\(5,\\ a,\Leftrightarrow x^2+3x-4+2x-x^2=5\\ \Leftrightarrow5x=9\Leftrightarrow x=\dfrac{9}{5}\\ b,\Leftrightarrow x^2-4x+4-x^2-x=3\\ \Leftrightarrow-5x=-1\Leftrightarrow x=\dfrac{1}{5}\)
Bài 5:
a: Ta có: \(\left(x-1\right)\left(x+4\right)+x\left(2-x\right)=5\)
\(\Leftrightarrow x^2+3x-4+2x-x^2=5\)
\(\Leftrightarrow5x=9\)
hay \(x=\dfrac{9}{5}\)
b: Ta có: \(\left(x-2\right)^2-x\left(x+1\right)=3\)
\(\Leftrightarrow x^2-4x+4-x^2-x=3\)
\(\Leftrightarrow-3x=-1\)
hay \(x=\dfrac{1}{3}\)
Bài 6:
a: \(\widehat{B}=\widehat{C}=\dfrac{180^0-40^0}{2}=70^0\)
b: Xét ΔOBC có \(\widehat{OBC}=\widehat{OCB}\)
nên ΔOBC cân tại O
\(a,\Delta ABC.cân.tại.A\Rightarrow\widehat{ABC}=\widehat{ACB}=\dfrac{180^0-\widehat{BAC}}{2}=70^0\)
\(b,\left\{{}\begin{matrix}\widehat{ABD}=\widehat{DBC}=\dfrac{1}{2}\widehat{ABC}\\\widehat{AEC}=\widehat{ECB}=\dfrac{1}{2}\widehat{ACB}\\\widehat{ABC}=\widehat{ACB}\end{matrix}\right.\Rightarrow\widehat{ABD}=\widehat{DBC}=\widehat{AEC}=\widehat{ECB}\\ \Rightarrow\Delta OBC.cân\)
\(c,\left\{{}\begin{matrix}\widehat{ACE}=\widehat{ABD}\left(cm.trên\right)\\AB=AC\left(gt\right)\\\widehat{BAC}.chung\end{matrix}\right.\Rightarrow\Delta AEC=\Delta ADB\left(g.c.g\right)\\ d,\Delta AEC=\Delta ADB\Rightarrow AE=AD\\ \Rightarrow\widehat{AED}=\dfrac{180^0-\widehat{BAC}}{2}=110^0\\ \Rightarrow\widehat{AED}=\widehat{ABC}\)
Mà 2 góc này ở vị trí đồng vị nên \(DE//BC\Rightarrow BEDC\) là hthang
Mà \(\widehat{ABC}=\widehat{ACB}\) nên \(BEDC\) là hthang cân
