Question

GPT:

a,\(\left(2x-1\right)^4+\left(2x-3\right)^4=0\)

b,\(a^4-4a^3+12^2-16a+8=0\)

c,\(a^4+12a^2+2=0\)

d,\(\left(x^2-4x\right)^2+2\left(x-2\right)^2-43=0\)

Answer 1

a) \(\left(2x-1\right)^4+\left(2x-3\right)^4=0\)

\(\Leftrightarrow\left(2x-1\right)^4+\left(2x-3\right)^4=0^4\)

\(\Leftrightarrow\left(2x-1\right)+\left(2x-3\right)=0\)

\(\Rightarrow\hept{\begin{cases}2x-1=0\\2x-3=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\left(0+1\right):2\\x=\left(0+3\right):2\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)

Answer 2

b) \(a^4-4a^3+12^2-16a+8=0\)

Answer 3

a)\(\left(2x-1\right)^4+\left(2x-3\right)^4=0\\ \)

do \(\left(2x-1\right)^4\ge0\)và \(\left(2x-3\right)^4\ge0\)

=> \(\hept{\begin{cases}\left(2x-1\right)^4=0\\\left(2x-3\right)^4=0\end{cases}\Rightarrow\hept{\begin{cases}2x-1=0\\2x-3=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{2}\end{cases}}}\).

Vậy x vừa = 1/2 vừa =3/2 => vô lý => không tồn tại nghiệm

Answer 4

d)\(\left(x^2-4x\right)^2+2\left(x-2\right)^2-43=0\\ \)

=> \(\left(x^2-4x\right)^2+2\left(x^2-4x+4\right)-43=0\)=> \(\left(x^2-4x\right)^2+2\left(x^2-4x\right)+8-43=0\)

=>\(\left(x^2-4x\right)^2+2\left(x^2-4x\right)-35=0\)=> \(\left(x^2-4x\right)^2+2\left(x^2-4x\right)+1-36=0\)

=> \(\left(x^2-4x+1\right)^2-36=0\)=>\(\left(x^2-4x+1-6\right)\left(x^2-4x+1+6\right)=0\)

=> \(\left(x^2-4x-5\right)\left(x^2-4x+7\right)=0\) đến đây giải được tiếp được ko 

Answer 5

c) \(a^4+12a^2+2=0\)

\(\Leftrightarrow\left(a^2\right)^2+2\cdot a^2\cdot6+6^2-34=0\)

\(\Leftrightarrow\left(a^2+6\right)^2=34\)

\(\Leftrightarrow\left(a^2+6\right)^2=\left(\pm\sqrt{34}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}a^2+6=\sqrt{34}\\a^2+6=-\sqrt{34}\end{cases}\Leftrightarrow\orbr{\begin{cases}a\in\left\{\pm\sqrt{\sqrt{34}-6}\right\}\\a\in\varnothing\end{cases}}}\)

Vậy....