\(y\times7,5+y+y\times91,4+y:10=257,8\)

\(y\times7,5+y\times1+y\times91,4+y\times0,1=257,8\)

\(y\times\left(7,5+1+91,4+0,1\right)=257,8\)

\(y\times100=257,8\)

\(y=257,8:100\)

\(y=2,578\)

y x 7,5 + y + y x 91,4 + y : 10 = 257,8

y x 7,5 + y + y x 91,4 + y x 0,1 = 257,8

y x (7,5 + 1 + 91,4 + 0,1) = 257,8

y x 100 = 257,8

y = 257,8 : 100

y = 2,578

4

-

52

 Ta có \(\left|\Omega\right|=C^5_{52}\)

 Gọi A là biến cố: "Có ít nhất 1 quân át." Khi đó xét biến cố \(\overline{A}:\) "Không có 1 quân át nào."

 Khi đó \(\left|\overline{A}\right|=C^5_{48}\) \(\Rightarrow P\left(\overline{A}\right)=\dfrac{C^5_{48}}{C^5_{52}}\) \(\Rightarrow P\left(A\right)=1-\dfrac{C^5_{48}}{C^5_{52}}\)

$x+x:0,5=4,5$

$x+x\times2=4,5$

$x\times(1+2)=4,5$

$x\times3=4,5$

$x=4,5:3$

$x=1,5$

1.5

1075

Bài 6

Ta có:

∠AMN + 130⁰ = 180⁰ (kề bù)

⇒ ∠AMN = 180⁰ - 130⁰ = 50⁰

∠ALK + ∠NLK = 180⁰ (kề bù)

⇒ ∠ALK = 180⁰ - ∠NLK

= 180⁰ - 140⁰

= 40⁰

Do KL // MN

⇒ ∠ANM = ∠ALK = 50⁰ (đồng vị)

∆AMN có:

∠A + ∠AMN + ∠ANM = 180⁰ (tổng ba góc trong ∆AMN)

⇒ ∠A = 180⁰ - (∠AMN + ∠ANM)

= 180⁰ - (50⁰ + 40⁰)

= 90⁰

Vậy x = ∠A = 90⁰

Thời gian ô tô đi hết quãng đường AB:

(11 giờ 20 phút - 1 giờ 30 phút - 6 giờ 30 phút) : 2 = 2 giờ 10 phút = 13/6 (giờ)

Độ dài quãng đường AB:

69,6 × 13/6 = 150,8 (km)

Do 25 < 37 nên 2/25 > 2/37

17/5 > 33/33

Sắp xếp theo thứ tự giảm dần:

17/5 > 33/33 > 2/25 > 2/37

$y:0,125+y:10+y\times1,9=202,3$

$y\times8+y\times0,1+y\times1,9=202,3$

$y\times(8+0,1+1,9)=202,3$

$y\times(8+2)=202,3$

$y\times10=202,3$

$y=202,3:10$

$y=20,23$

\(y:0,125+y:10+y\times1,9=202,3\)

\(y\times8+y\times01,+y\times1,9=202,3\)

\(y\times\left(8+0,1+1,9\right)=202,3\)

\(y\times10=202,3\)

\(y=202,3:10\)

\(y=20,23\)

\(1\dfrac{2}{24}\text{x}\text{ }5\dfrac{2}{5}\text{x}2\text{x}3\dfrac{7}{9}\text{x}2\dfrac{2}{17}\)

\(=\dfrac{13}{12}\text{x}\dfrac{27}{5}\text{x}2\text{x}\dfrac{34}{9}\text{x}\dfrac{36}{17}\)

\(=\dfrac{13}{12}\text{x}2\text{x}\dfrac{27}{9}\text{x}\dfrac{34}{17}\text{x}\dfrac{36}{5}\)

\(=\dfrac{13}{6}\text{x}3\text{x}2\text{x}\dfrac{36}{5}=13\text{x}\dfrac{36}{5}=\dfrac{468}{5}\)

\(2\dfrac{2}{17}\text{x}1\dfrac{1}{24}\text{x}5\dfrac{2}{5}\text{x}3\dfrac{7}{9}\text{x}2\)

\(=\dfrac{36}{17}\text{x}\dfrac{25}{24}\text{x}\dfrac{27}{5}\text{x}\dfrac{34}{9}\text{x}2\)

\(=\dfrac{36}{24}\text{x}\dfrac{25}{5}\text{x}\dfrac{27}{9}\text{x}\dfrac{34}{17}\text{x}2\)

\(=\dfrac{3}{2}\text{x}5\text{x}3\text{x}2\text{x}2=90\)