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\(\dfrac{-12}{25}.\left(\dfrac{3}{4}-x+\dfrac{6}{-11}-\dfrac{5}{6}\right)=0\)
\(\dfrac{3}{4}-x+\dfrac{-6}{11}-\dfrac{5}{6}=0\)
\(\dfrac{3}{4}-x+\dfrac{-91}{66}=0\)
\(\dfrac{3}{4}-x=0-\left(\dfrac{-91}{66}\right)\)
\(\dfrac{3}{4}-x=\dfrac{91}{66}\)
\(x=\dfrac{-83}{132}\)
a)khi x=8 thì y=15
=> 8.15=120
Vậy hai đại lượng x và y tỉ lệ nghịch với nhau theo hệ số tỉ lệ là 120.
b) biểu diễn x theo y:
y=a/c=> y=120/8=>a=8.15
c) khi x =6
=>y=120/6
=>y=20
Khi x=10
=>y=120:10
=>y=12.
1)Ta có: \(12,\left(1\right)=12+0,\left(1\right)=12+\frac{1}{9}=\frac{109}{9}\);
\(2,3\left(6\right)=2,3+\frac{1}{10}\times0,\left(6\right)=2,3+\frac{1}{10}\times6\times0,\left(1\right)=2,3+\frac{1}{10}\times6\times\frac{1}{9}=\frac{71}{30}\)\(4,\left(21\right)=4+21\times0,\left(01\right)=4+21\times\frac{1}{99}=\frac{139}{33}\)
\(\Rightarrow\)\(\left[\frac{109}{9}-\frac{71}{30}\right]\div\frac{139}{33}=\frac{9647}{4170}\)
2)Ta có: \(0,\left(12\right)=12\times0,\left(01\right)=12\times\frac{1}{99}=\frac{4}{33}\)
\(1,\left(6\right)=1+6\times0,\left(1\right)=1+6\times\frac{1}{9}=\frac{5}{3}\)
\(0,\left(4\right)=4\times0,\left(1\right)=4\times\frac{1}{9}=\frac{4}{9}\)
\(\Rightarrow\frac{4}{33}\div\frac{5}{3}=x\div\frac{4}{9}\Rightarrow x\div\frac{4}{9}=\frac{4}{55}\Rightarrow x=\frac{4}{55}\times\frac{4}{9}\Rightarrow x=\frac{16}{495}\)
a) \(\left|x\right|-15=6\Rightarrow\left|x\right|=21\Rightarrow\left[{}\begin{matrix}x=21\\x=-21\end{matrix}\right.\)
b) \(\left|x\right|+4=0\Rightarrow\left|x\right|=-4\Rightarrow x\in\varnothing\)
c) \(x^2-16=0\Rightarrow x^2=16=4^2\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
a) \(0,\left(31\right)+x=0,3\left(7\right)\\ \Rightarrow\dfrac{31}{99}+x=\dfrac{17}{45}\\ \Rightarrow x=\dfrac{17}{45}-\dfrac{31}{99}=\dfrac{32}{495}=0,0\left(64\right)\)
Vậy \(x=0,0\left(64\right)\)
b) \(0,\left(4\right)\cdot x=\dfrac{5}{6}\\ \Rightarrow\dfrac{4}{9}\cdot x=\dfrac{5}{6}\\ \Rightarrow x=\dfrac{5}{6}:\dfrac{4}{9}\\ \Rightarrow x=\dfrac{5}{6}\cdot\dfrac{9}{4}\\ \Rightarrow x=\dfrac{15}{8}=1,875\)
Vậy \(x=1,875\)
tìm x
\(x^6+x^4=0\\ =>x^4\left(x^2+1\right)=0\\ =>\hept{\begin{cases}x=0\\x^2=-1\left(L\right)\end{cases}}\)
Vậy x = 0
HT