Câu 1:
a: 5(3x+2)=4x+1
=>15x+10=4x+1
=>11x=-9
=>\(x=-\frac{9}{11}\)
b: (x-3)(x+4)=0
=>\(\left[\begin{array}{l}x-3=0\\ x+4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-4\end{array}\right.\)
c: ĐKXĐ: x∉{-1;2}
\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
=>\(\frac{2\left(x-2\right)-\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
=>3x-11=2(x-2)-(x+1)
=>3x-11=2x-4-x-1
=>3x-11=x-5
=>3x-x=-5+11
=>2x=6
=>x=3(nhận)
d: \(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\)
=>\(\left(\frac{x-1}{2013}-1\right)+\left(\frac{x-2}{2012}-1\right)+\left(\frac{x-3}{2011}-1\right)=\left(\frac{x-4}{2010}-1\right)+\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-6}{2008}-1\right)\)
=>\(\frac{x-2014}{2013}+\frac{x-2014}{2012}+\frac{x-2014}{2011}=\frac{x-2014}{2010}+\frac{x-2014}{2009}+\frac{x-2014}{2008}\)
=>x-2014=0
=>x=2014
e: \(\frac{1909-x}{91}+\frac{1907-x}{93}+\frac{1905-x}{95}+\frac{1903-x}{97}+4=0\)
=>\(\frac{1909-x}{91}+1+\frac{1907-x}{93}+1+\frac{1905-x}{95}+1+\frac{1903-x}{97}+1=0\)
=>\(\frac{2000-x}{91}+\frac{2000-x}{93}+\frac{2000-x}{95}+\frac{2000-x}{97}=0\)
=>2000-x=0
=>x=2000
Câu 2:
a: Xét ΔABC có \(\frac{AD}{AB}=\frac{AE}{AC}\left(\frac{4}{12}=\frac{5}{15}=\frac13\right)\)
nên DE//BC
b: Xét tứ giác BDEF có
BD//EF
BF//DE
Do đó: BDEF là hình bình hành
c: AE+EC=AC
=>CE=AC-AE=15-5=10(cm)
Xét ΔCAB có EF//AB
nên \(\frac{CE}{CA}=\frac{CF}{CB}\)
=>\(\frac{CF}{18}=\frac{10}{15}=\frac23\)
=>CF=12(cm)
BF+FC=BC
=>BF=18-12=6(cm)
Câu 3:
ĐKXĐ: a<>-1/3; a<>-3
\(\frac{3a-1}{3a+1}+\frac{a-3}{a+3}=2\)
=>\(\frac{\left(3a-1\right)\left(a+3\right)+\left(3a+1\right)\left(a-3\right)}{\left(3a+1\right)\left(a+3\right)}=2\)
=>(3a-1)(a+3)+(3a+1)(a-3)=2(3a+1)(a+3)
=>\(3a^2+9a-a-3+3a^2-9a+a-3=2\left(3a^2+9a+a+3\right)\)
=>\(6a^2-6=2\left(3a^2+10a+3\right)\)
=>\(6a^2+20a+6=6a^2-6\)
=>20a=-12
=>\(a=-\frac{12}{20}=-\frac35\) (nhận)
