Question

giúp em giải bài này với ạ

Bài 5. Tìm x biết

a) \((x+4)^3 - x^2(x+12) = 16\)

\((x+3)^3 - x(3x+1)^2 + (2x+1)(4x^2-2x+1) = 28\)

c) \(6(x+1)^2 + 2(x-1)(x^2 + x+1) - 2(x+1)^3 = 32\)

d) \((x-2)^3 - (x+5)(x^2-5x+25) + 6x^2 = 11\)

e) \((x+1)(x^2-x+1) - x(x-2)(x+2) = 7\)

Answer 1

a: \(\left(x+4\right)^3-x^2\left(x+12\right)=16\)

=>\(x^3+12x^2+48x+64-x^3-12x^2=16\)

=>48x+64=16

=>48x=16-64=-48

=>x=-1

b: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

=>\(x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1=28\)

=>\(9x^3+9x^2+27x+28-9x^3-6x^2-x=28\)

=>\(3x^2+26x\) =0

=>x(3x+26)=0

=>\(\left[\begin{array}{l}x=0\\ 3x+26=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-\frac{26}{3}\end{array}\right.\)

c: \(6\left(x+1\right)^2+2\left(x-1\right)\left(x^2+x+1\right)-2\left(x+1\right)^3=32\)

=>\(3\left(x+1\right)^2+\left(x-1\right)\left(x^2+x+1\right)-\left(x+1\right)^3=16\)

=>\(3\left(x^2+x+1\right)+x^3-1-x^3-3x^2-3x-1=16\)

=>\(3x^2+3x+3-3x^2-3x-2=16\)

=>1=16(vô lý)

=>Phương trình vô nghiệm

d: \(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)

=>\(x^3-6x^2+12x-8-x^3-125+6x^2=11\)

=>12x-133=11

=>12x=133+11=144

=>x=12

e: \(x\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x^2-3x+9\right)=1\)

=>\(x\left(x^2-4\right)-\left(x^3-3x^2+9x-3x^2+9x-27\right)=1\)

=>\(x^3-4x-\left(x^3-6x^2+18x-27\right)=1\)

=>\(x^3-4x-x^3+6x^2-18x+27-1=0\)

=>\(6x^2-22x+26=0\)

=>\(3x^2-11x+13=0\)

\(\Delta=\left(-11\right)^2-4\cdot3\cdot13=121-12\cdot13=121-156=-35<0\)

=>Phương trình vô nghiệm

f: \(\left(x+1\right)\left(x^2-x+1\right)-x\left(x-2\right)\left(x+2\right)=7\)

=>\(x^3+1-x\left(x^2-4\right)=7\)

=>\(x^3+1-x^3+4x=7\)

=>4x+1=7

=>4x=6

=>\(x=\frac64=\frac32\)