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a)\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2-\left(\frac{3}{5}\right)^2=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}+\frac{3}{5}\right)\left(2x+\frac{3}{5}-\frac{3}{5}\right)=0\)
\(\Leftrightarrow\left(2x+\frac{6}{5}\right).2x=0\)
\(\Leftrightarrow\left[\begin{matrix}x=-\frac{3}{5}\\x=0\end{matrix}\right.\)
Kết luận thôi
b) \(3.\left(3x-\frac{1}{2}\right)^3+\frac{1}{19}=0\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{19}:3\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{57}\)
\(\Leftrightarrow3x-\frac{1}{2}=\sqrt[3]{-\frac{1}{57}}\)
\(\Leftrightarrow3x=\sqrt[3]{-\frac{1}{57}}+\frac{1}{2}\)
\(\Leftrightarrow x=\frac{\sqrt[3]{-\frac{1}{57}}+\frac{1}{2}}{3}\)
Số hơi to
Kết luận thôi
Ta có: \(\overline{aaa}=1+2+...+n=\frac{n\left(n+1\right)}{2}\)
\(\Rightarrow\frac{n\left(n+1\right)}{2}=111a\Rightarrow n\left(n+1\right)=2.111a=2.3.37.a\)
Vì n(n+1) chia hết cho 37 nên một trong hai số chia hết cho 37
Mà \(\frac{n\left(n+1\right)}{2}\) là số có ba chữ số nên n và n+1 nhỏ hơn 74 => n=37 hoặc n+1=37
Nếu n=37 thì n+1=38 => \(\overline{aaa}=\frac{n\left(n+1\right)}{2}=\frac{37.38}{2}=703\) (loại)
Nếu n+1=37 thì n=36 => \(\overline{aaa}=\frac{n\left(n+1\right)}{2}=\frac{36.37}{2}=666\) (thỏa mãn)
Vậy n=36 và aaa = 666
d) \(\frac{x}{-9}=\left(\frac{2}{6}\right)^2\)
\(\Rightarrow\frac{x}{-9}=\frac{2}{6}.\frac{2}{6}\)
\(\Rightarrow\frac{x}{-9}=\frac{4}{36}\)
\(\Rightarrow\frac{x}{-9}=\frac{1}{9}\)
\(\Rightarrow\frac{-x}{9}=\frac{1}{9}\)
\(\Rightarrow-x=1\)
\(\Rightarrow x=1\)
e) \(\frac{a}{b}+\frac{3}{6}=0\)
\(\Rightarrow\frac{a}{b}=0-\frac{3}{6}\)
\(\Rightarrow\frac{a}{b}=0-\frac{1}{2}\)
\(\Rightarrow\frac{a}{b}=\frac{-1}{2}\)
\(\Rightarrow a=-1;b=2\)
Bài làm:
Ta có:
a) \(x^{n+1}\div5=5^n\)
\(\Leftrightarrow x^{n+1}=5^n.5\)
\(\Leftrightarrow x^{n+1}=5^{n+1}\)
\(\Rightarrow x=5\)
b) \(x^n.9=3^{n+2}\)
\(\Leftrightarrow x^n.3^2=3^{n+2}\)
\(\Leftrightarrow x^n=3^{n+2}\div3^2\)
\(\Leftrightarrow x^n=3^n\)
\(\Rightarrow x=3\)
- Chuyển vế phương trình:
- Khai căn hai vế:
- Giải từng trường hợp:
- Trường hợp 1: \(2x + \frac{1}{2} = \frac{1}{3}\)
- Trường hợp 2: \(2x + \frac{1}{2} = -\frac{1}{3}\)
Kết luận: Tập nghiệm \(x \in \{-\frac{1}{12}, -\frac{5}{12}\}\). [1]\(\frac{1}{9}-(2x+\frac{1}{2})^{2}=0\)
