Bài 12: Số thực

\(a,\sqrt{65}>\sqrt{63}\Leftrightarrow\sqrt{65}+1>\sqrt{63}+1\\ b,\sqrt{8}>\sqrt{7}\Leftrightarrow\dfrac{1}{\sqrt{8}}< \dfrac{1}{\sqrt{7}}\\ c,\sqrt{34,9}< \sqrt{36}=6\\ d,3\sqrt{25,5}=\sqrt{229,5}>\sqrt{196}=14\\ e,\left(2\sqrt{26}+4\right)^2=108+8\sqrt{26};13^2=169=61+108\\ 8\sqrt{26}=\sqrt{1664}< \sqrt{3721}=61\\ \Leftrightarrow2\sqrt{26}+4< 13\\ f,\left(\sqrt{24}+\sqrt{66}\right)^2=90+24\sqrt{11};16^2=256=90+166\\ 24\sqrt{11}=\sqrt{6336}< \sqrt{27556}=166\\ \Leftrightarrow90+24\sqrt{11}< 90+166\\ \Leftrightarrow\sqrt{24}+\sqrt{66}< 16\\ g,\dfrac{46-3\sqrt{49}}{4}=\dfrac{46-21}{4}=\dfrac{25}{4}=\sqrt{\dfrac{625}{16}}< \sqrt{\dfrac{800}{16}}=\sqrt{50}\)

\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\\ \Rightarrow\left\{{}\begin{matrix}x=12\cdot\dfrac{3}{2}=18\\y=12\cdot\dfrac{4}{3}=16\\z=12\cdot\dfrac{5}{4}=15\end{matrix}\right.\)

\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{x}{18}=\dfrac{y}{16}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{18}=\dfrac{y}{16}=\dfrac{z}{15}=\dfrac{x+y+z}{18+16+15}=\dfrac{49}{49}=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=1.18=18\\y=1.16=16\\z=1.15=15\end{matrix}\right.\)

e: \(2\sqrt{26}>9\)

nên \(2\sqrt{26}+4>13\)

27: \(\sqrt{\dfrac{81}{25}}-\dfrac{1}{5}\left(-3+\sqrt{16}\right)^{20}\cdot\left(-2\right)^3\)

\(=\dfrac{9}{5}-\dfrac{1}{5}\cdot\left(-8\right)\)

\(=\dfrac{9}{5}+\dfrac{8}{5}=\dfrac{17}{5}\)

28: \(\sqrt{\sqrt{16}}-\sqrt{\sqrt{81}}=2-3=-1\)

Giúp mình với

B nha em

B nha bạn

Ta có: góc xOm = góc xOa + góc aOm

          góc yOm = góc yPb + góc bOm

Mà góc xOm = yOm (cùng = 90⁰)

     góc xOa = góc yPb (gt)

=> góc aOm = góc bOm

Ta có: góc aOm = góc bOm (cmt)

          tia Om nằm giữa tia Oa và Ob

=> tia Om là tia phân giác góc aOb

Mik gửi ạ

@ann1234

\(\dfrac{a+b}{3}=\dfrac{b+c}{4}=\dfrac{c+a}{5}=\dfrac{a+b+b+c+c+a}{3+4+5}=\dfrac{2\left(a+b+c\right)}{12}=\dfrac{a+b+c}{6}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a+b}{3}=\dfrac{a+b+c}{6}\\\dfrac{b+c}{4}=\dfrac{a+b+c}{6}\\\dfrac{c+a}{5}=\dfrac{a+b+c}{6}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}3a+3b=3c\\2b+2c=4a\\a+c=b\end{matrix}\right.\)\(\Rightarrow a=b=c=0\)

Ko biết

Chọn B

Ta có:  \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\) ; \(\dfrac{1}{3^2}< \dfrac{1}{2.3}\) ; \(\dfrac{1}{4^2}< \dfrac{1}{3.4}\) ;....; \(\dfrac{1}{20^2}< \dfrac{1}{19.20}\)

⇒ M = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{20^2}\) \(< \) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}\)

⇒ M \(< \) \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

⇒ M \(< \) \(1-\dfrac{1}{20}\) \(< \)  \(1\)

Vậy M = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{20^2}\) \(< \) \(1\)