Câu 1: \(-7/6+15/6=8/6=4/3\)

Câu 2: \(x=7/12-1/6=>x=7/12-2/12=>x=5/12\) nhé =) 

câu 1

\(\dfrac{-7}{6}+\dfrac{15}{6}=\dfrac{-7+15}{6}=\dfrac{8}{6}=\dfrac{4}{3}\)

câu 2

\(x+\dfrac{1}{6}=\dfrac{7}{12}\)

\(x=\dfrac{7}{12}-\dfrac{1}{6}\)

\(x=\dfrac{7}{12}-\dfrac{2}{12}=\dfrac{7-2}{12}\)

\(x=\dfrac{5}{12}\)

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

\(\dfrac{1}{2}x=\dfrac{5}{10}+\dfrac{6}{10}\)

\(\dfrac{1}{2}x=\dfrac{11}{10}\\ x=\dfrac{22}{10}\)

\(x=\dfrac{11}{5}\)

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

\(\Rightarrow\dfrac{1}{2}x=\dfrac{11}{10}\)

\(\Rightarrow x=\dfrac{11}{5}\)

14 would be

15 would be

16 were

17 spoke

18 would visit

19 have been

20 driving

21 write - were

22 didn't have to

giúp j hở bn

giúp gì vậy?

a) Cho tam giác ABC vuông tại A. Biết AB = 6cm, AC = 8cm. Tính dô dài cạnh BC
Bài giải:
Áp dụng định lí Pi - ta - go vào t/giác ABC vuông tại A
Ta có : BC^2 = AB^2 + AC^2
=> BC^2 = 6^2 + 8^2 = 36 + 64 = 100
=> BC = 10
Vậy BC = 10 cm
bt mỗi câu a thôi bạn ơi 
 

mấy câu kia mình hỏi nhiều người rồi mà cx chẳng ra

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

\(\Leftrightarrow x\left(\dfrac{5-2}{2.5}+\dfrac{8-5}{5.8}+\dfrac{11-8}{8.11}+\dfrac{14-11}{11.14}\right)=\dfrac{1}{21}\)

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

\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\Leftrightarrow x.\dfrac{3}{7}=\dfrac{1}{21}\Leftrightarrow x=\dfrac{1}{9}\)

x=0.1111111111

`5/6-x=-3/7`

`=>x=5/6+3/7`

`=>x=53/42`

Vậy `x=53/42`.

`3x-2/5=1/2`

`=>3x=1/2+2/5`

`=>3x=9/10`

`=>x=3/10`

Vậy `x=3/10.`

a) Ta có: \(\dfrac{5}{6}-x=\dfrac{-3}{7}\)

nên \(x=\dfrac{5}{6}+\dfrac{3}{7}=\dfrac{35}{42}+\dfrac{18}{42}\)

hay \(x=\dfrac{53}{42}\)

b) Ta có: \(3x-\dfrac{2}{5}=\dfrac{1}{2}\)

nên \(3x=\dfrac{1}{2}+\dfrac{2}{5}=\dfrac{9}{10}\)

hay \(x=\dfrac{3}{10}\)

a) \(=3\left(xy-4\right)\)

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

c) \(=x^2-\left(y^2-12y+36\right)=x^2-\left(y-6\right)^2=\left(x-y+6\right)\left(x+y-6\right)\)

d) \(=\left(4p^2-36p+81\right)-25=\left(2p-9\right)^2-25=\left(2p-9-5\right)\left(2p-9+5\right)=4\left(p-7\right)\left(p-2\right)\)

Kẻ AK vuông góc SC

BC vuông góc (SAC)

=>BC vuông góc AK

=>d(A;(SBC))=AK

SC=căn SA^2+AC^2=2a

=>AK=a*a*căn 3/2a=a*căn 3/2

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