Vận tốc thanh niên 1 chạy :

\(600:\left(15x60\right)=\dfrac{2}{3}\left(m/s\right)\)

Vận tốc thanh niên 2 chạy :

\(230:32=\dfrac{115}{16}\left(m/s\right)\)

Vận tốc thanh niên 3 chạy :

\(450:\left(3x60\right)=\dfrac{5}{2}\left(m/s\right)\)

Bài 1:

1. $\frac{4}{15}+\frac{11}{15}=\frac{4+11}{15}=\frac{15}{15}=1$

2. $\frac{5}{6}+\frac{14}{12}=\frac{5}{6}+\frac{7}{6}=\frac{5+7}{6}=\frac{12}{6}=2$
3. $3+\frac{3}{5}=\frac{15}{5}+\frac{3}{5}=\frac{18}{5}$

4. $\frac{15}{13}-\frac{2}{13}=\frac{13}{13}=1$

5. $\frac{3}{4}-\frac{1}{3}=\frac{9}{12}-\frac{4}{12}=\frac{5}{12}$
 

Bài 1:

6.

$\frac{3}{5}-\frac{6}{15}=\frac{3}{5}-\frac{2}{5}=\frac{1}{5}$

7.

$\frac{18}{5}-3=\frac{18}{5}-\frac{15}{5}=\frac{3}{5}$

8.

$7-\frac{5}{7}=\frac{49}{7}-\frac{5}{7}=\frac{44}{7}$
9.

$\frac{6}{7}\times \frac{14}{36}=\frac{6\times 14}{7\times 36}=\frac{6\times 7\times 2}{7\times 7\times 6}=\frac{2}{6}=\frac{1}{3}$

10.

$3\times \frac{7}{9}=\frac{3\times 7}{3\times 3}=\frac{7}{3}$

f) \(\left(x-1\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy: \(S=\left\{1;2\right\}\)

a) Diện tích để làm bể kính là:

\(2\times\left(90+60\right)\times45+\left(90\times60\right)=18900\left(cm^2\right)=189\left(dm^2\right)\)

b) Đổi: \(245\left(l\right)=245\left(dm^3\right)\)

Thể tích của bể:

\(90\times60\times45=243000\left(cm^3\right)=243\left(dm^3\right)\)

Ta thấy: \(245< 243\) vậy thì nước sẽ tràn ra

a) Diện tích kính xung quanh bể :

\(\left(90+60\right)x2x45=13500\left(cm^2\right)\)

Cộng thêm diện tích kính đáy bể :

\(13500+90x60=18900\left(cm^2\right)\)

b) Thể tích bể :

\(90x60x45=243000\left(cm^2\right)=0,243\left(m^3\right)\)

Nếu người ta đổ \(245l=0,245m^3>0,243m^3\) 

⇒ Nước sẽ bị tràn khỏi bể

`@` `\text {Ans}`

`\downarrow`

`a)`

\(\dfrac{3}{2}\times\dfrac{4}{5}-x=\dfrac{2}{3}\)

\(\dfrac{6}{5}-x=\dfrac{2}{3}\)

\(x=\dfrac{6}{5}-\dfrac{2}{3}\)

\(x=\dfrac{18}{15}-\dfrac{10}{15}\)

\(x=\dfrac{8}{15}\)

Vậy, `x =`\(\dfrac{8}{15}\)

`b)`

\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)

\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)

\(x=\dfrac{40}{51}\div\dfrac{10}{3}\)

\(x=\dfrac{4}{17}\)

Vậy, \(x=\dfrac{4}{17}\)

`c)`

\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)

\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)

\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)

\(x=\dfrac{34}{7}\)

Vậy, `x = `\(\dfrac{34}{7}\)

a) \(\dfrac{3}{2}x\dfrac{4}{5}-x=\dfrac{2}{3}\Rightarrow\dfrac{6}{5}-x=\dfrac{2}{3}\Rightarrow x=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{18}{15}-\dfrac{10}{15}=\dfrac{8}{15}\)

b) \(x.3\dfrac{1}{3}=3\dfrac{1}{3}:4\dfrac{1}{4}\Rightarrow\dfrac{10}{3}.x=\dfrac{10}{3}:\dfrac{17}{4}\Rightarrow\dfrac{10}{3}.x=\dfrac{10}{3}.\dfrac{4}{17}\Rightarrow x=\dfrac{10}{3}.\dfrac{4}{17}:\dfrac{10}{3}=\dfrac{10}{3}.\dfrac{4}{17}.\dfrac{3}{10}=\dfrac{4}{17}\)

c) \(5\dfrac{2}{3}:x=3\dfrac{2}{3}-2\dfrac{1}{2}\Rightarrow\dfrac{17}{3}:x=\dfrac{11}{3}-\dfrac{5}{2}\Rightarrow\dfrac{17}{3}:x=\dfrac{22}{6}-\dfrac{15}{6}\Rightarrow\dfrac{17}{3}:x=\dfrac{7}{6}\Rightarrow x=\dfrac{17}{3}:\dfrac{7}{6}=\dfrac{17}{3}.\dfrac{7}{6}=\dfrac{119}{18}\)

a) \(4\dfrac{1}{2}+\dfrac{1}{2}:5\dfrac{1}{2}=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}=\dfrac{9}{2}+\dfrac{1}{2}x\dfrac{2}{11}=\dfrac{9}{2}+\dfrac{1}{11}=\dfrac{99}{22}+\dfrac{2}{22}=\dfrac{101}{22}\)

b) \(\dfrac{7}{6}:\left(\dfrac{1}{2}x\dfrac{3}{4}\right)-\dfrac{5}{8}=\dfrac{7}{6}:\dfrac{3}{8}-\dfrac{5}{8}=\dfrac{7}{6}x\dfrac{8}{3}-\dfrac{5}{8}=\dfrac{28}{9}-\dfrac{5}{8}=\dfrac{224}{72}-\dfrac{45}{72}=\dfrac{179}{72}\)

a) \(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}=\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}=\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}=\dfrac{40}{14}=\dfrac{20}{7}\)

b) \(3\dfrac{5}{6}+2\dfrac{1}{6}x6=\dfrac{23}{6}+\dfrac{13}{6}x6=\dfrac{23}{6}+\dfrac{78}{6}=\dfrac{101}{6}\)

96,75 + \(x\) - 47,8 = 07,54

96,75 + \(x\)            = 7,54 + 47,8

96,75 + \(x\)           = 55,34

             \(x\)           = 55,34 - 96,75

            \(x\)           = - 41,41

(3 -x ) . ( 4 - x ) . ( 5 - x ) = 0

⇒  3 - x = 0 hoặc 4 - x  = 0 hoặc 5-x = 0

⇒ x = 3 hoặc x = 4 hoặc x = 5

Vậy x = 3 hoặc x = 4 hoặc x = 5 .

\(\text{∘ Ans}\)

\(\downarrow\)

\(\left(3-x\right)\cdot\left(4-x\right)\cdot\left(5-x\right)=0\)

`TH1:`

`3 - x = 0`

`\Rightarrow x = 3-0`

`\Rightarrow x = 3`

`TH2:`

`4 - x = 0`

`\Rightarrow x = 4 - 0`

`\Rightarrow x = 4`

`TH3:`

`5 - x = 0`

`=> x = 5 - 0`

`=> x = 5`

Vậy, `x = {3; 4; 5}.`