• Mer. Gen 17th, 2024

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Disequazioni frazionarie di primo grado

Risolvere la seguente disequazione frazionaria: $\displaystyle \frac{1}{x^2-x}-\frac{1}{x}\ge\frac{2}{x-1}$ SOLUZIONE $\displaystyle \frac{1}{x(x-1)}-\frac{1}{x}\ge\frac{2}{x-1}$ $\displaystyle \frac{1}{x(x-1)}-\frac{1}{x}-\frac{2}{x-1}\ge0$ $\displaystyle \frac{1-1(x-1)-2x}{x(x-1)}\ge0$ $\displaystyle \frac{1-x+1-2x}{x(x-1)}\ge0$ $\displaystyle \frac{2-3x}{x(x-1)}\ge0$ Studio del…

Cubo di un binomio

Scomporre i seguenti polinomi: 1) $\displaystyle x^3-x^2+\frac{1}{3}x-\frac{1}{27}=\left(x-\frac{1}{3}\right)^3$ 2) $\displaystyle a^3-6a^2+12a-8=(a-2)^3$ 3) $\displaystyle 27a^6-54a^4+18a^2-8=(3a^2-2)^3$ 4) $\displaystyle 0,125t^3-3t^2+24t-64=(0,5t-4)^3$ 5) $\displaystyle x^{6n}-6x^{4n}+12x^{2n}-8=(x^{2n}-2)^3$ 6)…

Differenza di quadrati

Scomporre i seguenti polinomi: 1) $\displaystyle -a^4+b^2=(b+a^2)(b-a^2)$ 2) $\displaystyle \frac{1}{4}x^6-\frac{1}{9}y^2=\left(\frac{1}{2}x^3+\frac{1}{3}y\right)\left(\frac{1}{2}x^3-\frac{1}{3}y\right)$ 3) $\displaystyle x^{2n}-y^4=(x^n+y^2)(x^n-y^2)$ 4) $\displaystyle a^{2n+2}-1=a^{2(n+1)}-1=(a^{n+1}+1)(a^{n+1}-1)$ 5) $\displaystyle -81a^4+4^{-2}=-81a^4+\left(\frac{1}{4}\right)^2=$ $\displaystyle=\left(\frac{1}{4}+9a^2\right)\left(\frac{1}{4}-9a^2\right)=$…