Bjr,
1) Si j'ai bien compris il faut résoudre:
[tex] f(x) = \dfrac{5x^2}{x^2+16}\\f(x) \geq 1\\\\\dfrac{5x^2}{x^2+16} \geq 1\\\\\dfrac{5x^2}{x^2+16} -1 \geq 0\\\\\dfrac{5x^2-(x^2+16)}{x^2+16} \geq 0\\\\\dfrac{5x^2-x^2-16}{x^2+16} \geq 0 \\\\\dfrac{4x^2-16}{x^2+16} \geq 0 \\\\\dfrac{4(x^2-4)}{x^2+16} \geq 0 \\\\\\4(x^2-4) \geq 0 \\\quad \text{ou}\\x^2+16 \geq 0\\\\x\in \, ]-\infty; -2]\cup[2; + \infty]\\\text{ou}\\x \in \, \mathbb{R}\\\\x \in \left\{-2; 2\right\}\\\text{ou}\\x \in \varnothing \rightarrow \text{un carr\'e est toujours positif}\\\\\boxed{x\in\, ]-\infty; -2]\cup [2; -\infty[} [/tex]
2)
[tex] f(x) = 4x^2-(2x-1)^2 \rightarrow (a-b)^2 = a^2-2ab+b^2\\f(x) = 4x^2 - ((2x)^2-2\times 2x \times 1 + (1)^2)\\f(x) = 4x^2 -4x^2+4x -1\\f(x) = 4x -1 \rightarrow ax+b [/tex]
Bonne journée.
