Question

[Nguyễn Thị Lan Anh]

GIẢI CÁC PHƯƠNG TRÌNH SAU

A) \(13X^2+29X+17=0\)

B) \(X^2+1=X\)

C) \(X^2-1=X\)

Answer 1

[Nữ Thần Mặt Trăng]

a,\(13x^2+29x+17=0\)

<=>\(x^2+\frac{29}{13}x+\frac{17}{13}=0\)

<=>\(x^2+2.x.\frac{29}{26}+\left(\frac{29}{26}\right)^2+\frac{43}{676}=0\)

<=>\(\left(x+\frac{29}{26}\right)^2+\frac{43}{676}=0\)

Vì \(\left(x+\frac{29}{26}\right)^2\ge0\) => \(\left(x+\frac{29}{26}\right)^2+\frac{43}{676}>0\)

=>pt vô nghiệm

\(b,x^2+1=x\\ =>x^2-x+1=0\\ =>x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=0\\ =>\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=0\)

Vì \(\left(x+\frac{1}{2}\right)^2\ge0=>\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

=>pt vô nghiệm

\(c,x^2-1=x\\ =>x^2-x-1=0\\ =>x^2-2.x.\frac{1}{2}+\frac{1}{4}-\frac{5}{4}=0\\ =>\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=0\\ =>\left(x-\frac{1}{2}+\frac{\sqrt{5}}{2}\right)\left(x-\frac{1}{2}-\frac{\sqrt{5}}{2}=0\right)\)

\(=>\left(x-\frac{1-\sqrt{5}}{2}\right)\left(x-\frac{1+\sqrt{5}}{2}\right)=0\)

\(=>x_1=\frac{1-\sqrt{5}}{2};x_2=\frac{1+\sqrt{5}}{2}\)

Answer 2

[Lê Thành Vinh]

B)x²+1=x

<=>x²-x+1=0

<=>x²-x+\(\frac{1}{4}+\frac{3}{4}=0\)

<=>[\(x-\left(\frac{1}{2}\right)^2\)]\(+\frac{3}{4}=0\)

Vì [\(x-\left(\frac{1}{2}\right)^2\)]>=0 với mọi x nên[\(x-\left(\frac{1}{2}\right)^2\)]+\(\frac{3}{4}>=\frac{3}{4}\)>0 với mọi x

Vậy phương trình vô ngiệm