Câu 1 :
\(a,5\left(x+2\right)=2\left(x-4\right)\)
\(\Leftrightarrow5x+10=2x-8\)
\(\Leftrightarrow5x-2x=-8-10\)
\(\Leftrightarrow3x=-18\)
\(\Leftrightarrow x=-6\)
\(b,x\left(x+2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{3;2\right\}\)
\(c,\dfrac{2x-5}{4}-\dfrac{x+1}{3}=\dfrac{1}{2}\)
\(\Leftrightarrow3\left(2x-5\right)-4\left(x+1\right)=6\)
\(\Leftrightarrow6x-15-4x-4=6\)
\(\Leftrightarrow6x-4x=6+4+15\)
\(\Leftrightarrow2x=25\)
\(\Leftrightarrow x=\dfrac{25}{2}\)
Vậy \(S=\left\{\dfrac{25}{2}\right\}\)
\(d,\dfrac{3}{x-2}-\dfrac{6}{x+2}=\dfrac{-x}{x^2-4}\left(đkxđ:x\ne\pm2\right)\)
\(\Leftrightarrow3\left(x+2\right)-6\left(x-2\right)=-x\)
\(\Leftrightarrow3x+6-6x+12=-x\)
\(\Leftrightarrow3x-6x+x=-12-6\)
\(\Leftrightarrow-2x=-18\)
\(\Leftrightarrow x=9\left(nhận\right)\)
Vậy \(S=\left\{9\right\}\)
Câu 3 :
a, Xét ΔABD và ΔHBA có :
\(\widehat{A}=\widehat{H}=90^0\)
\(\widehat{B}:chung\)
\(\Rightarrow\Delta ABD\sim\Delta HBA\left(g-g\right)\)
b, Xét ΔADH và ΔDBC có :
\(\widehat{H}=\widehat{C}=90^0\)
\(\widehat{ADH}=\widehat{DBC}\left(AB//CD,slt\right)\)
\(\Rightarrow\Delta ADH\sim\Delta DBC\)
c, Ta có : \(\Delta ABD\sim\Delta HBA\left(cmt\right)\)
\(\Rightarrow\dfrac{AB}{BH}=\dfrac{BD}{AB}\)
\(\Rightarrow AB^2=BH.BD\)
d, Xét ΔABD vuông ở A , theo định lý Pi-ta-go ta được :
\(\Rightarrow BD=\sqrt{AB^2+AD^2}=\sqrt{12^2+9^2}=15\left(cm\right)\)
Ta có : \(\Delta ABD\sim\Delta HBA\left(cmt\right)\)
\(\Rightarrow\dfrac{AB}{BH}=\dfrac{BD}{AB}\)
hay \(\dfrac{12}{BH}=\dfrac{15}{12}\)
\(\Rightarrow BH=\dfrac{12.12}{15}=9,6\left(cm\right)\)
Mn giải giúp mik vs ạ 🥺 mình đang cần gấp. Cảm ơn mn nhiều
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lm giúp mik vs mn mik cần gấp