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b) Ta có : \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\Rightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}\left(x-\frac{1}{3}\right)^2=\left(\frac{1}{2}\right)^2\\\left(x-\frac{1}{3}\right)^2=\left(-\frac{1}{2}\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{2}\\x-\frac{1}{3}=-\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=-\frac{1}{6}\end{cases}}\)
b) \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{4}\\x-\frac{1}{3}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{12}\\x=\frac{1}{12}\end{cases}}\)
d) \(\frac{x+5}{2}=\frac{8}{x+5}\)
\(\Rightarrow\left(x+5\right)^2=16\)
\(\Rightarrow\orbr{\begin{cases}x+5=16\\x+5=-16\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=-21\end{cases}}}\)
a)
\(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-50\right)=255\\ x+x-1+x-2+...+x-50=255\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+50\right)\\ 51x-1275=255\\ 51x=1530\\ x=30\)
e)
\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\\ x+x+1+x+2+...+x+30=1240\\ \left(x+x+x+...+x\right)+\left(1+2+3+...+30\right)=1240\\ 31x+465=1240\\ 31x=775\\ x=25\)
f)
\(\left(x-1\right)+\left(x-2\right)+...+\left(x-19\right)+\left(x-20\right)=-610\\ x-1+x-2+...+x-19+x-20=-610\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+20\right)=-610\\ 20x-210=-610\\ 20x=-400\\ x=-20\)
a/ \(\left(x-23\right):14+25=42-1^{2002}\)
\(\Leftrightarrow\left(x-23\right):14+25=42-1\)
\(\Leftrightarrow\left(x-23\right):14+25=41\)
\(\Leftrightarrow\left(x-23\right):14=16\)
\(\Leftrightarrow x-23=192\)
\(\Leftrightarrow x=215\)
Vậy ....
b/ \(2^3.x+2002^0.x=995-15:3\)
\(\Leftrightarrow8.x+1.x=995-5\)
\(\Leftrightarrow x\left(8+1\right)=990\)
\(\Leftrightarrow9x=990\)
\(\Leftrightarrow x=110\)
Vậy ...
b/ \(x+2x+3x+.........+9x=459-3^2\)
\(\Leftrightarrow x\left(1+2+......+9\right)=459-9\)
\(\Leftrightarrow x.\dfrac{\left(9+1\right).\left[\left(9-1\right):1+1\right]}{2}=450\)
\(\Leftrightarrow x.45=450\)
\(\Leftrightarrow x=10\)
Vậy ...
\(3^x+4^2=19^0.\left(19^3.19^2\right)-3.1^{2005}\)
\(3^x+16=1.19^5-3.1\)
\(3^x+16=2476099-3\)
\(3^x+16=2476096\)
\(3^x=2476096-16\)
\(3^x=2476080\)
\(3^x=2^4.3^2.5.19.181\)
\(\Rightarrow x\in\left\{\varnothing\right\}\)
Vậy,...