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Bài 9:
a: \(A=-0.5x^2yz\cdot\left(-3\right)xy^3z=1.5x^3y^4z^2\)
b: Hệ số là 1,5
Bậc là 9
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{1}{4}+\dfrac{3}{4}:x=-2\)
\(\dfrac{3}{4}:x=-2-\dfrac{1}{4}=\dfrac{-8}{4}-\dfrac{1}{4}\)
\(\dfrac{3}{4}:x=\dfrac{-9}{4}\)
\(x=\dfrac{3}{4}:\dfrac{-9}{4}=\dfrac{3}{4}.\dfrac{-4}{9}\)
\(x=\dfrac{-1}{3}\)
b) \(\dfrac{3}{4}+2.\left(2x-\dfrac{2}{3}\right)=-2\)
\(2.\left(2x-\dfrac{2}{3}\right)=-2-\dfrac{3}{4}=\dfrac{-8}{4}-\dfrac{3}{4}\)
\(2.\left(2x-\dfrac{2}{3}\right)=\dfrac{-11}{4}\)
\(2x-\dfrac{2}{3}=\dfrac{-11}{4}:2=\dfrac{-11}{4}.\dfrac{1}{2}\)
\(2x-\dfrac{2}{3}=\dfrac{-11}{8}\)
\(2x=\dfrac{-11}{8}+\dfrac{2}{3}=\dfrac{-33}{24}+\dfrac{16}{24}\)
\(2x=\dfrac{-17}{24}\)
\(x=\dfrac{-17}{24}:2=\dfrac{-17}{24}.\dfrac{1}{2}\)
\(x=\dfrac{-17}{48}\)
c) \(\left(\dfrac{1}{2}+5x\right).\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}+5x=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{2}\\2x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{10}\\x=\dfrac{3}{2}\end{matrix}\right.\)
a, 1/4 + 3/4 : x = -2
3/4 : x = -2 - 1/4
3/4 : x = -9/4
x = 3/4 : -9/4
x = -1/3
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left|2x-1\right|+\frac{1}{3}=0\)
\(\Leftrightarrow\left|2x-1\right|=-\frac{1}{3}\)
=> vô lý
=> PT vô nghiệm
b) \(\left|x+2\right|+\left|x-3\right|=0\)
\(\Leftrightarrow\left|x+2\right|=-\left|x-3\right|\)
Vì \(\hept{\begin{cases}\left|x+2\right|\ge0\\-\left|x-3\right|\le0\end{cases}\left(\forall x\right)}\) nên dấu "=" xảy ra khi:
\(\left|x+2\right|=-\left|x-3\right|=0\Rightarrow\hept{\begin{cases}x=-2\\x=3\end{cases}}\) (vô lý)
=> PT vô nghiệm
/x-2/-/2x-3/-x=-2
/x-2/-/2x-3/=-2+x
/x-2/-/2x-3/=x-2
Nếu /x-2/=x-2 thì /2x-3/=0 =>2x-3=0 =>x=1,5
Nếu /x-2/=-(x-2) thì
-(x-2)-/2x-3/=x-2
-x+2-/2x-3/=x-2
Thực hiện quy tắc chuyển vế ta đc
-/2x-3/=2x-4
=>/2x-3/=-(2x-4)
Nếu 2x-3=-(2x-4)
2x-3=-2x+4
4x=7
x=7/4
Nếu 2x-3=2x-4
2x-2x=-4+3
0=-1 ( vô lí)
Vậy x có thể 7/4
Câu a