Question

ce înseamnă E? Și știți cum se face problema? ​

Answer 1

A)

$x^2+ \underbrace{ \;\; x \;\; }_{\sf 4x-3x} -12 = \overbrace{x^2+4x}^{\sf f.c.\;\; pe \;\; x}-3x-12 \Rightarrow \quad \quad \quad \quad \quad \quad \quad \\ \\x(x+4)\overbrace{-3x-12}^{\sf f.c.\;\; pe\;\; -3} \; \Rightarrow \overbrace{x(x+4) -3(x+4) }^{\sf Avem \; acelasi\; factor \; in \; paranteze} \xrightarrow{\sf f.c. \; pe \; (x+4)}\\ \\\red{\boxed{(x+4)(x-3)}} \Rightarrow \underline{ x^2+x-12=(x+4)(x-3) \; \mathcal{\forall}\; x \in \mathbb{R}}$

B)

$E(x)=\overbrace{(x-1)^2+(x+4)(x-3)-2(x^2-4) }^{\sf (a-b)^2\,=\,a^2-2ab+b^2 \Rightarrow (x-1)^2=x^2-2x+1} \quad \\ \\E(x)=\underline{x^2-2x+1}+\overbrace{\underbrace{\blue{(x+4)(x-3)}}_{\sf inmultim\;\; parantezele}}^{\sf fiecare\; termen\;intre \;ei}-2(x^2-4) \\ \\E(x)=x^2-2x+1+\underline{(x \cdot x - x \cdot 3 + 4 \cdot x - 4 \cdot 3)} -2(x^2-4) \\ \\E(x)=x^2-2x+1+\blue{x^2-3x+4x-12} - \underbrace{2(x^2-4)}_{\sf distribuim \; 2} \\ \quad E(x)=\green{x^2}-2x+1\green{\,+\,x^2}\blue{-3x+4x}-12-2x^2+8$

$E(x)=\underline{2x^2}-2x \green{\,+\,1}+\underline{x} \green{\,-12}-2x^2\green{\,+\,8}$

$E(x)=\overbrace{\red{2x^2}-2x \underline{\,-\,3}+x\red\,{-\,2x^2}}^{\sf - \; cu\; + \;se \;reduc }$

$E(x)=-2x-3+x=\boxed{-x-3}$

$\red{\boxed{E(x)=-x-3} \; \underline{\forall \; x \in \mathbb{R} }}$