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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)\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
=\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\frac{3^2}{5^2}\)
=\(x+\frac{1}{5}=\frac{3}{5}\)
\(x=\frac{2}{5}\)
b)\(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=\frac{24}{27}\)
=\(x=-\frac{35}{27}\)
g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
ủng hộ mk nha bn
bạn trả lời giúp
tk cho mk nhé mình đang bị âm nek
trả lời giúp
\(\left|2x-\frac{3}{4}\right|=\frac{17}{2}-\frac{-7}{4}\)
\(\left|2x-\frac{3}{4}\right|=\frac{34}{4}+\frac{7}{4}\)
\(\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)
\(TH1\):]
\(2x-\frac{3}{4}=\frac{41}{4}\)
\(2x=\frac{41}{4}+\frac{3}{4}\)
\(2x=11\)
\(x=11:2\)
\(TH2\)
\(2x-\frac{3}{4}=\frac{-41}{4}\)
\(2x=\frac{-41}{4}+\frac{3}{4}\)
\(2x=\frac{19}{2}\)
\(x=\frac{19}{2}:2\)
\(x=\frac{19}{4}\)
\(\frac{17}{2}-|2x-\frac{3}{4}=\frac{-7}{4}\)
\(|2x-\frac{3}{4}|=\frac{17}{2}-\frac{-7}{4}\)
\(|2x-\frac{3}{4}|=\frac{41}{4} \)
\(\rightarrow 2x-\frac{3}{4}=\frac{41}{4} hoặc 2x-\frac{3}{4}=\frac{-41}{4}\)
\(2x=\frac{41}{4}+\frac{3}{4}\) \(2x=\frac{-41}{4}+\frac{3}{4} \)
\(2x=\frac{44}{4}=11 \) \(2x=\frac{19}{2}\)
\(x=11:2=5,5\) \(x=\frac{19}{2}:2=\frac{19}{2}.\frac{1}{2}=\frac{19}{4}\)
\(Vậy...\)
\((x+\frac{1}{5})^2+\frac{17}{25}=\frac{26}{25}\)
\((x+\frac{1}{5})^2=\frac{26}{25}-\frac{17}{25}\)
\((x+\frac{1}{5})^2=\frac{9}{25}\)
\(\rightarrow(x+\frac{1}{5})^2=(\frac{3}{5})^2\)
\(\rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\rightarrow x = \frac{3}{5}-\frac{1}{5}\)
\(\rightarrow x=\frac{2}{5}\)
\(Vậy x = \frac{2}{5}\)
\(-1\frac{5}{27}-(3x-\frac{7}{9})^3=\frac{-24}{27}\)
\((3x-\frac{7}{9})^3=-1\frac{5}{27}-\frac{-24}{27}\)
\((3x-\frac{7}{9})^3=\frac{-32}{27}+\frac{24}{27}\)
\((3x-\frac{7}{9})^3=\frac{-8}{27}\)
\(\rightarrow(3x-\frac{7}{9})^3=(\frac{-2}{3})^3\)
\(\rightarrow 3x-\frac{7}{9}=\frac{-2}{3}\)
\(\rightarrow 3x=\frac{-2}{3}+\frac{7}{9}\)
\(\rightarrow 3x=\frac{1}{9}\)
\(\rightarrow x=\frac{1}{9}:3=\frac{1}{9}.\frac{1}{3}=\frac{1}{27}\)
\(Vậy x = \frac{1}{27}\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(< =>\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)
\(< =>\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(< =>\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(< =>\orbr{\begin{cases}\\\end{cases}}x+\frac{1}{5}=\frac{3}{5}\)
\(< =>\orbr{\begin{cases}\\\end{cases}x+\frac{1}{5}=\frac{-3}{5}}\)
\(< =>\orbr{\begin{cases}\\\end{cases}x=\frac{3}{5}-\frac{1}{5}}\)
\(< =>\orbr{\begin{cases}\\\end{cases}x=\frac{-3}{5}-\frac{1}{5}}\)
\(< =>\orbr{\begin{cases}\\\end{cases}x=\frac{2}{5}}\)
\( < =>\orbr{\begin{cases}\\\end{cases}x=\frac{-4}{5}}\)
mk chuc ban hoc tot nhe :))