\(3x^2-2x-8=0\\ \Leftrightarrow3x^2-2x=8\\ E=6x^2-4x+9\\ =3x^2+3x^2-2x-2x-8+17\\ =\left(3x^2-2x-8\right)+\left(3x^2-2x+17\right)\\ =3x^2-2x+17\\ =\left(3x^2-2x\right)+17=8+17=25\)

\(x+y=0\\ \Leftrightarrow y=-x\\ D=x^4-y^4+x^3y-xy^3\\ =\left(x^2+y^2\right)\left(x^2-y^2\right)+xy\left(x^2-y^2\right)\\ =\left(x^2+y^2+xy\right)\left(x^2-y^2\right)\\ =\left(x^2+\left(-x\right)^2+x.\left(-x\right)\right)\left(x^2-\left(-x\right)^2\right)\\ =\left(x^2+x^2-x^2\right)\left(x^2-x^2\right)\\ =x^2.0=0\)

d) <=>x²-5x-x+5=0

<=>x(x-5)-(x-5)=0

<=>(x-5)(x-1)=0

<=>x=5 hoặc x=1

thank nha

Mình giải mẫu pt đầu thôi nhé, những pt sau ttự.

1,\(x^4-\frac{1}{2}x^3-x^2-\frac{1}{2}x+1=0\)

Ta thấy x=0 ko là nghiệm.

Chia cả 2 vế cho x² >0:

pt\(\Leftrightarrow x^2-\frac{1}{2}x-1-\frac{1}{2x}+\frac{1}{x^2}=0\)

Đặt \(t=x-\frac{1}{x}\left(t\in R\right)\)

\(\Rightarrow x^2+\frac{1}{x^2}=t^2+2\)

pt\(\Leftrightarrow t^2-\frac{1}{2}t+1=0\)(vô n₀)

Vậy pt vô n₀.

#Walker

a: =x^4-3x^5+4x^8

b: =2x^3+2x^2+4x

c: =4x^2+8x-5

d: =2x+3x^2+7x^4

\(\Leftrightarrow x^4-2x^3-2x^2+5x^3-10x^2-10x-2x^2+4x+4=0\\ \Leftrightarrow\left(x^2-2x-2\right)\left(x^2+5x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2-2x-2=0\left(1\right)\\x^2+5x-2=0\left(2\right)\end{matrix}\right.\)

Ta có \(\Delta\left(1\right)=4+8=12;\Delta\left(2\right)=25+8=33\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2-2\sqrt{3}}{2}=1-\sqrt{3}\\x=\dfrac{2+2\sqrt{3}}{2}=1+\sqrt{3}\\x=\dfrac{-5+\sqrt{33}}{2}\\x=\dfrac{-5-\sqrt{33}}{2}\end{matrix}\right.\)

\(x^4+3x^3-14x^2-6x+4=0\\ \Leftrightarrow\left(x^4-2x^3-2x^2\right)+\left(5x^3-10x^2-10x\right)-\left(2x^2-4x-4\right)=0\\ \Leftrightarrow\left(x^2-2x-2\right)\left(x^2+5x-2\right)=0\\ \Leftrightarrow....\)

Biến đổi ta được:  1 x + 20 . T = 1 2 ⇒ T = x + 20 2