a: \(5^3\cdot2-100:4+2^3\cdot5\)

\(=125\cdot2-25+40\)

=200+15

=215

b: \(6^2:9+50\cdot2-3^3\cdot3\)

\(=36:3+100-81\)

=12+19=31

c: \(3^2\cdot5+2^3\cdot10-81:3\)

\(=9\cdot5+8\cdot10-27\)

=45+80-27

=98

d: \(5^{13}:5^{10}-25\cdot2^2\)

\(=5^3-25\cdot4\)

=125-100

=25

e: \(20:2^2+5^9:5^8\)

=20:4+5

=5+5=10

f: \(100:5^2+7\cdot3^2\)

\(=100:25+7\cdot9\)

=4+63

=67

g: \(84:4+3^9:3^7+5^0\)

=21+9+1

=31

h: \(29-\left[16+3\left(51-49\right)\right]\)

\(=29-16-3\cdot2\)

=13-6=7

a) 

\(5^3\cdot2-100:4+2^3\cdot5=\\ =125\cdot2-25+8\cdot5\\ =250-25+40\\ =290-25\\ =265\)

b) 

\(6^2:9+50\cdot2-3^3\cdot3\\ =36:9+100-3^{3-1}\\ =4+100-9\\ =100-5\\ =95\)

c) 

\(3^2\cdot5+2^3\cdot10-81:3\\ =9\cdot5+8\cdot10-27\\ =45+80-27\\ =125-27\\ =98\) 

d) 

\(5^{13}:5^{10}-25\cdot2^2\\ =5^{13-10}-25\cdot4\\ =5^3-100\\ =125-100\\ =25\)

Sửa: 

\(a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\\ \Leftrightarrow a^2+b^2+c^2+d^2+e^2\ge ab+ac+ad+ae\\ \Leftrightarrow\left(\dfrac{a^2}{4}+ab+b^2\right)+\left(\dfrac{a^2}{4}-ac+c^2\right)+\left(\dfrac{a^2}{4}-ad+d^2\right)+\left(\dfrac{a^2}{4}-ae+e^2\right)\ge0\\ \Leftrightarrow\left(\dfrac{a}{2}-b\right)^2+\left(\dfrac{a}{2}-c\right)^2+\left(\dfrac{a}{2}-d\right)^2+\left(\dfrac{a}{2}-e\right)^2\ge0\)  

Dấu: "=" xảy ra: \(\dfrac{a}{2}=b=c=d=e\)

\(\dfrac{2x-100}{4}=35\\ 2x-100=4\cdot35\\ 2x-100=140\\ 2x=140+100\\ 2x=240\\ x=\dfrac{240}{2}\\ x=120\)

tui k copy đâu nha

+) Cạnh hình vuông là 8m

=> Chu vi hình vuông là: 4 x 8 = 32 (m)

+) Cạnh hình vuông là 7cm

=> Chu vi hình vuông là: 4 x 7 = 28 (cm)

+) Cạnh hình vuông là 10m

=> Chu vi hình vuông là: 4 x 10 = 40 (m)

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

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

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

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

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

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

\(=0,5\times100\)

\(=50\)

Đây là lớp 5 ạ?????