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a) \(\frac{x}{27}=\frac{-2}{3,6}\Leftrightarrow x=\frac{-2.27}{3,6}=-15\)
b) \(\frac{4\frac{1}{4}}{2\frac{7}{8}}=\frac{x}{1,61}\Leftrightarrow\frac{4,25}{2,875}=\frac{x}{1,61}\Leftrightarrow x=\frac{4,25.1,61}{2,875}=2,38\)
a) \(\frac{x}{27}=-\frac{2}{3,6}\Rightarrow x=\frac{-2.27}{3,6}=-15\)
b) \(\frac{4\frac{1}{4}}{2\frac{7}{8}}=\frac{x}{1,61}\Rightarrow\frac{\frac{17}{4}}{\frac{23}{8}}=\frac{x}{1,61}\Leftrightarrow\frac{17}{4}\cdot\frac{8}{23}=\frac{x}{1,61}\)
\(\frac{34}{23}=\frac{x}{1,61}\Rightarrow x=\frac{1,61.34}{23}=2,38\)
mk thấy câu b) hơi khó ,mk lam giup bn
b) x/3 = y/3 = z/5
hay 2x/6 = 3y/9 z/5
ta có; ( 2x- 3y +z) / ( 6-9+5) = 6/2 =3
x = 3.2 =6
y = 3.2 =6
z = 5.2 =10
\(\frac{x}{11,9}=\frac{1,5}{3,5}\Rightarrow x=\frac{1,5.11,9}{3,5}\Rightarrow x=5,1\)
\(\frac{x}{0,2}=\frac{80}{x}\Rightarrow x^2=80.0,2\Rightarrow x^2=16\Rightarrow x=4;-4\)
\(\frac{2}{3}x:\frac{1}{5}=\frac{4}{3}:\frac{1}{4}\Rightarrow\frac{2}{3}x:\frac{1}{5}=\frac{16}{3}\Rightarrow\frac{2}{3}x=\frac{16}{15}\Rightarrow x=\frac{8}{5}\)
\(3:\frac{2}{5}x=\frac{1}{0,001}\Rightarrow\frac{2}{5}x=\frac{3}{1000}\Rightarrow x=\frac{3}{400}\)
\(\frac{x}{11,9}=\frac{1,5}{3,5}\Leftrightarrow x=\frac{11,9.1,5}{3,5}=5,1\)
Tương tự
a) \(\left(2x-1\right)^3=-27\)
\(\left(2x-1\right)^3=-3^3\)
\(2x-1=-3\)
\(2x=-3+1\)
\(2x=-2\)
\(x=-2:2\)
\(x=-1\)
b) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
*\(\Rightarrow2x-1=1\)
\(2x=1+1\)
\(2x=2\)
\(x=2:2\)
\(x=1\)
*\(\Rightarrow2x-1=-1\)
\(2x=-1+1\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
*\(\Rightarrow2x-1=0\)
\(2x=0+1\)
\(2x=1\)
\(x=1:2\)
\(x=\frac{1}{2}\)
Vậy \(x=\left\{1;0;\frac{1}{2}\right\}\)
\(\dfrac{2x+1}{-27}\)=\(\dfrac{-3}{2x+1}\) (1)
ĐKXĐ: x khác \(\dfrac{-1}{2}\)
(1) <=> (2x+1)2=81
=>\(\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x=4\left(tm\right)\\x=-5\left(tm\right)\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S={4;-5}
\(\dfrac{2x+1}{-27}=\dfrac{-3}{2x+1}\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1\right)=\left(-3\right)\left(-27\right)\)
\(\Leftrightarrow\left(2x+1\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)