Xét tứ giác ADBC có 

O là trung điểm của AB

O là trung điểm của CD

Do đó: ADBC là hình bình hành

Suy ra: AC//BD

b.

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 2\\x>\dfrac{9}{2}\end{matrix}\right.\\-\dfrac{1}{3}< x< 7\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{3}< x< 2\\\dfrac{9}{2}< x< 7\end{matrix}\right.\)

Hay \(S=\left(-\dfrac{1}{3};2\right);\left(\dfrac{9}{2};7\right)\)

d.

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\le-\dfrac{11}{5}\\x\ge7\end{matrix}\right.\\-\dfrac{1}{2}< x< 3\end{matrix}\right.\) \(\Rightarrow x\in\varnothing\) hay BPT vô nghiệm

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he he he he he he

bài 1:

bn lấy giá trị của √(4^2-3,9^2) là dc

bài 2

AB+BC=2√(3^2+4^2)=??

a) \(x-\dfrac{1}{4}=-\dfrac{1}{6}\Leftrightarrow x=-\dfrac{1}{6}+\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{12}\)

b) \(\dfrac{2}{5}x-\dfrac{1}{9}=\left(\dfrac{1}{3}\right)^2\Leftrightarrow\dfrac{2}{5}x=\dfrac{1}{9}+\dfrac{1}{9}\Leftrightarrow\dfrac{2}{5}x=\dfrac{2}{9}\Leftrightarrow x=\dfrac{5}{9}\)

c) \(\left|x+\dfrac{1}{3}\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3-\dfrac{1}{3}\\x=-3-\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)

\(\left(a\right)\)
\(x-\dfrac{1}{4}=\dfrac{-1}{6}\)
\(x=\dfrac{1}{12}\)
/\(x+\dfrac{1}{3}\)/=3
=>    \(x+\dfrac{1}{3}=3\)    hoặc  \(-x-\dfrac{1}{3}=3\)
=>    x=\(\dfrac{8}{3}\)    hoặc x= \(\dfrac{-10}{3}\)

a: Ta có: \(x-\dfrac{1}{4}=-\dfrac{1}{6}\)

\(\Leftrightarrow x=\dfrac{1}{4}-\dfrac{1}{6}\)

hay \(x=\dfrac{1}{12}\)

b: Ta có: \(\dfrac{2}{5}x-\dfrac{1}{9}=\left(\dfrac{1}{3}\right)^2\)

\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{1}{9}+\dfrac{1}{9}=\dfrac{2}{9}\)

hay \(x=\dfrac{5}{9}\)

c: Ta có: \(\left|x+\dfrac{1}{3}\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)

a) \(\dfrac{3}{7}+\left(-\dfrac{5}{2}\right)+\left(-\dfrac{3}{7}\right)=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{7}=-\dfrac{5}{2}\)

b) \(B=\dfrac{4^2.2^3}{2^6}=\dfrac{\left(2^2\right)^2.2^3}{2^6}=\dfrac{2^4.2^3}{2^6}=\dfrac{2^7}{2^6}=2\)

a, \(A=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{7}=-\dfrac{5}{2}\)

b, \(B=\dfrac{4^2.2^3}{2^6}=\dfrac{2^4.2^3}{2^6}=\dfrac{2^7}{2^6}=2\)

\(A=\dfrac{3}{7}+(-\dfrac{5}{2})+(-\dfrac{3}{7})\)

\(=(\dfrac{3}{7}-\dfrac{3}{7})-\dfrac{5}{2} \)

\(=-\dfrac{5}{2}\)

\(B=\dfrac{4^2.2^3}{2^6}\)

\(B=\dfrac{4^2}{2^3}\)

\(B=\dfrac{16}{8}\)

\(B=2\)

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