc -1 nên chỉ có một nghiệm duy nhất \(a_1=a_2=a_3=-1\). (2)

Từ (1) và (2) suy ra có 120 số -1, nên tổng của 120 số đó là \(120\cdot\left(-1\right)=-120\).