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\(\frac{x^{2n+1}}{x^{2n-1}}=49\)
\(\frac{x^2.x^{2n-1}}{x^{2n-1}}=49\)
\(x^2=49\)
x = 7 hoặc x = -7
Vậy ...
a)\(\dfrac{1}{3}^x=\dfrac{1}{729}\)
\(\dfrac{1}{3}^x=\dfrac{1}{3}^6\)
Suy ra x=6
vậy x=6
b)\(\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^x=\dfrac{1}{36}\)
\(\left(-\dfrac{1}{6}\right)^x\)=\(\dfrac{1}{6}^2\)
Do đó x=2
Vậy x=2
c)\(\dfrac{25}{5}^x=\dfrac{1}{125}\)
\(5^x=5^{-3}\)
Suy ra x=-3
Vậy x=-3
d)\(7^{2n-\dfrac{1}{49}}=-343\)
\(-7^{2n-\dfrac{1}{49}}=-7^3\)
Do đó \(2n-\dfrac{1}{49}=3\)
2n=\(\dfrac{148}{49}\)
n=\(\dfrac{74}{49}\)
vậy n=\(\dfrac{74}{49}\)
chúc bạn học tốt ạ
Ta có: \(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\Leftrightarrow\left(5x+1\right)^2=\left(\dfrac{6}{7}\right)^2\)
\(\Leftrightarrow5x+1=\dfrac{6}{7}\)
\(\Leftrightarrow5x=\dfrac{13}{7}\)
\(\Leftrightarrow x=\dfrac{13}{35}\)
Vậy \(x=\dfrac{13}{35}\)
(5x + 1)2 = \(\dfrac{36}{49}\)
<=> (5x + 1)2 = (\(\dfrac{6}{7}\))2
<=> 5x + 1 = \(\dfrac{6}{7}\)
<=> 5x = \(-\dfrac{1}{7}\)
<=> x = \(-\dfrac{1}{35}\)
@Võ Ngọc Tường Vy
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\Rightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2=\left(\frac{-6}{7}\right)^2\)
\(\Rightarrow\hept{\begin{cases}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{cases}}\Rightarrow\hept{\begin{cases}5x=\frac{-1}{7}\\5x=\frac{-13}{7}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{cases}}}\)
b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}^2\right)^3\)
\(\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\)
\(x-\frac{2}{9}=\frac{4}{9}\)
\(x=\frac{4}{9}+\frac{2}{9}\)
\(x=\frac{6}{9}=\frac{2}{3}\)
\(a.\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\frac{6^2}{7^2}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(\Rightarrow5x+1=\frac{6}{7}\)
\(\Rightarrow5x=\frac{6}{7}-1\)
\(\Rightarrow5x=-\frac{1}{7}\)
\(\Rightarrow x=-\frac{1}{7}:5\)
\(\Rightarrow x=-\frac{1}{35}\)
Vậy \(x=-\frac{1}{35}\)
\(b\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right).^6\)
\(\left(x-\frac{2}{9}\right)^3=\frac{\left(2^2\right)^3}{\left(3^2\right)^3}\)
\(\left(x-\frac{2}{9}\right)^3=\frac{4^3}{9^3}\)
\(\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\)
\(\Rightarrow x-\frac{2}{9}=\frac{4}{9}\)
\(\Rightarrow x=\frac{4}{9}+\frac{2}{9}\)
\(\Rightarrow x=\frac{6}{9}=\frac{2}{3}\)
Vậy \(x=\frac{2}{3}\)
\(\frac{2}{9}.5x+\frac{1}{2}-\frac{1}{18}=\frac{5}{36}\)
\(\frac{2}{9}.5x=\frac{5}{36}+\frac{1}{18}-\frac{1}{2}\)
\(\frac{2}{9}.5x=\frac{-11}{36}\)
\(5x=\frac{-11}{36}:\frac{2}{9}\)
\(5x=\frac{-11}{8}\)
\(x=\frac{-11}{8}:5\)
\(x=\frac{-11}{40}\)
Chú ý dấu chấm là dâu nhân nha