Giải mỗi phương trình sau: a) 0,5^2x^2+x-1=1/4

Trả lời

a) $0,5^{2x^2+x-1}=\frac{1}{4}\iff 0,5^{2x^2+x-1}=0,5^2$

$\iff 2x^2+x-3=0\iff \left[\begin{array}{l}x=-\frac{3}{2}\\ x=1\end{array}\right..$
Vậy phương trình có nghiệm $x\in \left\{-\frac{3}{2};1\right\}.$
b) $2^{x^2-6x-\frac{5}{2}}=16\sqrt{2}$
 $\iff 2^{x^2-6x-\frac{5}{2}}=2^4{.2}^{\frac{1}{2}}$
$\iff 2^{x^2-6x-\frac{5}{2}}=2^{\frac{9}{2}}$
$\iff x^2-6x-\frac{5}{2}=\frac{9}{2}$
$\iff x^2-6x-7=0\iff \left[\begin{array}{l}x=-1\\ x=7\end{array}\right..$
Vậy phương trình có nghiệm x ∈ {–1; 7}.
c) ${27}^{x^2-4x+4}=9^{x^2-4}$
$\iff 3^{3\left(x^2-4x+4\right)}=3^{2\left(x^2-4\right)}$
$\iff 3\left(x^2-4x+4\right)=2\left(x^2-4\right)$
$\iff 3x^2-12x+12=2x^2-8$
$\iff x^2-12x+20=0\iff \left[\begin{array}{l}x=2\\ x=10\end{array}\right..$
Vậy phương trình có nghiệm x ∈ {2; 10}.
d) $0,{05}^{x-3}={\left(2\sqrt{5}\right)}^x\iff {\left(\frac{1}{20}\right)}^{x-3}={\left(2\sqrt{5}\right)}^x$
$\iff {\left(\frac{1}{2\sqrt{5}}\right)}^{2\left(x-3\right)}={\left(2\sqrt{5}\right)}^x\iff {\left(2\sqrt{5}\right)}^{-2\left(x-3\right)}={\left(2\sqrt{5}\right)}^x$
⇔ –2(x – 3) = x ⇔ –3x = –6 ⇔ x = 2.
Vậy phương trình có nghiệm x = 2.
e) log3 3(x – 2) = –1 ⇔ 3(x – 2) = 3–1
$\iff 3\left(x-2\right)=\frac{1}{3}\iff x-2=\frac{1}{9}\iff x=\frac{19}{9}.$
Vậy phương trình có nghiệm $x=\frac{19}{9}.$
g) Ta có: log5 (x2 + 1) + log0,2 (4 – 5x – x2) = 0
$\iff {\log }_5\left(x^2+1\right)+{\log }_{5^{-1}}\left(4-5x-x^2\right)=0$
⇔ log5 (x2 + 1) – log5 (4 – 5x – x2) = 0
⇔ log5 (x2 + 1) = log5 (4 – 5x – x2)
$\iff \left\{\begin{array}{l}{x}^2+1=4–5x–x^2\\ x^2+1>0\end{array}\right.$
$\iff x^2+1=4–5x–x^2$(do x² + 1 > 0 ∀x ∈ ℝ)
$\iff 2x^2+5x–3=0$$\iff \left[\begin{array}{l}x=-3\\ x=\frac{1}{2}\end{array}\right..$
Vậy phương trình có nghiệm $x\in \left\{-3;\frac{1}{2}\right\}.$