Chứng minh nếu F(x) = ax + b= G(x) = MX + n với mọi x thì a = m, b = n

Bài 1: Bài này tớ làm không đảm bảo đúng 100% nên nếu có gì sai sót mong bạn thông cảm

a) Nếu F(x) = G(x)

\(\Rightarrow ax+b-mx-n=0\)

\(\Rightarrow x\left(a-m\right)+\left(b-n\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x\left(a-m\right)=0\\b-n=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-m=0\\b=n\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=m\\b=n\end{matrix}\right.\)

b) Nếu F(x) = G(x)

\(\Rightarrow ax^2+bx+c-mx^2-nx-p=0\)

\(\Rightarrow x^2\left(a-m\right)+x\left(b-n\right)+\left(c-p\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x^2\left(a-m\right)=0\\x\left(b-n\right)=0\\c-p=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-m=0\\b-n=0\\c-p=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=m\\b=n\\c=p\end{matrix}\right.\)

Bài 2:

a) \(A\left(x\right)=0\)

\(\Leftrightarrow2\left(\dfrac{1}{3}x-\dfrac{1}{2}\right)-\dfrac{1}{2}\left(3-x\right)=0\)

\(\Leftrightarrow2.\dfrac{1}{3}x-2.\dfrac{1}{2}-\dfrac{1}{2}.3+\dfrac{1}{2}x=0\)

\(\Leftrightarrow\dfrac{2}{3}x-1-\dfrac{3}{2}+\dfrac{1}{2}x=0\)

\(\Leftrightarrow\dfrac{7}{6}x-\dfrac{5}{2}=0\)

\(\Leftrightarrow\dfrac{7}{6}x=\dfrac{5}{2}\)

\(\Leftrightarrow x=\dfrac{15}{7}\)

b) Nếu B (x) = 0

\(\Leftrightarrow\left(2x-5\right)\left(x^2-\dfrac{9}{16}\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\x^2-\dfrac{9}{16}=0\\x^2+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=5\\x^2=\dfrac{9}{16}\\x^2=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{3}{4};x=-\dfrac{3}{4}\\x=1;x=-1\end{matrix}\right.\)

c) Nếu C(x) = 0

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{2};x=-\sqrt{2}\end{matrix}\right.\)

d) Nếu D(x) = 0

\(\Leftrightarrow9x^2+16=0\)

\(\Leftrightarrow9x^2=-16\)

\(\Leftrightarrow x^2=-\dfrac{16}{9}\)

Vậy không tồn tại x thỏa mãn

e) Nếu M(x) = 0

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

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

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)