Bài 7:
a: 7x+4=2x-6
=>7x-2x=-6-4
=>5x=-10
=>\(x=-\dfrac{10}{5}=-2\)
b: \(\dfrac{3x-4}{5}-\dfrac{x+3}{4}=\dfrac{x+1}{10}\)
=>\(\dfrac{4\left(3x-4\right)-5\left(x+3\right)}{20}=\dfrac{2x+2}{20}\)
=>\(4\left(3x-4\right)-5\left(x+3\right)=2x+2\)
=>12x-16-5x-15=2x+2
=>7x-21=2x+2
=>5x=23
=>\(x=\dfrac{23}{5}\)
c: ĐKXĐ: \(x\notin\left\{2;4\right\}\)
\(\dfrac{x-3}{x-2}-\dfrac{x-2}{x-4}=-1\)
=>\(\dfrac{\left(x-3\right)\left(x-4\right)-\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1\)
=>\(x^2-7x+12-\left(x^2-4x+4\right)=-\left(x^2-6x+8\right)\)
=>\(x^2-7x+12-x^2+4x-4+x^2-6x+8=0\)
=>\(x^2-9x+16=0\)
=>\(x^2-9x+20,25-4,25=0\)
=>\(\left(x-4,5\right)^2=4,25\)
=>\(x-4,5=\pm\dfrac{\sqrt{17}}{2}\)
=>\(x=\dfrac{\pm\sqrt{17}+9}{2}\)
Bài 5:
a: 3x-5=4x+1
=>3x-4x=5+1
=>-x=6
=>x=-6
b: \(\dfrac{x+5}{3}-\dfrac{x-2}{2}=\dfrac{x-1}{4}\)
=>\(\dfrac{4\left(x+5\right)-6\left(x-2\right)}{12}=\dfrac{3\left(x-1\right)}{12}\)
=>4(x+5)-6(x-2)=3(x-1)
=>4x+20-6x+12=3x-3
=>-2x+32=3x-3
=>-5x=-35
=>\(x=\dfrac{35}{5}=7\)
c: ĐKXĐ: \(x\in R\backslash\left\{1;-1\right\}\)
\(\dfrac{2x+1}{x-1}=\dfrac{5\left(x-1\right)}{x+1}\)
=>\(\dfrac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{5\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\)
=>\(5\left(x-1\right)^2=\left(2x+1\right)\left(x+1\right)\)
=>\(5x^2-10x+5-\left(2x^2+3x+2\right)=0\)
=>\(3x^2-13x+3=0\)
=>\(\left[{}\begin{matrix}x=\dfrac{13+\sqrt{133}}{6}\left(nhận\right)\\x=\dfrac{13-\sqrt{133}}{6}\left(loại\right)\end{matrix}\right.\)
Bài 8:
Gọi độ dài quãng đường AB là x(km)
(Điều kiện: x>0)
Thời gian người đó đi từ A đến B là \(\dfrac{x}{12}\left(giờ\right)\)
Thời gian người đó đi từ B về A là \(\dfrac{x}{12-3}=\dfrac{x}{9}\left(giờ\right)\)
Thời gian về nhiều hơn thời gian đi là 1h20p=4/3h nên ta có:
\(\dfrac{x}{9}-\dfrac{x}{12}=\dfrac{4}{3}\)
=>\(\dfrac{x}{36}=\dfrac{4}{3}\)
=>\(x=36\cdot\dfrac{4}{3}=12\cdot4=48\left(nhận\right)\)
vậy: Độ dài quãng đường AB là 48km