\((2x+\frac{1}{2})^{2}=\frac{1}{9}\)
\(2x+\frac{1}{2}=\sqrt{\frac{1}{9}}\quad \text{hoc}\quad 2x+\frac{1}{2}=-\sqrt{\frac{1}{9}}\)
\(2x+\frac{1}{2}=\frac{1}{3}\quad \text{hoc}\quad 2x+\frac{1}{2}=-\frac{1}{3}\)
\(2x=\frac{1}{3}-\frac{1}{2}\)
\(2x=\frac{2}{6}-\frac{3}{6}=-\frac{1}{6}\)
\(x=-\frac{1}{6}\div 2=-\frac{1}{12}\)
\(2x=-\frac{1}{3}-\frac{1}{2}\)
\(2x=-\frac{2}{6}-\frac{3}{6}=-\frac{5}{6}\)
\(x=-\frac{5}{6}\div 2=-\frac{5}{12}\)
-Với \(2x + \frac{1}{2} = \frac{1}{3}\)
\(\Rightarrow 2x = -\frac{1}{6}\)
\(x = -\frac{1}{12}\)
-Với \(2x + \frac{1}{2} = -\frac{1}{3}\)
Vậy \(x\in \left\{-\frac{1}{12};-\frac{5}{12}\right\}\)\(2x = -\frac{5}{6}\)
\(x = -\frac{5}{12}\)
\(\frac19-\left(2x+\frac12\right)^2=0\)
\(\lrArr\left(2x+\frac12\right)^2=\frac19\)
\(\lrArr\left[\begin{array}{l}2x+\frac12=\frac13\\ 2x+\frac12=\frac{-1}{3}\end{array}\right.\)
\(\lrArr\left[\begin{array}{l}2x=\frac{-1}{6}\\ 2x=\frac{-5}{6}\end{array}\right.\)
\(\lrArr\left[\begin{array}{l}x=\frac{-1}{12}\\ x=\frac{-5}{12}\end{array}\right.\)
\(\left(2x+\frac{1}{2}\right)^{2}=\frac{1}{9}\)
\(2x + \frac{1}{2} = \pm \frac{1}{3}\)
Với \(2x + \frac{1}{2} = \frac{1}{3}\)
\(\Rightarrow 2x = -\frac{1}{6}\)
\(x = -\frac{1}{12}\)
Với \(2x + \frac{1}{2} = -\frac{1}{3}\)
\(2x = -\frac{5}{6}\)
\(x = -\frac{5}{12}\)
Vậy \(x\in \left\{-\frac{1}{12};-\frac{5}{12}\right\}\)\(\frac19-\left(2x+\frac12\right)^2=0\)
\(\left(2x+\frac12\right)^2=\frac19\)
\(Th_1:2x+\frac12=\left(\frac13\right)^2\)
\(\Rightarrow2x=-\frac16\)
\(\Rightarrow x=-\frac{1}{12}\)
\(Th_2:2x+\frac12=\left(-\frac13\right)^2\)
\(\Rightarrow2x=-\frac56\)
\(\Rightarrow x=-\frac{5}{12}\)
Vậy \(x\in\left\lbrace-\frac{1}{12};-\frac{5}{12}\right\rbrace\)
\[\]
Bài giải:
\(\frac19-\left(2x+\frac12\right)^2=0\)
\((2x + \frac{1}{2})^2 = \frac{1}{9}\)
\((2x + \frac{1}{2})^2 = (\frac{1}{3})^2\)
TH1:
\(2x + \frac{1}{2} = \frac{1}{3}\)
\(2x = \frac{1}{3} - \frac{1}{2}\)
\(2x = \frac{2}{6} - \frac{3}{6}\)
\(2x = -\frac{1}{6}\)
\(x = -\frac{1}{6} : 2\)
\(x = -\frac{1}{12}\)
TH2:
\(2x + \frac{1}{2} = -\frac{1}{3}\)
\(2x = -\frac{1}{3} - \frac{1}{2}\)
\(2x = -\frac{2}{6} - \frac{3}{6}\)
\(2x = -\frac{5}{6}\)
\(x = -\frac{5}{6} : 2\)
\(x = -\frac{5}{12}\)
Vậy \(x \in \{-\frac{1}{12}; -\frac{5}{12}\}\) .