Tìm x, biết: \(\frac{2x-1}{-27}=\frac{3}{1-2x}\)
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Ta có: \(\frac{x-1}{5}=\frac{y-2}{3}=\frac{z-1}{4}\) \(\Leftrightarrow\frac{2x-2}{10}=\frac{3y-6}{9}=\frac{2z-2}{8}\)
Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{2x-2}{10}=\frac{3y-6}{9}=\frac{2z-2}{8}=\frac{2x-2-3y+6-2z+2}{10-9-8}=\frac{-27+6}{-7}=\frac{-21}{-7}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{x-1}{5}=3\\\frac{y-2}{3}=3\\\frac{z-1}{4}=3\end{cases}\Rightarrow}\hept{\begin{cases}x-1=15\\y-2=9\\z-1=12\end{cases}\Rightarrow}\hept{\begin{cases}x=16\\y=11\\z=13\end{cases}}\)
Vậy...
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
a, \(\frac{1}{\left(2x-3\right)^2}=9\Leftrightarrow\left(\frac{1}{2x-3}\right)^2=3^2\Leftrightarrow\frac{1}{2x-3}=3\Leftrightarrow1=6x-9\Leftrightarrow x=\frac{5}{3}\)
b, \(\frac{1}{\left(x-1\right)^3}=-\frac{1}{27}\Leftrightarrow\left(\frac{1}{x-1}\right)^3=\left(\frac{-1}{3}\right)^3\Leftrightarrow\frac{1}{x-1}=\frac{1}{-3}\Leftrightarrow x-1=-3\Leftrightarrow x=-2\)
c, \(\left(x-1\right)^2=\left(2x-5\right)^2\Leftrightarrow\orbr{\begin{cases}x-1=2x-5\\x-1=-2x+5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}}\)
\(\frac{2x-1}{-27}=\frac{3}{1-2x}\)
\(\Rightarrow\left(2x-1\right).\left(1-2x\right)=3.\left(-27\right)\)
\(-4x^2=-81\) ( chỗ này bn tự phân tích ra nha!)
\(x^2=\frac{-81}{-4}=\frac{81}{4}=\left(\frac{9}{2}\right)^2=\left(-\frac{9}{2}\right)^2\)
=> x = 9/2 hoặc x = -9/2