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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
\(x+\dfrac{1}{6}=\dfrac{7}{12}\)
\(x=\dfrac{7}{12}-\dfrac{1}{6}\)
\(x=\dfrac{7}{12}-\dfrac{2}{12}=\dfrac{7-2}{12}\)
\(x=\dfrac{5}{12}\)
\(\dfrac{1}{2}x=\dfrac{1}{2}+\dfrac{3}{5}\\ \)
\(\dfrac{1}{2}x=\dfrac{5}{10}+\dfrac{6}{10}\)
\(\dfrac{1}{2}x=\dfrac{11}{10}\\ x=\dfrac{22}{10}\)
\(x=\dfrac{11}{5}\)
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
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
\(\Leftrightarrow x\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\)
\(\Leftrightarrow x\left(\dfrac{5-2}{2.5}+\dfrac{8-5}{5.8}+\dfrac{11-8}{8.11}+\dfrac{14-11}{11.14}\right)=\dfrac{1}{21}\)
\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)
\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\Leftrightarrow x.\dfrac{3}{7}=\dfrac{1}{21}\Leftrightarrow x=\dfrac{1}{9}\)
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
`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 \(x=\dfrac{3}{10}\)