Đáp án:
Giải thích các bước giải:
Bài $1$:
Xét $ΔKEM$ và $ΔCFM$ có:
$BM=CM$ (GT)
$\widehat{BME}$=$\widehat{FMC}$ (2 góc đối đỉnh)
$EM=FM$ (GT)
⇒$ΔKEM=ΔCFM$ (C-G-C)
⇒$\widehat{KEM}$=$\widehat{CFM}$ (2 góc tương ứng)
Mà $\widehat{KEM}$=$\widehat{AEH}$ (đối đỉnh)
⇒$\widehat{AEH}$=$\widehat{CFM}$
⇒$\widehat{CFM}$+$\widehat{CAM}$=$\widehat{AEH}$+$\widehat{CAM}$
⇒$\widehat{CFA}$+$\widehat{CAF}$=90 độ
⇒$FC⊥AC$ (đpcm)
Bài $2$:
a,$2x(x-3)-3(x-3)=0$
$(x-3)(2x-3)=0$
\(\left[ \begin{array}{l}x-3=0\\2x-3=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=3\\x=\frac{3}{2}\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=3\\x=\frac{3}{2}\end{array} \right.\)
b,$x^2(x-1)+4(1-x)=0$
$x^2(x-1)-4(x-1)=0$
$(x^2-4)(x-1)=0$
\(\left[ \begin{array}{l}x-1=0\\x^2-4=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=1\\x=±2\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=1\\x=±2\end{array} \right.\)
c,$2x(x-5)+(x-5)^2=0$
$2x^2-10x+x^2-10x+25=0$
$3x^2+25-20x=0$
$(x-5)(3x-5)=0$
\(\left[ \begin{array}{l}x-5=0\\3x-5=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=5\\x=\frac{5}{3}\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=5\\x=\frac{5}{3}\end{array} \right.\)
d,$(2x-1)=(4-3x)^2$
$2x-1=9x^2-24x+16$
$2x-1-9x^2+24x-16=0$
$9x^2+26x-17=0$
$(x-1)(9x-7)=0$
\(\left[ \begin{array}{l}x-1=0\\9x-7=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=1\\x=\frac{17}{9}\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=1\\x=\frac{17}{9}\end{array} \right.\)
e,$2x(3-4x)-5x^2(4x-3)=0$
$2x(3-4x)+5x^2(3-4x)=0$
$x(2+5x^2)(3-4x)=0$
\(\left[ \begin{array}{l}x=0\\3-4x=0;2-5x=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=0;x=\frac{2}{5}\\x=\frac{3}{4}\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=0;x=\frac{2}{5}\\x=\frac{3}{4}\end{array} \right.\)
