Question

Îmi explică si mie cineva ,va rog , cum se fac exercițiile de tipul acesta?( exercițiul 3)

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$1. Convertim \:\ pe \:\ \left(6-\sqrt{35}\right)^x la \:\ baza \:\ \left(6+\sqrt{35}\right) \:\ si \:\ avem:\\ => \left(6-\sqrt{35}\right)^x=\left(\left(6+\sqrt{35}\right)^{-1}\right)^x = 2\sqrt{35}\\\\\\ 2. \boxed{\left(a^b\right)^c=a^{bc}} => \\ \left(\left(6+\sqrt{35}\right)^{-1}\right)^x=\left(6+\sqrt{35}\right)^{-1x} \\\\  \left(6+\sqrt{35}\right)^x-\left(6+\sqrt{35}\right)^{-1\cdot \:x}=2\sqrt{35}\\\\3. Rescriem \:\ ecuatia \:\ in \:\ felul \:\ urmator: \left(6+\sqrt{35}\right)^x=u$

$u-\left(u\right)^{-1}=2\sqrt{35} \\\boxed{\sqrt{35}+6,\:u=\sqrt{35}-6}\\u-u^{-1}=2\sqrt{35} => u-\frac{1}{u}=2\sqrt{35}\\\\ 3. Multiplicam \:\ in \:\ amblele \:\ parti \:\ cu \:\ u =>\\\\uu-\frac{1}{u}u=2\sqrt{35}u => Simplificare: u^2-1=2\sqrt{35}\\\\4. Acum \:\ rezolvam \:\ in \:\ felul \:\ urmator:\\u^2-1-2\sqrt{35}u=2\sqrt{35}u-2\sqrt{35}u =>\\\boxed{u^2-2\sqrt{35}u-1=0}\\u=\frac{-\left(-2\sqrt{35}\right)-\sqrt{\left(-2\sqrt{35}\right)^2-4\cdot \:1\left(-1\right)}}{2\cdot \:1}:\quad \sqrt{35}-6$

$u=\frac{-\left(-2\sqrt{35}\right)-\sqrt{\left(-2\sqrt{35}\right)^2-4\cdot \:1\left(-1\right)}}{2\cdot \:1}:\quad \sqrt{35}-6\\\\Pentru \:\ \sqrt{35}+6 \:\ \boxed{x = 1}\\Pentru \sqrt{35}-6 \:\ nu \:\ avem \:\ solutie \:\ pentru\:\ x\in \mathbb{R}\\\\Deci \:\ raspunsul \:\ corect \:\ este \:\ \boxed{B}$