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a) \(\left|x\right|=\dfrac{3}{7}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-3}{7}\\x=\dfrac{3}{7}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{3}{7};\dfrac{-3}{7}\right\}\)
b) \(\left|x\right|=0\)
\(\Rightarrow x=0\)
Vậy x=0
c) \(\left|x\right|=-8,7\)
Vì \(\left|x\right|\ge0\)
\(\Rightarrow x=\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có
\(\left|x-1,7\right|=2,3\\ \Rightarrow\left[{}\begin{matrix}x-1,7=2.3\\x-1.7=-2,3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-0,6\end{matrix}\right.\)
Vậy....
b, Ta có :
\(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{3}=0\\ \Rightarrow\left|x+\dfrac{3}{4}\right|=\dfrac{1}{3}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=-\dfrac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
(x + 7)(x - 2) > 0
TH1: \(\left\{{}\begin{matrix}x+7>0\\x-2>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>-7\\x>2\end{matrix}\right.\) \(\Rightarrow x>2\)
TH2: \(\left\{{}\begin{matrix}x+7< 0\\x-2< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< -7\\x< 2\end{matrix}\right.\) \(\Rightarrow x< -7\)
2.
\(\dfrac{37-x}{x+13}=\dfrac{3}{7}\) \(\Rightarrow7\left(37-x\right)=3\left(x+13\right)\)
\(\Leftrightarrow259-7x=3x+39\)
\(\Leftrightarrow259-39=3x+7x\)
\(\Leftrightarrow220=10x\Rightarrow x=22\)
3.
\(\dfrac{x-3}{x+8}< 0\)
TH1: \(\left\{{}\begin{matrix}x-3< 0\\x+8>0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x< 3\\x>-8\end{matrix}\right.\) => -8 < x < 3
TH2: \(\left\{{}\begin{matrix}x-3>0\\x+8< 0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x>3\\x< -8\end{matrix}\right.\) (loại)
Vậy -8 < x < 3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{37-x}{x+13}=\dfrac{3}{7}\)
\(\Leftrightarrow7\left(37-x\right)=3\left(x+13\right)\)
\(\Leftrightarrow259-7x=3x+39\)
\(\Leftrightarrow-7x-3x=39-259\)
\(\Leftrightarrow-10x=-220\)
\(\Leftrightarrow x=22\)
Vậy x = 22
![](https://rs.olm.vn/images/avt/0.png?1311)
a x+y=x.y=x:y
x=x.y-y=y(x-1)
Suy ra x:y=y(x-1):y=x-1
Mà x:y=x+y
x+y-(x-1)=0 . x+y-x+1=0. y+1=0
y=0-1=-1
x=0,5=1/2
b x-y=x.y
Suy ra x=x.y+y=y(x+1)
Lại có x:y=y(x+1):y=x+1
x-y=x+1 suy ra y =-1 x=-1/2=-0,5
![](https://rs.olm.vn/images/avt/0.png?1311)
x+y+x=0
=) x+y=-z
(=) (x+y)^3 = (-z)^3
(=) x^3+3x^2y+3xy^2+y = -z^3
(=) x^3+y^3+z^3 = -3x^2y- 3xy^2
= x^3+y^3+z^3= -3xy(x+y)
(=) x^3+y^3+z^3 = -3xy(-z)
=) x^3+y^3+z^3 = 3xyz
Cần chứng minh :
x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
Có :
x3 + y3 + z3 - 3xyz
= (x + y)3 - 3xy(x + y) + z3 - 3xyz
= (x + y)3 + z3 - 3xy.(x + y + z)
= (x + y + z).[(x + y)2 - (x + y).z + z2) - 3xy(x + y + z)
= (x + y + z).[x2 + 2xy + y2 - zx - yz + z2) - 3xy(x + y + z)
= (x + y + z).(x2 + y2 + z2 + 2xy - 3xy - yz - zx)
= (x + y + z).(x2 + y2 + z2 xy - yz - zx) (Điều cần chứng minh)
=> (x + y + z).(x2 + y2 + z2 xy - yz - zx) = 0 (vì x + y + z = 0)
=> x3 + y3 + z3 - 3xyz = 0
=> x3 + y3 + z3 = 3xyz
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow\)(37-x)5=(x+13)3
\(\Rightarrow\)185-5x=3x+39
\(\Rightarrow\)185-39=3x+5x
\(\Rightarrow\)8x=146
x=18,25