(x+1/3)2=4/9
ghi hẳn ra hộ mik nhs
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Gọi \(J=CE\cap AB\), \(F=BD\cap AC\) , \(H=CE\cap BD\)
Có \(\widehat{EAB}=\widehat{ECB}=\dfrac{1}{2}sđ\stackrel\frown{EB}\)
\(\widehat{CAD}=\widehat{DBC}=\dfrac{1}{2}sđ\stackrel\frown{DC}\)
\(\Rightarrow\widehat{EAB}+\widehat{CAD}=\widehat{ECB}+\widehat{DBC}=180^0-\widehat{BHC}\) (*)
Lại có \(\widehat{AJC}+\widehat{AFB}=180^0\) => Tứ giác AJHF nội tiếp đường tròn
\(\Rightarrow180^0=\widehat{BAC}+\widehat{JHF}=\widehat{BAC}+\widehat{BHC}\)
\(\Rightarrow180^0-\widehat{BHC}=\widehat{BAC}\) (2*)
Từ (*); (2*) => \(\widehat{EAB}+\widehat{CAD}=\widehat{BAC}\)
\(\Leftrightarrow\widehat{EAB}+\widehat{BAC}+\widehat{CAD}=2\widehat{BAC}\)
\(\Leftrightarrow\widehat{EAD}=2\alpha\)
Ý C
Có \(sđ\stackrel\frown{BD}=\widehat{BOD}=40^0\)
Có \(\widehat{BED}=\dfrac{1}{2}\left(sđ\stackrel\frown{BD}+sđ\stackrel\frown{AC}\right)\)
\(\Leftrightarrow\)\(60^0=\dfrac{1}{2}\left(40^0+sđ\stackrel\frown{AC}\right)\) \(\Leftrightarrow sđ\stackrel\frown{AC}=80^0\)
Ý B
B
`sdBC=1/2(sdBD+sdAC)`
`=>sdAC=2sdBC-sdBD`
`<=>sdAC=120^o-40^o=80^o`
vì x;y chia hết cho 42 => y thuộc tập hợp 2;4;6;8;0 (x;y thuoc N ) => 2014x0 chia hết cho 42
2014x2 chia hết cho 42
2014x4 chia hết cho 42
2014x6 chia hết cho 42
2014x8 chia hết cho 42
Ps; mk giải đầu tiên nhé **** mk di
\(\left(|x|+2\right)\cdot\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}|x|+2=0\\x^2-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}|x|=0+2=2\\x^2=0+4=4\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}|x|=2\\x^2=4\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x^2=+-2\end{cases}}\)
\(a,\dfrac{7}{12}+\dfrac{3}{4}\times\dfrac{2}{9}=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{7}{12}+\dfrac{2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
\(b,\dfrac{8}{9}-\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{8}{9}-\dfrac{4}{15}\times\dfrac{5}{2}=\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{8}{9}-\dfrac{6}{9}=\dfrac{2}{9}\)
\(a,\dfrac{1}{2}-\dfrac{1}{3}-\left(-\dfrac{5}{4}\right)=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{5}{4}=\dfrac{1\times6-1\times4+5\times3}{12}=\dfrac{6-4+15}{12}=\dfrac{17}{12}\\ b,\dfrac{5}{4}-\dfrac{1}{2}-\dfrac{7}{8}=\dfrac{5\times2-1\times4-7}{8}=\dfrac{10-4-7}{8}=-\dfrac{1}{8}\\ c,\dfrac{1}{5}-\dfrac{1}{2}+\dfrac{9}{10}=\dfrac{1\times2-1\times5+9}{10}=\dfrac{2-5+9}{10}=\dfrac{6}{10}=\dfrac{3}{5}\\ d,\dfrac{5}{4}-\dfrac{1}{3}+\dfrac{7}{6}=\dfrac{5\times3-1\times4+7\times2}{12}=\dfrac{15-4+14}{12}=\dfrac{25}{12}\)
\(\sqrt{2x+5}\) xác định khi \(2x+5\ge0\Rightarrow2x\ge-5\Rightarrow x\ge-\dfrac{5}{2}\)
\(\sqrt{2x+5}\le0\Leftrightarrow2x+5\le0\Leftrightarrow2x\le-5\Leftrightarrow x\ge\dfrac{-5}{2}\)
\(\Rightarrow\) Đáp án: A
ĐKXĐ: \(2x-3\ge0\\ \Rightarrow2x\ge0+3\\ \Rightarrow2x\ge3\\ \Rightarrow x\ge\dfrac{3}{2}\left(A\right)\)