cho 2 góc kề nhau , xoy = 130 độ , yoz = 60 đọ . tính xoz
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\text{∘ Ans}\)
\(\downarrow\)
\(A=\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-...-\dfrac{1}{6}-\dfrac{1}{2}\)
`=`\(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)
`=`\(\dfrac{8}{9}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\right)\)
`=`\(\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
`=`\(\dfrac{8}{9}-\left[1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\left(\dfrac{1}{8}-\dfrac{1}{8}\right)-\dfrac{1}{9}\right]\)
`=`\(\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)
`=`\(\dfrac{8}{9}-\dfrac{8}{9}=0\)
Vậy, ` A = 0.`
\(A=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=\)
\(A=\dfrac{8}{9}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right)=\)
\(A=\dfrac{8}{9}-\left(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{9-8}{8.9}\right)\)
\(A=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(A=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)=0\)
a) -1/24 - [ 1/4 - ( 1/2 - 7/8 )]
= -1/24 - [ 1/4 +3/8 ]
= -1/24 - 5/8
= -2/3.
a) -1/24 - [ 1/4 - ( 1/2 - 7/8 )]
= -1/24 - [ 1/4 +3/8 ]
= -1/24 - 5/8
= -2/3.
\(...=1-\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{4}-...-\dfrac{1}{98}+\dfrac{1}{99}\)
\(=\dfrac{1}{99}\) (Bạn xem lại đề)
\(...-\dfrac{1}{24}-\left[\dfrac{1}{4}--\dfrac{3}{8}\right]=-\dfrac{1}{24}-\dfrac{5}{8}=-\dfrac{2}{3}\)
\(-\dfrac{1}{24}-\left[\dfrac{1}{4}-\left(\dfrac{1}{2}-\dfrac{7}{8}\right)\right]\)
\(=-\dfrac{1}{24}-\left[\dfrac{1}{4}-\left(\dfrac{4}{8}-\dfrac{7}{8}\right)\right]\)
\(=-\dfrac{1}{24}-\left(\dfrac{1}{4}+\dfrac{3}{8}\right)\)
\(=-\dfrac{1}{24}-\left(\dfrac{2}{8}+\dfrac{3}{8}\right)\)
\(=-\dfrac{1}{24}-\dfrac{5}{8}\)
\(=-\dfrac{1}{24}-\dfrac{15}{24}\)
\(=\dfrac{-16}{24}\)
\(=-\dfrac{2}{3}\)
...\(a=\left[\left(\left(50-1\right):1+1\right):2\right]\left(50+1\right)=25.51=1275\)
\(...a1=\left[\left(\left(98-35\right):3+1\right):2\right]\left(35+98\right)=11.133=1463\)
\(\dfrac{2}{15}:\left(-\dfrac{29}{5}\right).\dfrac{29}{12}=\dfrac{2}{15}.\left(-\dfrac{5}{29}\right).\dfrac{29}{12}=-\dfrac{1}{18}\)
2/15 : (-5 4/5 ) . 2 5/12
= 2/15 : (-29/5 ) .29/12
= -2/87 . 29/12
= -1/18
a)\(...-\dfrac{22}{7}.\dfrac{6}{55}.\left(-\dfrac{7}{12}\right)=\dfrac{22}{7}.\dfrac{6}{55}.\dfrac{7}{12}=\dfrac{1}{5}\)
b) \(\dfrac{18}{39}.\left(-\dfrac{13}{8}\right):\left(-\dfrac{27}{4}\right)=-\dfrac{3}{4}.\left(-\dfrac{4}{27}\right)=\dfrac{1}{9}\)
`@` `\text {Ans}`
`\downarrow`
`a)`
\(1\dfrac{3}{5}\div\left(-5\dfrac{5}{7}\right)\)
`=`\(\dfrac{8}{3}\div\left(-\dfrac{30}{7}\right)\)
`=`\(\dfrac{8}{3}\cdot\left(-\dfrac{7}{30}\right)=-\dfrac{28}{45}\)
`b)`
\(-1\dfrac{1}{8}\cdot\dfrac{4}{51}\cdot\left(-11\dfrac{1}{3}\right)=-\dfrac{9}{8}\cdot\dfrac{4}{51}\cdot\left(-\dfrac{34}{3}\right)=-\dfrac{3}{34}\cdot\left(-\dfrac{34}{3}\right)=1\)
a)...\(\dfrac{8}{5}:\left(-\dfrac{40}{7}\right)=\dfrac{8}{5}.\left(-\dfrac{7}{40}\right)=-\dfrac{7}{25}\)
b) \(...-\dfrac{9}{8}.\dfrac{4}{51}\left(-\dfrac{34}{3}\right)=1\)
Ta có \(\widehat{xOy}\) và \(\widehat{yOz}\) là hai góc kề nhau nên:
\(\widehat{xOy}+\widehat{yOz}=\widehat{xOz}\)
\(\Rightarrow\widehat{xOz}=130^o+60^o=190^o\)
\(\widehat{xoz}=130+60=190^o\)