Bài 1:
a: 15-5x=0
=>5x=15
=>\(x=\dfrac{15}{5}=3\)
b: \(\left(3x-5\right)^2-\left(x+1\right)\left(9x-1\right)=4\)
=>\(9x^2-30x+25-\left(9x^2-x+9x-1\right)=4\)
=>\(9x^2-30x+25-9x^2-8x+1=4\)
=>-38x=4-1-25=-3-25=-22
=>\(x=\dfrac{22}{38}=\dfrac{11}{19}\)
c: \(\dfrac{2\left(3x+1\right)}{4}-5=\dfrac{3\left(3x-1\right)}{5}-\dfrac{3x+2}{10}\)
=>\(\dfrac{3x+1}{2}-5=\dfrac{3\left(3x-1\right)}{5}-\dfrac{3x+2}{10}\)
=>\(\dfrac{5\left(3x+1\right)}{10}-\dfrac{50}{10}=\dfrac{6\left(3x-1\right)}{10}-\dfrac{3x+2}{10}\)
=>\(15x+5-50=18x-6-3x-2\)
=>15x-45=15x-8
=>-45=-8(vô lý)
Vậy: phương trình vô nghiệm
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