a) \(64^x:16^x=256\)

\(\Rightarrow\left(2^6\right)^x:\left(2^4\right)^x=2^8\)

\(\Rightarrow2^{6x}:2^{4x}=2^8\)

\(\Rightarrow2^{6x-4x}=2^8\)

\(\Rightarrow2^{2x}=2^8\)

\(\Rightarrow2x=8\)

\(\Rightarrow x=4\)

b) \(\dfrac{-2401}{7^x}=-7\)

\(\Rightarrow\dfrac{-7^4}{7^x}=-7\)

\(\Rightarrow-7^{4-x}=-7\)

\(\Rightarrow7^{4-x}=7\)

\(\Rightarrow4-x=1\)

\(\Rightarrow x=4-1\)

\(\Rightarrow x=3\)

c) \(\dfrac{64}{\left(-4\right)^x}=-256\)

\(\Rightarrow\left(-4\right)^x=\dfrac{64}{-256}\)

\(\Rightarrow\left(-4\right)^x=-4\)

\(\Rightarrow\left(-4\right)^x=\left(-4\right)^1\)

\(\Rightarrow x=1\)

\(a) 64^x:16^x=256\\\Rightarrow (64:16)^x=256\\\Rightarrow 4^x=4^4\\\Rightarrow x=4\\---\)

\(b,\dfrac{-2401}{7^x}=-7\)

\(\Rightarrow7^x=-2401:\left(-7\right)\)

\(\Rightarrow7^x=343\)

\(\Rightarrow7^x=7^3\)

\(\Rightarrow x=3\)

\(c,\dfrac{64}{\left(-4\right)^x}=-256\)

\(\Rightarrow\left(-4\right)^x=64:\left(-256\right)\)

\(\Rightarrow\left(-4\right)^x=-\dfrac{1}{4}\)

\(\Rightarrow\left(-4\right)^x=\left(-4\right)^{-1}\)

\(\Rightarrow x=-1\)

#\(Toru\)

a) \(\left|x\right|-\frac{7}{6}=\frac{9}{15}\)

=> \(\left|x\right|=\frac{9}{15}+\frac{7}{6}=\frac{53}{30}\)

=> \(\orbr{\begin{cases}x=\frac{53}{30}\\x=-\frac{53}{30}\end{cases}}\)

b) \(\left|x-\frac{4}{3}\right|=\frac{1}{6}\)

=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{1}{6}\\x-\frac{4}{3}=-\frac{1}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{7}{6}\end{cases}}\)

c) \(\left|x-\frac{4}{3}\right|-\frac{1}{3}=\frac{1}{2}\)

=> \(\left|x-\frac{4}{3}\right|=\frac{1}{2}+\frac{1}{3}\)

=> \(\left|x-\frac{4}{3}\right|=\frac{5}{6}\)

=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{5}{6}\\x-\frac{4}{3}=-\frac{5}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{6}\\x=\frac{1}{2}\end{cases}}\)

d) \(\frac{8}{3}-\left|\frac{7}{9}-x\right|=-\frac{1}{5}\)

=> \(\left|\frac{7}{9}-x\right|=\frac{43}{15}\)

=> \(\orbr{\begin{cases}\frac{7}{9}-x=\frac{43}{15}\\\frac{7}{9}-x=-\frac{43}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{94}{45}\\x=\frac{164}{45}\end{cases}}\)

e) \(\left|x-\left(\frac{1}{4}\right)^2\right|-\frac{25}{64}=0\)

=> \(\left|x-\frac{1}{16}\right|=\frac{25}{64}\)

=> \(\orbr{\begin{cases}x-\frac{1}{16}=\frac{25}{64}\\x-\frac{1}{16}=-\frac{25}{64}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{29}{64}\\x=-\frac{21}{64}\end{cases}}\)

f) \(\left(x-\frac{1}{4}\right)^2+\frac{17}{64}=\frac{21}{32}\)

=> \(\left(x-\frac{1}{4}\right)^2=\frac{25}{64}\)

=> \(\left(x-\frac{1}{4}\right)^2=\left(\frac{5}{8}\right)^2\)

=> \(\orbr{\begin{cases}x-\frac{1}{4}=\frac{5}{8}\\x-\frac{1}{4}=-\frac{5}{8}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{3}{8}\end{cases}}\)

\(9^7.81:9^5=9^7.9^2:9^5=9^{7+2-5}=9^4\\ x^{12}:x.x^8=x^{12-1+8}=x^{19}\\ 16.2^4:8=2^4.2^4:2^3=2^{4+4-3}=2^5\\ 64.4^5:16=4^3.4^5:4^2=4^{3+5-2}=4^6\\ 3^{12}.3:3^8=3^{12+1-8}=3^5\\ 7^9.7^{12}:2015^0=7^{9+12}:1=7^{19}\)

5 + ( x + 27 ) = 64

a) \(\Rightarrow x+27=59\Rightarrow x=32\)

b) \(\Rightarrow x-2=39\Rightarrow x=41\)

c) \(\Rightarrow x+5=-322\Rightarrow x=-327\)

d) \(\Rightarrow5x=35\Rightarrow x=7\)

e) \(\Rightarrow4\left(x-5\right)=56\Rightarrow x-5=14\Rightarrow x=19\)

f) \(\Rightarrow15+x=37\Rightarrow x=22\)

g) \(\Rightarrow7\left(13-x\right)=35\Rightarrow13-x=5\Rightarrow x=8\)

h) \(\Rightarrow10\left(x+1\right)=100\Rightarrow x+1=10\Rightarrow x=9\)

\(3\left(x+2\right)^2+\left(2x-3\right)^2-7\left(x-4\right)\left(x+4\right)=64\)

\(\Leftrightarrow3\left(x^2+4x+4\right)+\left(4x^2-12x+9\right)-7\left(x^2-16\right)=64\)

\(\Leftrightarrow3x^2+12x+12+4x^2-12x+9-7x^2+112=64\)

\(\Leftrightarrow12+9+112=64\)(vô lí)

Vậy pt vô nghiệm

 TL:

\(\Leftrightarrow3\left(x^2+4x+4\right)+4x^2-6x+9-7x^2+112=64\)

\(\Leftrightarrow6x+133=64\)  

\(\Leftrightarrow6x=-69\) 

\(\Leftrightarrow x=\frac{-23}{2}\) 

Vậy....

Nguyễn Văn Tuấn AnhSai từ dòng đầu tiên

\(\left(2x-3\right)^2=4x^2-12x+9\ne4x^2-6x+9\)