Ta có: \(\hat{xOz}+\hat{yOz}=180^0\) (hai góc kề bù)

=>\(\hat{xOz}=180^0-50^0=130^0\)

Om là phân giác của góc xOz

=>\(\hat{xOm}=\frac12\cdot\hat{xOz}=\frac{130^0}{2}=65^0\)

Ta có: \(\hat{xOm}+\hat{yOm}=180^0\) (hai góc kề bù)

=>\(\hat{yOm}=180^0-65^0=115^0\)

lớp mấy

a) \(\left(2x+3\right)^2=\frac{9}{144}\)

\(\Leftrightarrow\left(2x+3\right)^2=\left(\frac{1}{4}\right)^2=\left(-\frac{1}{4}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}2x+3=\frac{1}{4}\\2x+3=\frac{-1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{-11}{4}\\2x=\frac{-13}{4}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-11}{8}\\x=\frac{-13}{8}\end{cases}}}\)

Vậy ...

b) Ta có: \(\left(3x-1\right)^3=\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)

\(\Leftrightarrow3x-1=\frac{-2}{3}\Leftrightarrow3x=\frac{1}{3}\Leftrightarrow x=\frac{1}{9}\)

Vậy ....

c) \(x^{10}=25x^8\Leftrightarrow x^{10}:x^8=25\Leftrightarrow x^2=25\Leftrightarrow x=\left\{5;-5\right\}\)

Vậy ...

d) \(\frac{x^7}{81}=27\Leftrightarrow x^7=27.81=2187\)

Mà 3⁷ = 2187 => x⁷ = 3⁷ => x = 3

Vậy ....

e) \(\frac{x^8}{9}=729\Leftrightarrow x^8=729.9=6561\)

Mà 3⁸ = (-3)⁸ = 6561

=> x⁸ = 3⁸ = (-3)⁸

=> x = {-3;3}

Vậy ...

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{80}-\frac{1}{81}=1-\frac{1}{81}=\frac{80}{81}\)

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\)

\(A=1-\frac{1}{81}\)

\(A=\frac{80}{81}\)

Cái này là toán lớp 6 nha bn

Ủng hộ mk nha ^_-

Sai rồi.

Vậy giải giùm câu c) đi. Mà u là Suzue hả?

Bài 1:

\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)

Ta thấy:

\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)

\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)

\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)

\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)

\(\Rightarrow10x+\frac{10}{11}=0\)

\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)

Bài 2:

Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)

\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)

Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)

c )

\(1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}}=1+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{3}{2}}}}=1+\frac{1}{1+\frac{1}{1+\frac{5}{3}}}=1+\frac{1}{1+\frac{1}{\frac{8}{3}}}=1+\frac{1}{\frac{11}{8}}=\frac{19}{11}\)

cái này là đố vui hả