\(x^2+\sqrt[]{x}\) là gì bạn ?

rút gọn

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Vì \(\widehat{aOc}\) và \(\widehat{cOb}\) kề bù với nhau

\(\Rightarrow\widehat{aOb}=\widehat{aOc}+\widehat{cOb}\)

\(\Leftrightarrow\widehat{aOc}=\widehat{aOb}-\widehat{cOb}\)

\(\Leftrightarrow\widehat{aOc}=100^o-30^o\)

\(\Leftrightarrow\widehat{aOc}=70^o\)

Vậy \(\widehat{aOc}=70^o\)

Vì góc aOc và góc cOb kề bù với nhau

=> aOb=aOc+cOb

=>aOc=aOb-cOb

<=>aOc=100^o-30^o

<=>aOc=70^o

Vậy góc aOc= 70độ

a) \(A=\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.10}+\dfrac{1}{143}\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)+\dfrac{1}{143}\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{100}\right)+\dfrac{1}{143}=\dfrac{1}{2}.\dfrac{99}{100}+\dfrac{1}{143}=\dfrac{99}{200}+\dfrac{1}{143}=\dfrac{99.143+200.1}{200.143}=\dfrac{14157+200}{28600}=\dfrac{14357}{28600}\)

b) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=14950\)

\(\Rightarrow x+x+...+x+\left(1+2+...+99\right)=14950\)

\(\Rightarrow100x+\left(\left(99+1\right):2\right).99:2=14950\)

\(\Rightarrow100x+2475=14950\Rightarrow100x=12475\Rightarrow x=\dfrac{12475}{100}=\dfrac{499}{4}\)

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Ừ mình sai đề sorry

\(\left(48.0,75-240:10\right)-\left(16.0,5-36:9\right)\)

\(=\left(48.\dfrac{3}{4}-24\right)-\left(16.\dfrac{1}{2}-4\right)=\left(36-24\right)-\left(8-4\right)=12-4=8\)

(36 - 24) - ( 8 - 4)
= 12 - 4
= 8

`1/2+1/2+1/3+1/3`

`=(1/2+1/2)+(1/3+1/3)`

`=2/2+2/3`

`=1+2/3`

`=3/3+2/3`

`=5/3`

\(\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{1}{2}\times2+\dfrac{1}{3}\times2=\left(\dfrac{1}{2}+\dfrac{1}{3}\right)\times2=\dfrac{5}{6}\times2=\dfrac{5}{3}\)