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Ta có:

∠M + ∠N + ∠P + ∠Q = 360⁰ (tổng các góc trong tứ giác MNPQ)

⇒ ∠M + ∠N + ∠P + (∠P + 10⁰) = 360⁰

⇒ ∠M + ∠N + (∠N + 10⁰) + (∠N + 10⁰ + 10⁰) = 360⁰

⇒ ∠M + (∠M + 10⁰) + (∠M + 10⁰ + 10⁰) + (∠M + 10⁰ + 10⁰ + 10⁰)

⇒ ∠M + ∠M + 10⁰ + ∠M + 20⁰ + ∠M + 30⁰ = 360⁰

⇒ 4∠M + 60⁰ = 360⁰

⇒ 4∠M = 360⁰ - 60⁰

⇒ 4∠M = 300⁰

⇒ ∠M = 300⁰ : 4

⇒ ∠M = 75⁰

⇒ ∠N = 75⁰ + 10⁰ = 85⁰

⇒ ∠P = 85⁰ + 10⁰ = 95⁰

⇒ ∠Q = 95⁰ + 10⁰ = 105⁰

\(\widehat{M}+\widehat{N}+\widehat{P}+\widehat{Q}=360^o\)

\(\widehat{M}+\widehat{M}+10+\widehat{M}+20+\widehat{M}+30=360\)

\(4\widehat{M}=360-60=300\Rightarrow M=75^o\)

D = \(\dfrac{1}{1\times1981}\) + \(\dfrac{1}{2\times1982}\)+...+ \(\dfrac{1}{25\times2005}\)

D =\(\dfrac{1}{1980}\times\)( \(\dfrac{1980}{1\times1981}\)+ \(\dfrac{1980}{2\times1982}\)+....+ \(\dfrac{1980}{25\times2005}\))

D = \(\dfrac{1}{1980}\) \(\times\)(\(\dfrac{1}{1}\) - \(\dfrac{1}{1981}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{1982}\)+....+ \(\dfrac{1}{25}\) \(\times\) \(\dfrac{1}{2005}\))

D= \(\dfrac{1}{1980}\)[( \(\dfrac{1}{1}\) + \(\dfrac{1}{2}\) +....+ \(\dfrac{1}{25}\)) - ( \(\dfrac{1}{1981}\)+ \(\dfrac{1}{1982}\)+...+ \(\dfrac{1}{2005}\))]

E =\(\dfrac{1}{25}\times\)( \(\dfrac{1}{1\times26}\)+ \(\dfrac{1}{2\times27}\)+...+ \(\dfrac{1}{1980\times2005}\))

E =  \(\dfrac{1}{25}\). (\(\dfrac{25}{1\times26}\) + \(\dfrac{25}{2\times27}\)+....+ \(\dfrac{25}{1980\times2005}\))

E = \(\dfrac{1}{25}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{26}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{27}\)+...+\(\dfrac{1}{1980}\)-\(\dfrac{1}{2005}\))

E=\(\dfrac{1}{25}\)[\(\dfrac{1}{1}\)+...+ \(\dfrac{1}{25}\)+ (\(\dfrac{1}{26}\)+...+\(\dfrac{1}{1980}\)) - (\(\dfrac{1}{26}\)+...+\(\dfrac{1}{1980}\)) - (\(\dfrac{1}{1981}\)+..\(\dfrac{1}{2005}\))]

E = \(\dfrac{1}{25}\) .[\(\dfrac{1}{1}\)+\(\dfrac{1}{2}\)+...+\(\dfrac{1}{25}\) - (\(\dfrac{1}{1981}\)+\(\dfrac{1}{1982}\)+...+ \(\dfrac{1}{2005}\))]

\(\dfrac{D}{E}\) = \(\dfrac{\dfrac{1}{1980}}{\dfrac{1}{25}}\) = \(\dfrac{5}{396}\)

`7(x-1/2)^2=9`

`(x-1/2)^2=9/7`

\(=>\left[{}\begin{matrix}x-\dfrac{1}{2}=\sqrt{\dfrac{9}{7}}\\x-\dfrac{1}{2}=-\sqrt{\dfrac{9}{7}}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{3}{\sqrt{7}}+\dfrac{1}{2}\\x=-\dfrac{3}{\sqrt{7}}+\dfrac{1}{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{6+\sqrt{7}}{2\sqrt{7}}\\x=\dfrac{-6+\sqrt{7}}{2\sqrt{7}}\end{matrix}\right.\)

7.(x-\(\dfrac{1}{2}\))²=9

7.x+\(\dfrac{1}{4}\) =9

7.x=\(\dfrac{37}{4}\)

x=\(\dfrac{37}{28}\)

2 bài cũm đc 

\(Bài.7:\\ a,\dfrac{\left(-3\right)^{10}.15^5}{25^3.\left(-9\right)^7}=\dfrac{3^{10}.3^5.5^5}{\left(5^2\right)^3.\left(3^2\right)^6.3.\left(-3\right)}\\ =\dfrac{3^{15}.5^5}{-5^6.3^{14}}=-\dfrac{3}{5}\\ b,2^3+3.\left(\dfrac{1}{9}\right)^0-2^{-2}.4+\left[\left(-2\right)^2:\dfrac{1}{2}\right].8\\ =8+3.1-\dfrac{1}{4}.4+\left[4:\dfrac{1}{2}\right].8\\ =8+3-1+8.8\\ =11-1+64=10+64=74\)

Bài 10:

\(a,3^{35}=\left(3^7\right)^5=2187^5\\ 5^{20}=\left(5^4\right)^5=625^5\\ Vì:2187^5>625^5\left(Vì:2187>625\right)\\ \Rightarrow3^{35}>5^{20}\\ b,2^{32}=\left(2^4\right)^8=16^8\\ Vì:37^8>16^8\left(Do:37>16\right)\\ \Rightarrow37^8>2^{32}\)