\(9\cdot3^2\cdot\dfrac{1}{81}\cdot27=3^2\cdot\dfrac{3^2}{3^4}\cdot3^3=3^3\)

Sửa đề:

\(H=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\\ H=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\\ H=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\\ H=\dfrac{1}{2}-\dfrac{1}{10}\\ H=\dfrac{2}{5}\)

Vậy \(H=\dfrac{2}{5}\)

\(H=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(H=\dfrac{1}{90}+\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{72}\right)\)

\(H=\dfrac{1}{90}+\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\right)\)

\(H=\dfrac{1}{90}+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\right)\)

\(H=\dfrac{1}{90}+\left(\dfrac{1}{2}-\dfrac{1}{8}\right)\)

\(H=\dfrac{1}{90}+\left(\dfrac{4}{8}-\dfrac{1}{8}\right)\)

\(H=\dfrac{1}{90}+\dfrac{3}{8}\)

\(H=\dfrac{49}{360}\)

\(\dfrac{9.3^2.1}{81.27}\)

\(=\dfrac{81.1}{81.27}\)

\(=\dfrac{3^4.1}{3^4.3^3}\)

Bạn tự vẽ hình nhé
a) Có IA + IB = AB (tính chất cộng đoạn thẳng)
=> IB = AB - IA = 8 - 4 = 4 cm
Mà IA = 4 cm
=> IA = IB (=4 cm)
kết hợp I nằm giữa AB
=> I là trung điểm của AB (đpcm)
b) Vì E là trung điểm của IA 
=> IE = \(\dfrac{IA}{2}\) = \(\dfrac{4}{2}\) = 2 cm
Vì F là trung điểm của IB
=> IF = \(\dfrac{IB}{2}\) = \(\dfrac{4}{2}\) = 2 cm
Có EF = IF + IE (tính chất cộng đoạn thẳng)
=> EF = 2 + 2 = 4 cm

\(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+\dfrac{3x}{8.11}+\dfrac{3x}{11.14}=\dfrac{1}{21}\)

\(3x.\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}\right)=\dfrac{1}{21}\)

\(3x\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)

\(3x.\left(\dfrac{1}{2}-\dfrac{1}{14}\right)\)                                                  \(=\dfrac{1}{21}\)

\(3x.\dfrac{3}{7}\)                                                                 \(=\dfrac{1}{21}\)

\(3x\)                                                                     \(=\dfrac{1}{21}:\dfrac{3}{7}\)

\(3x=\dfrac{1}{9}\)

\(x=\dfrac{1}{9}:3\)

\(x=\dfrac{1}{27}\)

1

x(3/2.5 + 3/5.8 + 3/8.11 + 3/11.14)=1/21

x(1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11- 1/14)=1/21

x(1/2 - 1/14)=1/21

x . 6/14=1/21

x=1/21 : 6/14=1/21 . 14/6=2/3

\(A=\dfrac{3.5.7.11.13.37-10101}{1212120+40404}\)

\(A=\dfrac{5.11.3.7.11.13-3.7.11.13}{120.10101+4.10101}\)

\(A=\dfrac{3.7.11.13.\left(5.11-1\right)}{10101\cdot\left(120+4\right)}\)

\(A=\dfrac{10101.54}{10101.124}\)

\(A=\dfrac{54}{124}=\dfrac{27}{62}\)

\(\Rightarrow\) Vậy \(A=\dfrac{27}{62}\)

Bài 2

a) 5/3 - x = 2 1/3

5/3 - x = 7/3

x = 5/3 - 7/3

x = -2/3

b) 3,5 - 1/2 x = -5/4

1/2 x = 3,5 - (-5/4)

1/2 x = 19/4

x = 19/4 : 1/2

x = 19/2

c) 4/(2 - x) - 2/3 = 0

4/(2 - x) = 2/3

2(2 - x) = 3.4

2(2 - x) = 12

2 - x = 12 : 2

2 - x = 6

x = 2 - 6

x = -4

d) 0,25 + 7,5% x = 2 5/6

3/40 x = 17/6 - 0,25

3/40 x = 31/12

x = 31/12 : 3/40

x = 310/9

Bài 4

a) Số học sinh xếp loại tốt:

120 . 4/15 = 32 (học sinh)

Số học sinh xếp loại khá:

32 : 80% = 40 (học sinh)

Số học sinh xếp loại đạt:

120 - 32 - 40 = 48 (học sinh)

b) Tỉ số phần trăm của số học sinh xếp loại khá so với cả khối:

40 . 100% : 120 ≈ 33,33%

Bài 5:

1: \(\dfrac{2}{5}+\dfrac{3}{5}x=\dfrac{9}{20}\)

=>\(\dfrac{3}{5}x=\dfrac{9}{20}-\dfrac{2}{5}=\dfrac{9}{20}-\dfrac{8}{10}=\dfrac{1}{10}\)

=>\(x=\dfrac{1}{10}:\dfrac{3}{5}=\dfrac{1}{10}\cdot\dfrac{5}{3}=\dfrac{5}{30}=\dfrac{1}{6}\)

2: \(\dfrac{1}{4}+\dfrac{1}{3}:\left(3x\right)=-5\)

=>\(\dfrac{1}{3}:\left(3x\right)=-5-\dfrac{1}{4}=-\dfrac{21}{4}\)

=>\(3x=-\dfrac{1}{3}:\dfrac{21}{4}=-\dfrac{1}{3}\cdot\dfrac{4}{21}=-\dfrac{4}{63}\)

=>\(x=-\dfrac{4}{63}:3=-\dfrac{4}{189}\)

