Question

[yoonaa]

Mn giải giúp mik vs

Answer 1

[Lương Đại]

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 \(S=\left\{9\right\}\)

Answer 2

[Lương Đại]

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 \(\dfrac{12}{BH}=\dfrac{15}{12}\)

\(\Rightarrow BH=\dfrac{12.12}{15}=9,6\left(cm\right)\)