Question

[Nguyễn An Na]

giúp tớ với ạ, tớ cần gấp ạ

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(x^4+x^3+4x^2+5x+25=0\)

=>\(x^4+3x^3+5x^2-2x^3-6x^2-10x+5x^2+15x+25=0\)

=>\(x^2\left(x^2+3x+5\right)-2x\left(x^2+3x+5\right)+5\left(x^2+3x+5\right)=0\)

=>\(\left(x^2+3x+5\right)\left(x^2-2x+5\right)=0\)

mà \(x^2+3x+5=\left(x+\frac32\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\forall x\)

và \(x^2-2x+5=\left(x-1\right)^2+4\ge4>0\forall x\)

nên x∈∅

b: \(x^4+3x^3-14x^2-6x+4=0\)

=>\(x^4-2x^3-2x^2+5x^3-10x^2-10x-2x^2+4x+4=0\)

=>\(x^2\left(x^2-2x-2\right)+5x\left(x^2-2x-2\right)-2\left(x^2-2x-2\right)=0\)

=>\(\left(x^2-2x-2\right)\left(x^2+5x-2\right)=0\)

TH1: \(x^2-2x-2=0\)

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

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

=>\(x-1=\pm\sqrt3\)

=>\(x=1\pm\sqrt3\)

TH2: \(x^2+5x-2=0\)

=>\(x^2+5x+\frac{25}{4}-\frac{33}{4}=0\)

=>\(\left(x+\frac52\right)^2=\frac{33}{4}\)

=>\(x+\frac52=\pm\frac{\sqrt{33}}{2}\)

=>\(x=-\frac52\pm\frac{\sqrt{33}}{2}\)

c: \(x^4+5x^3-14x^2-20x+16=0\)

=>\(x^4+6x^3-4x^2-x^3-6x^2+4x-4x^2-24x+16=0\)

=>\(x^2\left(x^2+6x-4\right)-x\left(x^2+6x-4\right)-4\left(x^2+6x-4\right)=0\)

=>\(\left(x^2+6x-4\right)\left(x^2-x-4\right)=0\)

TH1: \(x^2+6x-4=0\)

=>\(x^2+6x+9=13\)

=>\(\left(x+3\right)^2=13\)

=>\(x+3=\pm\sqrt{13}\)

=>\(x=-3\pm\sqrt{13}\)

TH2: \(x^2-x-4=0\)

=>\(x^2-x+\frac14=\frac{17}{4}\)

=>\(\left(x-\frac12\right)^2=\frac{17}{4}\)

=>\(x-\frac12=\pm\frac{\sqrt{17}}{2}\)

=>\(x=\frac12\pm\frac{\sqrt{17}}{2}\)

d: \(3x^4+2x^3-13x^2-4x+12=0\)

=>\(3x^4-3x^3+5x^3-5x^2-8x^2+8x-12x+12=0\)

=>\(3x^3\left(x-1\right)+5x^2\left(x-1\right)-8x\left(x-1\right)-12\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(3x^3+5x^2-8x-12\right)=0\)

=>\(\left(x-1\right)\left(3x^3+6x^2-x^2-2x-6x-12\right)=0\)

=>\(\left(x-1\right)\left\lbrack3x^2\left(x+2\right)-x\left(x+2\right)-6\left(x+2\right)\right\rbrack=0\)

=>\(\left(x-1\right)\left(x+2\right)\left(3x^2-x-6\right)=0\)

TH1: x-1=0

=>x=1

TH2: x+2=0

=>x=-2

TH3: \(3x^2-x-6=0\)

=>\(x^2-\frac12x-2=0\)

=>\(x^2-2\cdot x\cdot\frac14+\frac{1}{16}-\frac{33}{16}=0\)

=>\(\left(x-\frac14\right)^2=\frac{33}{16}\)

=>\(x-\frac14=\pm\frac{\sqrt{33}}{4}\)

=>\(x=\frac14\pm\frac{\sqrt{33}}{4}\)