a) 248 x 36 + 4 x 64 x 62

= 2 x 124 x 36 + (2 x 62) x 2 x 64

= 72 x 124 + 124 x 128

= 124 x (72 + 128)

= 124 x 200

= 124 x 2 x 100 

= 248 x 100

= 24800 

b) 567 x 65 + 567 x 35 + 5670

= 567 x 65 + 567 x 35 + 5670

= 567 x (65 + 35) + 5670

= 567 x 100 + 5670 

= 56700 + 5670

= 62370

a)

\(248\times36+4\times64\times62\\ =248\times36+ \left(4\times62\right)\times64\\ =248\times36+248\times64\\ =248\times\left(36+64\right)\\ =248\times100\\ =24800\)

Theo đề bài ta có:

+) \(AB+AC+BC=37\left(dm\right)\)

+) \(AB=\dfrac{2}{3}AC\); \(BC=\dfrac{4}{5}AC\)

Thay vào \(AB+AC+BC=37\) được:

\(\dfrac{2}{3}AC+AC+\dfrac{4}{5}AC=\dfrac{37}{15}AC=37\)

\(\Rightarrow AC=37:\dfrac{37}{15}=15\left(dm\right)\)

Thay vào ​​\(AB=\dfrac{2}{3}AC;BC=\dfrac{4}{5}AC\)​​ được:

\(\left\{{}\begin{matrix}AB=\dfrac{2}{3}\cdot15=10\left(dm\right)\\BC=\dfrac{4}{5}\cdot15=12\left(dm\right)\end{matrix}\right.\)

Vậy...​

Xin lỗi mình ghi thiếu!

Diện tích tam giác ABC là:

\(\dfrac{AB\cdot BC}{2}=\dfrac{10\cdot12}{2}=60\left(dm^2\right)\)

Vậy diện tích tam giác ABC là 60 dm²

67 + 135 + 33

= ( 67 + 33 ) + 135

= 100 + 135

= 235

.

997 + 86 + 98 + 9999

= (997 + 3) + (98 + 2) + (9999 + 1) + (86 - 3 - 2 - 1)

= 1000 + 100 + 10000 + 80

= 11180

.

37 . 38 + 62 . 37

= 37 . (38 + 62)

= 37 . 100

= 3700

53 . 39 + 47 . 39 - 53 . 21 - 47 . 21

= 39 . (53 + 47) - 21 . (53 + 47)

= 39 . 100 - 21 . 100

= 100 . (39 - 21)

= 100 . 18

= 1800

.

47 . 29 - 13 . 29 - 24 . 29

= 29 . (47 - 13 - 24)

= 29 . 10

= 290

.

1754 : 17 - 74 : 17 + 20 : 17

= (1754 - 74 + 20) : 17

= (1774 - 74) : 17

= 1700 : 17

= 100

\(\dfrac{x-2y}{z-y}=-5\Rightarrow\dfrac{x-2y}{y-z}=5\\ \Rightarrow x-2y=5\left(y-z\right)\\ \Rightarrow x-2y=5y-5z\\ \Rightarrow x+5z=7y\)

Ta có: 

\(\dfrac{1}{7}\cdot\dfrac{x-2z}{y-z}=\dfrac{x-2z}{7\left(y-z\right)}=\dfrac{x-2z}{7y-7z}\\ =\dfrac{x-2z}{x+5z-7z}=\dfrac{x-2z}{x-2z}=1\)

\(\Rightarrow\dfrac{x-2z}{y-z}=1:\dfrac{1}{7}=7\)

Đặt \(A=1\cdot2+2\cdot3+3\cdot4+...+98\cdot99\) 

\(3A=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot4\cdot3+...+98\cdot99\cdot3\)

\(3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+98\cdot99\cdot\left(100-97\right)\)

\(3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+98\cdot99\cdot100-97\cdot98\cdot99\)

\(3A=\left(1\cdot2\cdot3-1\cdot2\cdot3\right)+\left(2\cdot3\cdot4-2\cdot3\cdot4\right)+\left(3\cdot4\cdot5-3\cdot4\cdot5\right)+...+\left(97\cdot98\cdot99-97\cdot98\cdot99\right)+98\cdot99\cdot100\)

\(3A=98\cdot99\cdot100\)

\(A=\dfrac{98\cdot99\cdot100}{3}\)

\(A=323400\)

Vậy \(A=323400\)

323400

\(\dfrac{12}{121}< \dfrac{12}{120}=\dfrac{1}{10}=\dfrac{35}{350}< \dfrac{35}{333}\)

1) \(x^2-4=x^2-2^2=\left(x-2\right)\left(x+2\right)\)

2) \(1-4x^2=1-\left(2x\right)^2=\left(1-2x\right)\left(1+2x\right)\)

3) \(4x^2-9=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)

4) \(9-25x^2=3^2-\left(5x\right)^2=\left(3-5x\right)\left(3+5x\right)\)

5) \(4x^2-25=\left(2x\right)^2-5^2=\left(2x-5\right)\left(2x+5\right)\)

6) \(9x^2-36=\left(3x\right)^2-6^2=\left(3x-6\right)\left(3x+6\right)\)

7) \(\left(3x\right)^2-y^2=\left(3x+y\right)\left(3x-y\right)\)

8) \(x^2-\left(2y\right)^2=\left(x+2y\right)\left(x-2y\right)\)

9) \(\left(2x\right)^2-y^2=\left(2x-y\right)\left(2x+y\right)\)

\(\dfrac{1}{2}\times38+50\%\times22+\dfrac{50}{100}+39\times0,5\)

\(=0,5\times38+0,5\times22+0,5\times1+39\times0,5\)

\(=0,5\times\left(38+22+1+39\right)\)

\(=0,5\times100\)

\(=50\)

Đây là lớp 5 ạ?????