Question

Tìm x biết:

a) \(-0,6^0+\frac{1}{2}.2-3x=-\frac{1}{4}\)

b) \(2^{x-2}+22=3.2^x\)

c) \(\left(x-1\right)^2=\sqrt{\left(-\frac{9}{16}\right)^2}\)

d) \(\sqrt{x^2+2}=12-\sqrt{25}\)

 

Answer 1

a) \(-0,6^0+\frac{1}{2}.2-3x=-\frac{1}{4}\)

 

\(\Leftrightarrow-1+1-3x=-\frac{1}{4}\Leftrightarrow-3x=-\frac{1}{4}\Leftrightarrow3x=\frac{1}{4}\Leftrightarrow x=\frac{1}{4}:3=\frac{1}{12}\)

 

b)\(2^{x-2}+22=3.2^x\Leftrightarrow3.2^x-2^{x-2}=22\Leftrightarrow2^{x-2}\left(3.2^2-1\right)=22\)

\(\Leftrightarrow2^{x-2}.11=22\Leftrightarrow2^{x-2}=2\Leftrightarrow x-2=1\Leftrightarrow x=3\)

c) \(\left(x-1\right)^2=\sqrt{\left(-\frac{9}{16}\right)^2}\Leftrightarrow\left(x-1\right)^2=\frac{9}{16}\Leftrightarrow\left(x-1\right)^2=\left(\frac{3}{4}\right)^2\)

TH1: x - 1 = 3/4 => x = 3/4 + 1  => x = 7/4

Th2: x - 1 = - 3/4 => x  = -3/4 +1 => x = 1/4

d) \(\Leftrightarrow\sqrt{x^2+2}=12-5=7\Leftrightarrow x^2+2=7^2\Leftrightarrow x^2=49-2\Leftrightarrow x^2=47\)

\(x=\sqrt{47};x=-\sqrt{47}\)