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Bài 3:
a: \(x=\dfrac{1}{4}-\dfrac{2}{13}=\dfrac{13}{52}-\dfrac{8}{52}=\dfrac{5}{52}\)
b: \(\dfrac{x}{3}=\dfrac{2}{3}-\dfrac{1}{7}\)
nên \(x\cdot\dfrac{1}{3}=\dfrac{14}{21}-\dfrac{3}{21}=\dfrac{11}{21}\)
hay \(x=\dfrac{11}{21}:\dfrac{1}{3}=\dfrac{11}{7}\)
a: \(=\dfrac{48}{36}+\dfrac{7}{5}\cdot\dfrac{20}{21}=\dfrac{4}{3}+\dfrac{4}{3}=\dfrac{8}{3}\)
b: \(=\dfrac{7}{3}-\dfrac{1}{3}\cdot\left[-\dfrac{3}{2}+\dfrac{2}{3}+\dfrac{4}{10}\cdot5\right]\)
\(=\dfrac{7}{3}-\dfrac{1}{3}\cdot\dfrac{7}{6}=\dfrac{7}{3}-\dfrac{7}{18}=\dfrac{42-7}{18}=\dfrac{35}{18}\)
c: \(=\left(29+\dfrac{1}{4}\right):\dfrac{9}{4}=\dfrac{117}{4}\cdot\dfrac{4}{9}=\dfrac{117}{9}=13\)
d: \(=\left(4-\dfrac{4}{5}\right)\cdot\dfrac{11}{8}-\dfrac{8}{5}\cdot4\)
\(=\dfrac{16}{5}\cdot\dfrac{11}{8}-\dfrac{32}{5}\)
\(=\dfrac{22}{5}-\dfrac{32}{5}=-\dfrac{10}{5}=-2\)
12 : {390 : [500 - (125 + 35 . 7)]}
= 12 : {390 : [500 - (125 + 245)]}
= 12 : [390 : (500 - 370)]
= 12 : (390 : 130)
= 12 : 3
= 4
12 : { 390 : [ 500 - ( 125 + 35. 7 ) ] }
= 12 : { 390 : [ 500 - ( 125 + 245 ) ] }
= 12 : { 390 : [ 500 - 370 ] }
= 12 : { 390 : 130 }
= 12 : 3
= 4
a, x(x+1)=0
=>\(\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
b,(x-1)(x+3)=0
=>\(\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
- 12 + 3(-x + 7) = -12 -3x - 21 = -3x - 33 = -3(x + 11)
\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\left(2x-15\right)^3\left(2x-15-1\right)\left(2x-15+1\right)=0\)
\(\Rightarrow\left(2x-15\right)^3\left(2x-16\right)\left(2x-14\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=15\\2x=16\\2x=14\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
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