Giải thích các bước giải:
1. a, 3x-5=7⇔3x=12⇔x=4
b,16-8x=0⇔16=8x⇔x=2
c,5x-3=16-8x⇔5x+8x=16+3⇔13x=19⇔x=19/13
d, 5(8+3x)+2(3x-8)=0⇔40+15x+6x-16=0⇔21x=-24⇔x=-8/7
e, -4(x-3)=6x+(x-3)⇔5(x-3)+6x=0⇔5x-15+6x=0⇔11x=15⇔x=15/11
2. a, (x+4)/5-x+4=x/3-(x-2)/2⇔(x+4-5x+20)/5=(2x-3x+6)/6⇔(24-4x)/5=(6-x)/6
⇔6(24-4x)=5(6-x)⇔144-24x=30-5x⇔114=19x⇔x=114/19
b, $\frac{4-5x}{6}$ =$\frac{2(-x+1)}{2}$⇔4-5x=6(-x+1)⇔4-5x=-6x+6⇔x=2
c, $\frac{-(x-3)}{2}$-2=$\frac{5(x+2)}{4}$⇔-2(x-3)-8=5(x+2)⇔-2x+6-8=5x+10⇔7x=-12⇔x=-12/7
d, $\frac{x-1}{2}$+$\frac{3(x+1)}{8}$=$\frac{11-5x}{3}$ ⇔$\frac{4x-4+3x+3}{8}$=$\frac{11-5x}{3}$
⇔$\frac{7x-1}{8}$=$\frac{11-5x}{3}$⇔3(7x-1)=8(11-5x)⇔21x-3=88-40x⇔61x=91⇔x=91/61
e, $\frac{x+1}{2}$+$\frac{x+3}{4}$=3-$\frac{x+2}{3}$
⇔6(x+1)+3(x+3)=36-4(x+2)⇔6x+6+3x+9=36-4x-8⇔13x=13⇔x=1
3. a, (8-3x)(-3x+5)-0⇔\(\left[ \begin{array}{l}8-3x=0\\-3x+5=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=8/3\\x=5/3\end{array} \right.\)
b,(16-8x)(2-6x)=0⇔\(\left[ \begin{array}{l}16-8x=0\\2-6x=0\end{array} \right.\)⇔\(\left[ \begin{array}{l}x=2\\x=1/3\end{array} \right.\)
c, (x+4)(6x-12)=0⇔\(\left[ \begin{array}{l}x+4=0\\6x-12=0\end{array} \right.\)⇔\(\left[ \begin{array}{l}x=-4\\x=2\end{array} \right.\)
d, (11-33x)(x+11)=0⇔\(\left[ \begin{array}{l}11-33x=0\\x+11=0\end{array} \right.\)⇔\(\left[ \begin{array}{l}x=3\\x=-11\end{array} \right.\)
e, (x-3)(2x-1)=(2x-1)(2x+3)⇔(2x-1)(x-3-2x-3)=0⇔(2x-1)(-x-6)=0⇔\(\left[ \begin{array}{l}2x-1=0\\-x-6=0\end{array} \right.\)⇔\(\left[ \begin{array}{l}x=1/2\\x=-6\end{array} \right.\)
4. a, AN=AC-NC=12-8=4; ta có: $\frac{AR}{AB}$ =3/9=1/3
$\frac{AN}{AC}$=4/12=1/3
⇒$\frac{AR}{AB}$ =$\frac{AN}{AC}$ ⇒RN||BC (theo Talet đảo)
b,do RJ || BC ⇒ $\frac{RJ}{IB}$=$\frac{AR}{AB}$ =1/3 (1)
$\frac{NR}{CB}$=1/3⇒$\frac{NR}{2IB}$=1/3⇒$\frac{NR}{IB}$=2/3 (2)
chia (1) và (2): $\frac{RJ}{NR}$=1/2
