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b: 4(x+1)^2-9(x-1)^2=0
=>(2x+2)^2-(3x-3)^2=0
=>(2x+2-3x+3)(2x+2+3x-3)=0
=>(-x+5)(5x-1)=0
=>x=1/5 hoặc x=5
c: (x-1)^3+x^3+(x+1)^3=(x+2)^3
=>x^3-3x^2+3x-1+x^3+x^3+3x^2+3x+1=x^3+6x^2+12x+8
=>3x^3+6x-x^3-6x^2-12x-8=0
=>2x^3-6x^2-6x-8=0
=>x^3-3x^2-3x-4=0
=>x^3-4x^2+x^2-4x+x-4=0
=>(x-4)(x^2+x+1)=0
=>x-4=0
=>x=4
b: \(x+\dfrac{5}{6}=\dfrac{3}{8}\)
=>\(x=\dfrac{3}{8}-\dfrac{5}{6}\)
=>\(x=\dfrac{9}{24}-\dfrac{20}{24}=-\dfrac{11}{24}\)
c: \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{5}{6}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{6}+\dfrac{1}{2}=\dfrac{8}{6}=\dfrac{4}{3}\)
=>2x=4
=>x=4/2=2
d: \(\dfrac{x}{7}=\dfrac{6}{-21}\)
=>\(\dfrac{x}{7}=\dfrac{-2}{7}\)
=>x=-2
e: \(\left(\dfrac{7}{3}x-0,6\right):\dfrac{3^2}{5}=1\)
=>\(\dfrac{7}{3}x-0,6=\dfrac{3^2}{5}=1,8\)
=>\(\dfrac{7}{3}x=2,4\)
=>\(x=2,4:\dfrac{7}{3}=2.4\cdot\dfrac{3}{7}=\dfrac{7.2}{7}=\dfrac{36}{35}\)
f: \(\dfrac{x}{45}=\dfrac{5}{6}+\dfrac{-29}{30}\)
=>\(\dfrac{x}{45}=\dfrac{25}{30}-\dfrac{29}{30}=-\dfrac{4}{30}=-\dfrac{2}{15}\)
=>\(x=-\dfrac{2}{15}\cdot45=-6\)
g: \(\left(4,5-2x\right)\cdot\left(-\dfrac{1^4}{7}\right)=\dfrac{11}{14}\)
=>\(4,5-2x=\dfrac{11}{14}:\dfrac{-1}{7}=\dfrac{-11}{2}\)
=>\(2x=4,5+\dfrac{11}{2}=\dfrac{20}{2}=10\)
=>x=10/2=5
h: \(-\dfrac{2}{7}+\dfrac{4}{7}x=\dfrac{5}{7}\)
=>\(\dfrac{4}{7}x=\dfrac{5}{7}+\dfrac{2}{7}=\dfrac{7}{7}\)
=>4x=7
=>\(x=\dfrac{7}{4}\)
