diện tích xung quanh của hình hộp CN là:

(20+4)x2 x10 = 480(m2)

đs...

\(7\left(x-2\right)+3\left(x+4\right)=5\)

⇒\(7x-14+3x+12=5\)

⇔\(10x-2=5\)

⇔\(10x=7\)

⇔\(x=\dfrac{7}{10}\)

E=\(\dfrac{\dfrac{2}{15}-\dfrac{2}{21}+\dfrac{2}{39}}{\dfrac{5}{20}-\dfrac{5}{28}+\dfrac{5}{52}}\)

E=\(\dfrac{2.\dfrac{1}{3}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{13}\right)}{5.\dfrac{1}{4}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{13}\right)}\)

E=\(\dfrac{2.\dfrac{1}{3}}{5.\dfrac{1}{4}}\)

E=8/15

Ta có: 

\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)

\(\Rightarrow90^o+\widehat{B}+40^o=50^o\)

Theo giả thiết:

\(AD\) là phân giác của \(\widehat{A}\)

\(\Rightarrow\widehat{BAD}=45^o\)

Ta có: 

\(\widehat{BAH}+\widehat{AHB}+\widehat{ABH}=180^o\)

\(\Rightarrow90^o+\widehat{AHB}+50^o=180^o\)

\(\Rightarrow\widehat{AHB}=40^o\)

Ta có:

\(\widehat{BAH}+\widehat{HAD}=45^o\)

\(\Rightarrow40^o+\widehat{HAD}=45^o\)

\(\Rightarrow\widehat{HAD}=5^o\).

1.

(x-2,5).2-3,5=4

(x-2,5).2      =4+3,5

(x-2,5).2      =7,5

x-2,5           =7,5:2

x-2,5           =3,75

x                 =3,75+2,5

x                 =6,25

Vậy x = 6,25

2.

(2x+4,2)-3,6=5

2x+4,2        =5+3,6

2x+4,2        =8,6

2x               =8,6-4,2

2x               =4,4

x                 =4,4:2

x                 =2,2

vậy x = 2,2

99:10=9(dư 9)

Đ/S 99

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

\(\dfrac{3}{4}x+\dfrac{4}{5}x=\dfrac{4}{5}+3\)

\(\dfrac{31x}{20}=\dfrac{19}{5}\)

\(x\) = \(\dfrac{19}{5}:\dfrac{31}{20}\)

\(x\) = 76/31

Bảng số La Mã không có số 0

|x+1|=\(\dfrac{8}{3}\)-\(\dfrac{2}{3}\)

|x+1|=2

TH1: x+1=2                   TH2: x+1=-2

        x=2-1                             x=-2-1

        x=1                                x=-3

vậy x={1;-3}

\(\dfrac{8}{3}-\left|x+1\right|=\dfrac{2}{3}\)

\(\left|x+1\right|=\dfrac{8}{3}-\dfrac{2}{3}\)

\(\left|x+1\right|=2\)

\(x+1=2\)  hoặc \(x+1=-2\)

TH1 : \(x+1=2\)

\(x=2-1\)

\(x=1\)

TH2 : \(x+1=-2\)

\(x=-2-1\)

\(x=-3\)

Vậy x = 1 hoặc x = -3