a. \(4\dfrac{1}{4}-2\dfrac{5}{8}+2\dfrac{3}{5}=\dfrac{17}{4}-\dfrac{21}{8}+\dfrac{13}{5}=\dfrac{169}{40}\)

b. \(4\dfrac{4}{9}:2\dfrac{2}{3}+3\dfrac{1}{6}=\dfrac{40}{9}:\dfrac{8}{3}+\dfrac{19}{6}=\dfrac{5}{3}+\dfrac{19}{6}=\dfrac{29}{6}\)

c. \(3\dfrac{1}{5}+2\dfrac{3}{5}-2\dfrac{4}{5}=\dfrac{16}{5}+\dfrac{13}{5}-\dfrac{14}{5}=\dfrac{16+13-14}{5}=\dfrac{15}{5}=3\)

d. \(5\dfrac{1}{7}-2\dfrac{4}{5}:1\dfrac{1}{5}=\dfrac{36}{7}-\dfrac{14}{5}:\dfrac{6}{5}=\dfrac{36}{7}-\dfrac{7}{3}=\dfrac{59}{21}\)

e. \(2\dfrac{3}{5}+1\dfrac{1}{4}.2\dfrac{2}{3}=\dfrac{13}{5}+\dfrac{5}{4}.\dfrac{8}{3}=\dfrac{13}{5}+\dfrac{10}{3}=\dfrac{89}{15}\)

g. \(4\dfrac{1}{3}.1\dfrac{1}{2}+5\dfrac{2}{7}=\dfrac{13}{3}.\dfrac{3}{2}+\dfrac{37}{7}=\dfrac{13}{2}+\dfrac{37}{7}=\dfrac{165}{14}\)

I.

1B

2C

3A

4D

5C

3-(-10)=3+10=13

3-(-10)

=3+10

=13

3-(-10)

=3+10

=13

a) \(\Leftrightarrow x^2=\sqrt{4}\)

\(\Leftrightarrow x^2=2\Leftrightarrow x=\pm2\)

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

\(\Leftrightarrow\left|\dfrac{1}{2}x+1\right|=9\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+1=9\\\dfrac{1}{2}x+1=-9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=-16\end{matrix}\right.\)

c) \(\Leftrightarrow\sqrt{2x}-4\sqrt{2x}+16\sqrt{2x}=52\left(đk:x\ge0\right)\)

\(\Leftrightarrow13\sqrt{2x}=52\Leftrightarrow\sqrt{2x}=4\Leftrightarrow2x=16\Leftrightarrow x=8\left(tm\right)\)

f: Ta có: \(\sqrt{\dfrac{50-25x}{4}}-8\sqrt{2-x}+\sqrt{18-9x}=-10\)

\(\Leftrightarrow\sqrt{2-x}\cdot\dfrac{5}{2}-8\sqrt{2-x}+3\sqrt{2-x}=-10\)

\(\Leftrightarrow\sqrt{2-x}=4\)

\(\Leftrightarrow2-x=16\)

hay x=-14

\(b,\Leftrightarrow\left\{{}\begin{matrix}m+2=1\\m\ne2\end{matrix}\right.\Leftrightarrow m=-1\\ c,\text{PT giao Ox: }y=0\Leftrightarrow\left(m+2\right)x-m=0\\ \text{Thay }x=2\Leftrightarrow2m+4-m=0\\ \Leftrightarrow m=-4\\ d,\text{PT giao Ox và Oy: }\\ y=0\Leftrightarrow x=\dfrac{m}{m+2}\Leftrightarrow A\left(\dfrac{m}{m+2};0\right)\Leftrightarrow OA=\left|\dfrac{m}{m+2}\right|\\ x=0\Leftrightarrow y=-m\Leftrightarrow B\left(0;-m\right)\Leftrightarrow OB=\left|m\right|\\ \Delta OAB\text{ cân }\Leftrightarrow OA=OB\Leftrightarrow\left|\dfrac{m}{m+2}\right|=\left|m\right|\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{m}{m+2}=m\\\dfrac{m}{m+2}=-m\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m\left(m+1\right)=0\\m\left(m+3\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=0\\m=-1\\m=-3\end{matrix}\right.\)

a: Xét tứ giác OBAC có 

\(\widehat{OBA}+\widehat{OCA}=180^0\)

Do đó: OBAC là tứ giác nội tiếp

câu hỏi là j em ơi

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