Toán a) 4x ² +1 =4x b) (x ²+1) ² + 3x(x ²+1) + 2x ² =0. 10/10/2021 By Madeline a) 4x ² +1 =4x b) (x ²+1) ² + 3x(x ²+1) + 2x ² =0.
Đáp án: $a) x=\dfrac{1}{2}$ $b) x=-1$ Giải thích các bước giải: $a) 4x^{2}+1=4x\\\Leftrightarrow 4x^{2}+1-4x=0\\\Leftrightarrow 4x^{2}-4x+1=0\\\Leftrightarrow (2x-1)^{2}=0\\\Leftrightarrow 2x-1=0\\\Leftrightarrow 2x=1\\\Leftrightarrow x=\dfrac{1}{2}\\b) (x^{2}+1)^{2}+3x(x^{2}+1)+2x^{2}=0\\\Leftrightarrow x^{4}+2x^{2}+1+3x^{3}+3x+2x^{2}=0\\\Leftrightarrow x^{4}+4x^{2}+1+3x^{3}+3x=0\\\Leftrightarrow x^{4}+3x^{3}+4x^{2}+3x+1=0\\\Leftrightarrow x^{4}+x^{3}+2x^{3}+2x^{2}+2x^{2}+2x+x+1=0\\\Leftrightarrow x^{3}(x+1)+2x^{2}(x+1)+2x(x+1)+1(x+1)=0\\\Leftrightarrow (x+1)(x^{3}+2x^{2}+2x+1)=0\\\Leftrightarrow (x+1)[(x+1)(x^{2}-x|1)+2x(x+1)]=0\\\Leftrightarrow (x+1)(x+1)(x^{2}-x+1+2x)=0\\\Leftrightarrow (x+1)(x+1)(x^{2}+x+1)=0\\\Leftrightarrow (x+1)^{2}(x^{2}+x+1)=0\\\Leftrightarrow \left[ \begin{array}{l}(x+1)^{2}=0\\x^{2}+x+1=0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=-1\\x\notin \mathbb{R}\end{array} \right.$ Trả lời
Đáp án:
Giải thích các bước giải:
Đáp án:
$a) x=\dfrac{1}{2}$
$b) x=-1$
Giải thích các bước giải:
$a) 4x^{2}+1=4x\\\Leftrightarrow 4x^{2}+1-4x=0\\\Leftrightarrow 4x^{2}-4x+1=0\\\Leftrightarrow (2x-1)^{2}=0\\\Leftrightarrow 2x-1=0\\\Leftrightarrow 2x=1\\\Leftrightarrow x=\dfrac{1}{2}\\b) (x^{2}+1)^{2}+3x(x^{2}+1)+2x^{2}=0\\\Leftrightarrow x^{4}+2x^{2}+1+3x^{3}+3x+2x^{2}=0\\\Leftrightarrow x^{4}+4x^{2}+1+3x^{3}+3x=0\\\Leftrightarrow x^{4}+3x^{3}+4x^{2}+3x+1=0\\\Leftrightarrow x^{4}+x^{3}+2x^{3}+2x^{2}+2x^{2}+2x+x+1=0\\\Leftrightarrow x^{3}(x+1)+2x^{2}(x+1)+2x(x+1)+1(x+1)=0\\\Leftrightarrow (x+1)(x^{3}+2x^{2}+2x+1)=0\\\Leftrightarrow (x+1)[(x+1)(x^{2}-x|1)+2x(x+1)]=0\\\Leftrightarrow (x+1)(x+1)(x^{2}-x+1+2x)=0\\\Leftrightarrow (x+1)(x+1)(x^{2}+x+1)=0\\\Leftrightarrow (x+1)^{2}(x^{2}+x+1)=0\\\Leftrightarrow \left[ \begin{array}{l}(x+1)^{2}=0\\x^{2}+x+1=0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=-1\\x\notin \mathbb{R}\end{array} \right.$