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\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
\(a,\frac{27}{x+1}=\frac{x+1}{3}\)
\(\Leftrightarrow27.3=\left(x+1\right)\left(x+1\right)\)
\(\Leftrightarrow81=\left(x+1\right)^2\)
\(\Rightarrow\orbr{\begin{cases}\left(x+1\right)^2=9^2\\\left(x+1\right)^2=\left(-9\right)^2\end{cases}\Rightarrow\orbr{\begin{cases}x+1=9\\x+1=-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-10\end{cases}}}\)
Vậy \(x\in\left\{8;-10\right\}\)
a)\(\frac{27}{x+1}=\frac{x+1}{3}\)
Ta có : 27 . 3 = (x + 1)(x + 1)
=> 81 = (x + 1)2
=> 92 = (x + 1)2
=> \(\orbr{\begin{cases}x+1=9\\x+1=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=8\\x=-10\end{cases}}\)
Bài 1 : \(a,\left|x-3,5\right|=7,5\)
\(\Rightarrow\orbr{\begin{cases}x-3,5=7,5\\x-3,5=-7,5\end{cases}}\Rightarrow\orbr{\begin{cases}x=11\\x=-4\end{cases}}\)
\(b,\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{cases}}\)
\(c,3,6-\left|x-0,4\right|=0\)
\(\Rightarrow\left|x-0,4\right|=3,6\)
\(\Rightarrow\orbr{\begin{cases}x-0,4=3,6\\x-0,4=-3,6\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-3,2\end{cases}}\)
\(d,\left|x-\frac{1}{2}\right|-\frac{1}{3}=1\)
\(\Rightarrow\left|x-\frac{1}{2}\right|=\frac{4}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{4}{3}\\x-\frac{1}{2}=-\frac{4}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{6}\\x=-\frac{5}{6}\end{cases}}\)
1) \(x-\left|1\frac{1}{6}\right|=\frac{5}{21}\)
\(\Rightarrow x-\frac{5}{21}=\left|1\frac{1}{6}\right|\)
\(\Rightarrow x-\frac{5}{21}=\frac{7}{6}\)
\(\Rightarrow x=\frac{7}{6}+\frac{5}{21}=\frac{49}{42}+\frac{10}{42}=\frac{59}{42}\)
2) \(x+\left|-1\frac{2}{3}\right|=\left|-\frac{3}{4}\right|\)
\(\Rightarrow x+\left|-1\frac{2}{3}\right|=\frac{3}{4}\)
\(\Rightarrow x-\frac{3}{4}=-\left|-1\frac{2}{3}\right|\)
\(\Rightarrow x-\frac{3}{4}=-1\frac{2}{3}\)
\(\Rightarrow x-\frac{3}{4}=-\frac{5}{3}\)
\(\Rightarrow x=-\frac{5}{3}+\frac{3}{4}=-\frac{11}{12}\)
3) \(\left|x-\frac{1}{3}\right|=\frac{5}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{1}{3}=\frac{5}{2}\\x-\frac{1}{3}=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}+\frac{1}{3}=\frac{17}{6}\\x=-\frac{5}{2}+\frac{1}{3}=-\frac{13}{6}\end{matrix}\right.\)
4) \(\left|x+\frac{2}{3}\right|=0\)
\(\Rightarrow x+\frac{2}{3}=0\)
\(\Rightarrow x=0-\frac{2}{3}=-\frac{2}{3}\)
5) \(\left|x+2\right|=\frac{1}{3}-\frac{1}{5}\)
\(\Rightarrow\left|x+2\right|=\frac{2}{15}\)
\(\Rightarrow\left[{}\begin{matrix}x+2=\frac{2}{15}\\x+2=-\frac{2}{15}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{2}{15}-2=-\frac{28}{15}\\x=-\frac{2}{15}-2=-\frac{32}{15}\end{matrix}\right.\)
6) \(\left|x-4\right|=\frac{1}{5}-\left(\frac{1}{2}-\frac{5}{4}\right)\)
\(\Rightarrow\left|x-4\right|=\frac{19}{20}\)
\(\Rightarrow\left[{}\begin{matrix}x-4=\frac{19}{20}\\x-4=-\frac{19}{20}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{19}{20}+4=\frac{99}{20}\\x=-\frac{19}{20}+4=\frac{61}{20}\end{matrix}\right.\)
7) \(\left|x-\frac{5}{4}\right|=-\frac{1}{3}\)
Vì \(\left|x-\frac{5}{4}\right|\ge0\)
=> Không có giá trị x thỏa mãn với điều kiện trên
b, \(\left(x+1\right)^2=1\)
Vì \(1^2=1=>\left(x+1=1\right)\)
=> x=0
câu a không có máy tính nhác làm
a) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(=>\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(=>x-\frac{1}{2}=\frac{1}{3}\)
\(=>x=\frac{5}{6}\)
b) \(\left(x+1\right)^2=1\)
\(=>\orbr{\begin{cases}x+1=-1\\x+1=1\end{cases}=>\orbr{\begin{cases}x=-2\\x=0\end{cases}}}\)