\(\dfrac{1}{5}+\dfrac{4}{5}xX=\dfrac{7}{3}\\ \dfrac{4}{5}xX=\dfrac{7}{3}-\dfrac{1}{5}\\ \dfrac{4}{5}xX=\dfrac{32}{15}\\ X=\dfrac{32}{15}:\dfrac{4}{5}\\ \text{X}=\dfrac{8}{3}\)

8 phần 3

a) \(3^2.5+2^3.10-81:3\\ =9.5+8.10-27\\ =45+80-27=98\)

b) \(5^{13}:5^{10}-25.2^2\\ =5^3-25.4\\ =125-100=25\)

c) \(20:2^2+5^9:5^8\\ =20:4+5^1\\ =5+5=10\)

d) \(100:5^2+7.3^2\\ =100:25+7.9\\ =4+63=67\)

e) \(84:4+3^9:3^7+5^0\\ =21+3^2+1\\ =21+9+1=31\)

98

25

10

67

31

3x-7=x-7

=> 3x-x=7-7

=> 2x=0

=> x=0:2

=> x=0

x=0

lạ vải vậy mấy tụi nhóc con !

Ta có:

\((x+3)\vdots(x+2)\\\Rightarrow (x+2)+1\vdots(x+2)\\\Rightarrow 1\vdots (x+2)\\\Rightarrow x+2\inƯ(1)\\\Rightarrow x+2\in\{1;-1\}\\\Rightarrow x\in\{-1;-3\}\)

Vì (x+3) > (x+2) 1 đơn vị

⇒ Ta có 2 ⋮ 1 và 0 ⋮ -1

+) x + 3 = 2                    x + 2 = 1

          x = 2 - 3                     x = 1 - 2

          x = -1                         x = -1

+) x + 3 = 0                 x + 2 = -1      

          x = 0 - 3                  x = -1 - 2

          x = -3                      x = -3

Vậy x ϵ { -1 ; -3 }

$3x+12=2x-10$

$\Rightarrow 3x-2x=-10-12$

$\Rightarrow x=-22$

x=-22

$[(4-x).3+(-17).(-3)]:3-2^2=-2.(-7)$

$\Rightarrow [(4-x).3+17.3]:3-4=14$

$\Rightarrow [3.(4-x+17)]:3=14+4$

$\Rightarrow 21-x=18$

$\Rightarrow x=21-18$

$\Rightarrow x=3$

=3

+, Phân số $\frac{51}{61}$ đã tối giản.

+, $\frac{112}{648}=\frac{112:8}{648:8}=\frac{14}{81}$

`51/61` là phân số tối giản

\(\dfrac{112}{648}=\dfrac{112:2}{648:2}=\dfrac{56}{324}=\dfrac{56:2}{324:2}=\dfrac{28}{162}=\dfrac{28:2}{162:2}=\dfrac{14}{81}\)

(3 + x).2 - 47 = -147

(3 + x).2 = -147 + 47 

(3 + x).2= - 100 

3 + x = -100 : 2 

3 + x = -50

x = -50 - 3 

x = -53

\(\left(3+x\right)\cdot2-47=-147\)

=>\(2\left(x+3\right)=-147+47=-100\)

=>x+3=-50

=>x=-53

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2023\cdot2024}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\)

\(=1-\dfrac{1}{2024}=\dfrac{2023}{2024}\)

\(B=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{85\cdot89}\)

\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{85}-\dfrac{1}{89}\)

\(=1-\dfrac{1}{89}=\dfrac{88}{89}\)

\(C=\dfrac{7}{10\cdot11}+\dfrac{7}{11\cdot12}+...+\dfrac{7}{69\cdot70}\)

\(=7\left(\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}+...+\dfrac{1}{69\cdot70}\right)\)

\(=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)

\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=7\cdot\dfrac{6}{70}=\dfrac{42}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)

\(D=\dfrac{1}{18}+\dfrac{1}{54}+...+\dfrac{1}{990}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+...+\dfrac{3}{30\cdot33}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2023\cdot2024}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\)

\(=1-\dfrac{1}{2024}\)

\(=\dfrac{2023}{2024}\)

\(B=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{85\cdot89}\)

\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{85}-\dfrac{1}{89}\)

\(=1-\dfrac{1}{89}\)

\(=\dfrac{88}{89}\)

\(C=\dfrac{7}{10\cdot11}+\dfrac{7}{11\cdot12}+...+\dfrac{7}{69\cdot70}\)

\(=7\left(\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}+...+\dfrac{1}{69\cdot70}\right)\)

\(=7\cdot\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)

\(=7\cdot\left(\dfrac{1}{10}-\dfrac{1}{70}\right)\)

\(=7\cdot\dfrac{6}{70}\)

\(=\dfrac{3}{5}\)

\(D=\dfrac{1}{18}+\dfrac{1}{54}+...+\dfrac{1}{990}\)

\(=\dfrac{1}{3\cdot6}+\dfrac{1}{6\cdot9}+\dfrac{1}{9\cdot12}+...+\dfrac{1}{30\cdot33}\)

\(=\dfrac{1}{3}\cdot\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+...+\dfrac{3}{30\cdot33}\right)\)

\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{6}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\cdot\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\\ =\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)