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

\(\Leftrightarrow x^4+4x^3+4x^2+x^2-4x+4=0\)

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

Vì \(x^2\left(x+2\right)^2\ge0\forall x;\left(x-2^2\right)\ge0\forall x\)

\(\Rightarrow x^2\left(x+2\right)^2+\left(x-2\right)^2\ge0\)

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

\(\Rightarrow\hept{\begin{cases}x\left(x+2\right)=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=0;x=-2\\x=2\end{cases}}\)

Mà ko cùng một lúc tồn tại 2 giá trị của x

\(\Rightarrow\)Phương trình vô nghiệm

Vậy ...

khó quá ?????

mik chưa học đến lớp 8 nên ko biết

\(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)
\(\Leftrightarrow-5x^2-9x+2=0\)
\(\Leftrightarrow-5x^2-10x+x+2=0\)
\(\Leftrightarrow-5x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(-5x+1\right)\left(x+2\right)=0 \)
\(\Leftrightarrow\left(-5x+1\right)=0\) Hoặc \(x+2=0\)
\(\Leftrightarrow x=\frac{1}{5}\)Hoặc \(x=-2\)
 

(x+2)(3-4x)=x²+4x+4

<=>(x+2)(3-4x)=(x+2)²

<=>(x+2)(3-4x)-(x+2)²=0

<=>(x+2)(3-4x-x-2)=0

<=>(x+2)(1-5x)=0

<=>x+2=0 hoặc 1-5x=0

<=>x=-2 hoặc x=1/5

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

\(\Leftrightarrow\left(x-2\right)\left(4x+3\right)=\left(x-2\right)^2\)

\(\Leftrightarrow\left(x-2\right)\left(4x+3\right)-\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x+3-x+2\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{-5}{3}\end{cases}}\)

Vậy....

pt <=> x^4+8x-4x^3-5 = 0

<=> (x^4-x^3)-(3x^3-3x)+(5x-5) = 0

<=> x^3.(x-1)-3.x.(x-1).(x+1)+5.(x-1) = 0

<=> (x-1).(x^3-3x^2-3x+5) = 0

<=> (x-1).[(x^3-x^2)-(2x^2-2x)-(5x-5)] = 0

<=> (x-1)^2.(x^2-2x-5) = 0

<=> x-1=0 hoặc x^2-2x-5=0

<=> x=1 hoặc x = \(1+-\sqrt{6}\)

Vậy ...............

Tk mk nha

\(-x^5+4x^4=-12x^3\)

\(\Leftrightarrow x^5-4x^4-12x^3=0\)

\(\Leftrightarrow x^3\left(x^2-4x-12\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-4x-12=0\left(1\right)\end{cases}}\)

\(Pt\left(1\right)\Leftrightarrow\left(x+2\right)\left(x-6\right)=0\)

              \(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=6\end{cases}}\)

Vậy \(x\in\left\{-2;0;6\right\}\)

hok tốt

ngu vc