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a) \(\left(3x-\frac{1}{2}\right)^2=\frac{1}{121}=\left(\frac{1}{11}\right)^2\)
=> \(\orbr{\begin{cases}3x-\frac{1}{2}=\frac{1}{11}\\3x-\frac{1}{2}=-\frac{1}{11}\end{cases}}\)
=> \(\orbr{\begin{cases}3x=\frac{13}{22}\\3x=\frac{9}{22}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{13}{66}\\x=\frac{3}{22}\end{cases}}\)
b) \(\left(5-3x\right)^3=\left(-\frac{1}{27}\right)=\left(-\frac{1}{3}\right)^3\)
=> \(5-3x=-\frac{1}{3}\)
=> \(3x=\frac{16}{3}\)
=> \(x=\frac{16}{3}:3=\frac{16}{9}\)
c) 5x + 5x+2 = 650
=> 5x + 5x . 52 = 650
=> 5x(1 + 52) = 650
=> 5x . 26 = 650
=> 5x = 25
=> 5x = 52 => x = 2
d) 3x-1 + 5.3x-1 = 126
=> (1 + 5).3x-1 = 126
=> 6.3x-1 = 126
=> 3x-1 = 21
=> 3x-1 =3.7
tới đây là không xử lí được x luôn :)
a,\(\left(3x-\frac{1}{2}\right)^2=\frac{1}{121}=\left(\frac{1}{11}\right)^2=\left(-\frac{1}{11}\right)^2\)
\(< =>\orbr{\begin{cases}3x-\frac{1}{2}=\frac{1}{11}\\3x-\frac{1}{2}=-\frac{1}{11}\end{cases}}< =>\orbr{\begin{cases}3x=\frac{1}{11}+\frac{1}{2}\\3x=-\frac{1}{11}+\frac{1}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}3x=\frac{2}{22}+\frac{11}{22}=\frac{13}{22}\\3x=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{13}{22}:3=\frac{13}{22}.\frac{1}{3}=\frac{13}{66}\\x=\frac{9}{22}:3=\frac{9}{22}.\frac{1}{3}=\frac{9}{66}=\frac{3}{22}\end{cases}}\)
b,\(\left(5-3x\right)^2=-\frac{1}{27}=\left(-\frac{1}{3}\right)^3\)
\(< =>5-3x=-\frac{1}{3}< =>-3x=-\frac{1}{3}-5=-\frac{16}{3}\)
\(< =>3x=\frac{16}{3}< =>x=\frac{16}{3}:3=\frac{16}{3}.\frac{1}{3}=\frac{16}{9}\)
c,\(5^x+5^{x+2}=650< =>5^x+5^x.25=650\)
\(< =>5^x\left(25+1\right)=5^x=\frac{650}{36}=25< =>x=2\)
bạn nào giúp câu d
a) \(\frac{1}{9}=\frac{x}{27}\)
\(\Rightarrow x=\frac{1}{9}\cdot27\)
\(\Rightarrow x=3\)
b) \(\frac{4}{x}=\frac{8}{6}\)
\(\Rightarrow x=4:\frac{8}{6}\)
\(\Rightarrow x=3\)
c) \(\frac{x}{3}-\frac{1}{2}=\frac{1}{5}\)
\(\Rightarrow\frac{x}{3}=\frac{1}{5}+\frac{1}{2}\)
\(\Rightarrow x=\frac{7}{10}\cdot3\)
\(\Rightarrow x=\frac{21}{10}=2,1\)
\(a,\frac{1}{9}\)=\(\frac{3}{27}\)
\(b,\frac{4}{3}\)=\(\frac{8}{6}\)
\(c,\frac{x}{3}\)-\(\frac{1}{2}=\frac{1}{5}\)
\(\frac{x}{3}=\frac{1}{5}+\frac{1}{2}\)
\(\frac{x}{3}=\frac{7}{10}\)
\(\)
a) x - 2/3 = -5/12
=> x = -5/12 + 2/3
=> x = 1/4
b) \(\frac{x+5}{3}=\frac{5}{9}\)
=> \(\frac{3\left(x+5\right)}{9}=\frac{5}{9}\)
=> 3(x + 5) = 5
=> x + 5 = 5 : 3
=> x + 5 = 5/3
=> x = 5/3 - 5
=> x = 20/3
c) \(\frac{x-2}{27}=\frac{3}{x-2}\)
=> (x - 2)(x - 2) = 27 . 3
=> (x - 2)2 = 81
=> (x - 2)2 = 92
=> \(\orbr{\begin{cases}x-2=9\\x-2=-9\end{cases}}\)
=> \(\orbr{\begin{cases}x=11\\x=-7\end{cases}}\)
Vậy ...
c) Giải:
\(\frac{1-x}{3}=\frac{27}{1-x}\\ \Leftrightarrow\left(1-x\right)\left(1-x\right)=27.3\\ \Rightarrow\left(1-x\right)^2=81\\ \Rightarrow\left(1-x\right)^2=\pm9^2\\ 1-x=\pm9\)
+) 1-x=9
x=1-9
x=-8
+) 1-x=-9
x=1-(-9)
x=10
Vậy \(x\in\left\{-8;10\right\}\)
Chúc bạn học tốt!
\(\Rightarrow\left(x-1\right)\left(1-x\right)=3.27\)
\(\Rightarrow\left(x-1\right)\left(-1\right)\left(x-1\right)=81\)
\(\Rightarrow\left(x-1\right)^2=-81\)
Mặt khác \(\left(x-1\right)^2\ge0\) với mọi x
=> \(x\in\varnothing\)