c) Ta có: \(\widehat{ABD}=\widehat{DBC}\)( BD là phân giác )\(\Rightarrow90^0-\widehat{ABD}=90^0-\widehat{DBC}\Rightarrow\widehat{BIH}=\widehat{ADI}\Rightarrow\widehat{AID}=\widehat{ADI}\Rightarrow\Delta ADI\) cân tại A\(\Rightarrow AI=AD\Rightarrow\dfrac{AB}{AI}=\dfrac{AB}{AD}\)

Xét Δ ABI và Δ CBD có:

\(\widehat{BAI}=\widehat{BCD}\left(\Delta ABC\sim\Delta HBA\right)\)

\(\dfrac{AB}{AI}=\dfrac{BC}{CD}\left(=\dfrac{AB}{AD}\right)\)

\(\Rightarrow\Delta ABI\sim\Delta CBD\left(c.g.c\right)\)

d) Xét ΔABH có:

BI là tia phân giác của \(\widehat{ABH}\)

\(\Rightarrow\dfrac{IH}{IA}=\dfrac{BH}{AB}\left(1\right)\)( tính chất tia phân giác)

Xét ΔABC có:

BD là tia phân giác của \(\widehat{ABC}\)

\(\Rightarrow\dfrac{AD}{DC}=\dfrac{AB}{BC}\left(2\right)\)( tính chất tia phân giác)

Ta có: \(\dfrac{BH}{AB}=\dfrac{AB}{BC}\left(\Delta ABC\sim\Delta HBA\right)\left(3\right)\)

Từ \(\left(1\right),\left(2\right),\left(3\right)\Rightarrow\dfrac{IH}{IA}=\dfrac{AD}{DC}\left(đpcm\right)\)

\(2x^2+13x=36+7\sqrt{x^3-24x+32}\Leftrightarrow2x^2+13x-36=7\sqrt{x^3-24x+32}\Leftrightarrow\left(2x^2+13x-36\right)^2=\left(7\sqrt{x^3-24x+32}\right)^2\Leftrightarrow4x^4+169x^2+1296+2\left(26x^3-72x^2-468x\right)=49\left(x^3-24x+32\right)\Leftrightarrow4x^4+3x^3+25x^2+240x-272=0\Leftrightarrow\left(x-1\right)\left(x+4\right)\left(4x^2-9x+68\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

Xét tứ giác ABCD có:

\(\widehat{BAC}+\widehat{ABC}+\widehat{ACB}+\widehat{ADC}=360\) độ

\(\Rightarrow4x+2x+2x+x=360\) độ

\(\Rightarrow9x=360\) độ \(\Rightarrow x=40\) độ
\(\Rightarrow\left\{{}\begin{matrix}\widehat{BAC}=4x=160\\\widehat{ABC}=2x=80\\\widehat{ACB}=2x=80\\\widehat{ADC}=x=40\end{matrix}\right.\)( đơn vị là độ)

Bài 1

a) \(3x\left(x^2-4x^3+5\right)-\left(3x^2+15x-7\right)=3x^3-12x^4+15x-3x^2-15x+7=-12x^4+3x^3-3x^2+7\)b)\(\left(2x+1\right)^2+\left(5-4x\right)\left(x+3\right)=4x^2+4x+1+\left(5x+15-4x^2-12x\right)=-3x+16\)

Bài 2

a)\(\left(x-1\right)\left(4x-3\right)-4x\left(x-5\right)=5\Rightarrow\left(4x^2-7x+3\right)-\left(4x^2-20x\right)=5\Rightarrow13x=2\Rightarrow x=\dfrac{2}{13}\)

b)\(\left(x-1\right)^2-81=0\Rightarrow\left(x-1-9\right)\left(x-1+9\right)=0\Rightarrow\left[{}\begin{matrix}x-10=0\\x+8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)

a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)

c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)