3: \(-\dfrac{21}{13}x+\dfrac{1}{3}=-\dfrac{2}{3}\)

=>\(-\dfrac{21}{13}x=-\dfrac{2}{3}-\dfrac{1}{3}=-\dfrac{3}{3}=-1\)

=>\(x=1:\dfrac{21}{13}=\dfrac{13}{21}\)

4: \(\dfrac{3}{4}x-\dfrac{1}{2}=\dfrac{3}{7}\)

=>\(\dfrac{3}{4}x=\dfrac{1}{2}+\dfrac{3}{7}=\dfrac{7+6}{14}=\dfrac{13}{14}\)

=>\(x=\dfrac{13}{14}:\dfrac{3}{4}=\dfrac{13}{14}\cdot\dfrac{4}{3}=\dfrac{52}{42}=\dfrac{26}{21}\)

5: \(\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{3}{10}-\dfrac{1}{5}\)

=>\(\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{10}\)

=>\(\dfrac{2}{3}x=\dfrac{1}{10}-\dfrac{1}{2}=\dfrac{1}{10}-\dfrac{5}{10}=-\dfrac{4}{10}=-\dfrac{2}{5}\)

=>\(x=-\dfrac{2}{5}:\dfrac{2}{3}=-\dfrac{2}{5}\cdot\dfrac{3}{2}=-\dfrac{3}{5}\)

6: \(\dfrac{2}{3}+\dfrac{1}{3}x=-\dfrac{1}{2}\)

=>\(\dfrac{1}{3}x=-\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{-7}{6}\)

=>\(x=-\dfrac{7}{6}:\dfrac{1}{3}=-\dfrac{7}{6}\cdot3=-\dfrac{7}{2}\)

7: \(\dfrac{11}{12}x+\dfrac{3}{4}=-\dfrac{1}{6}\)

=>\(\dfrac{11}{12}x=-\dfrac{1}{6}-\dfrac{3}{4}=\dfrac{-2}{12}-\dfrac{9}{12}=-\dfrac{11}{12}\)

=>x=-1

8: \(-\dfrac{5}{6}-\dfrac{2}{3}x=\dfrac{7}{12}+\dfrac{-1}{3}\)

=>\(-\dfrac{5}{6}-\dfrac{2}{3}x=\dfrac{7}{12}-\dfrac{4}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)

=>\(\dfrac{2}{3}x=-\dfrac{5}{6}-\dfrac{1}{4}=\dfrac{-10}{12}-\dfrac{3}{12}=-\dfrac{13}{12}\)

=>\(x=-\dfrac{13}{12}:\dfrac{2}{3}=-\dfrac{13}{12}\cdot\dfrac{3}{2}=\dfrac{-39}{24}=\dfrac{-13}{8}\)

9: \(2\dfrac{2}{3}x+8\dfrac{2}{3}=3\dfrac{1}{3}\)

=>\(\dfrac{8}{3}x=3+\dfrac{1}{3}-8-\dfrac{2}{3}=-5-\dfrac{1}{3}=-\dfrac{16}{3}\)

=>\(x=-\dfrac{16}{3}:\dfrac{8}{3}=-\dfrac{16}{3}\cdot\dfrac{3}{8}=-\dfrac{16}{8}=-2\)

10: \(\dfrac{3}{4}-\dfrac{1}{4}:\left(x:\dfrac{5}{2}\right)=-3\)

=>\(\dfrac{1}{4}:\left(x:\dfrac{5}{2}\right)=\dfrac{3}{4}+3=\dfrac{15}{4}\)

=>\(x:\dfrac{5}{2}=\dfrac{1}{4}:\dfrac{15}{4}=\dfrac{1}{15}\)

=>\(x=\dfrac{1}{15}\cdot\dfrac{5}{2}=\dfrac{1}{6}\)

1: Khối lượng của quả dưa là:

\(\dfrac{7}{2}:\dfrac{3}{4}=\dfrac{7}{2}\cdot\dfrac{4}{3}=\dfrac{14}{3}\left(kg\right)\)

Câu 2:

a: Trên cùng một nửa mặt phẳng bờ chứa tia Ox, ta có: \(\widehat{xOy}< \widehat{xOz}\)

nên tia Oy nằm giữa hai tia Ox và Oz

=>\(\widehat{xOy}+\widehat{yOz}=\widehat{xOz}\)

=>\(\widehat{yOz}+40^0=120^0\)

=>\(\widehat{yOz}=80^0\)

b: Ta có: \(\widehat{xOy}+\widehat{xOt}=180^0\)(hai góc kề bù)

=>\(\widehat{xOt}+40^0=180^0\)

=>\(\widehat{xOt}=140^0\)

c: Om là phân giác của góc yOz

=>\(\widehat{yOm}=\widehat{zOm}=\dfrac{\widehat{yOz}}{2}=\dfrac{80^0}{2}=40^0\)

Vì \(\widehat{zOm}< \widehat{zOx}\)

nên tia Om nằm giữa hai tia Oz và Ox

=>\(\widehat{mOz}+\widehat{mOx}=\widehat{xOz}=120^0\)

=>\(\widehat{xOm}=120^0-40^0=80^0\)

Vì \(\widehat{xOy}+\widehat{yOm}=40^0+40^0=80^0=\widehat{xOm}\)

và \(\widehat{xOy}=\widehat{yOm}\left(=40^0\right)\)

nên Oy là phân giác của góc xOm

bạn trl 1 câu cũng được nhé làm được câu nào trl câu .mik cũng sẽ tick cho các bạn nếu bạn nào giúp mình trl lời mà trl câu nào cũng được