Question

tìm nghiệm

g(x)=(2x-3)(x+1/4)

h(x) = x^2 - 7

Answer 1

Đặt \(g\left(x\right)=0\)

\(\Rightarrow\left(2x-3\right)\left(x+\dfrac{1}{4}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\x+\dfrac{1}{4}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=3\\x=-\dfrac{1}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Vậy nghiệm của \(g\left(x\right)\) là \(\dfrac{3}{2};-\dfrac{1}{4}\)

Đặt \(h\left(x\right)=0\)

\(\Rightarrow x^2-7=0\)

\(\Rightarrow x^2=7\)

\(\Rightarrow x=\pm\sqrt{7}\)

Vậy nghiệm của \(h\left(x\right)\) là \(\pm\sqrt{7}\)

Answer 2

Đặt G(x)=0

=>(2x-3)(x+1/4)=0

=>x=3/2 hoặc x=-1/4

Đặt H(x)=0

=>x²=7

hay \(x\in\left\{\sqrt{7};-\sqrt{7}\right\}\)