ĐỀ bài em sai nhé

Cho \(f\left(x\right)=ax^{2^{ }}+bx+c\)

suy ra \(f\left(x_0\right)=0\Rightarrow f\left(x_0\right)=ax_0^{2^{ }}+bx_0+c=0\)

\(g\left(x\right)=cx^{2^{ }}+bx+a\Rightarrow g\left(\frac{1}{x_0}\right)=c.\left(\frac{1}{x_0}\right)^2+b.\frac{1}{x_0}+a\)

\(\Rightarrow g\left(\frac{1}{x_0}\right)=\frac{c}{x_0^2}+\frac{b}{x_0}+a=\frac{c+bx_0+ax^2_0}{x_0^2}=\frac{f\left(x_0\right)}{x_0^2}=0\) (với x₀ khác 0) 

Bài này đơn giản lắm bạn! Lưu ý mk thay đổi x0 thành m cho dễ ghi nha

Ta có \(f\left(m\right)=am^2+bm+c=0\)

Lại có \(g\left(\frac{1}{m}\right)=c\cdot\frac{1}{m^2}+b\cdot\frac{1}{m}+a=\frac{c}{m^2}+\frac{bm}{m^2}+\frac{am^2}{m^2}=\frac{am^2+bm+c}{m^2}=0\left(ĐPCM\right)\)

Viết đề còn sai =.=

g(x) = cx² + bx + a

\(f\left(x_0\right)=ax^2_0+bx_0+c=0\)

\(\Rightarrow g\left(\frac{1}{x_0}\right)=\frac{c}{x^2_0}+\frac{b}{x_0}+a=\frac{c+bx_0+ax_0^2}{x_0^2}=\frac{0}{x_0^2}=0\)

dap an bang o dung ko 

f(x0)=?.

2.f(x)=x^2+4x+10=x^2+4x+4+6=(x+2)^2+6

Mà(x+2)^2>=0=>(x+2)^2+6>0=>f(x) vô nghiệm

ahhii

Cho \(F\left(x\right)=ax^{2^{ }}+bx+c\)

suy ra \(F\left(x_0\right)=0\Rightarrow F\left(x_0\right)=ax_0^{2^{ }}+bx_0+c=0\)

\(G\left(x\right)=cx^{2^{ }}+bx+a\Rightarrow G\left(\frac{1}{x_0}\right)=c.\left(\frac{1}{x_0}\right)^2+b.\frac{1}{x_0}+a\)

\(\Rightarrow G\left(\frac{1}{x_0}\right)=\frac{c}{x_0^2}+\frac{b}{x_0}+a=\frac{c+bx_0+ax^2_0}{x_0^2}=\frac{f\left(x_0\right)}{x_0^2}=0\) (với x₀ khác 0)