Tìm x
X^3+3x^2+1+28=0
Làm nhanh nhé :)))
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a: \(A=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}-\dfrac{12x^2}{\left(x-3\right)\left(x+3\right)}\right)\)
\(=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{-x^2-6x-9+x^2-6x+9-12x^2}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-\left(x+1\right)}{x\left(x-3\right)}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{-12x^2-12x}\)
\(=\dfrac{-\left(x+1\right)\cdot\left(x+3\right)}{-12x^2\left(x+1\right)}=\dfrac{x+3}{12x^2}\)
b: Ta có: |2x-1|=5
=>2x-1=5 hoặc 2x-1=-5
=>x=-2
Thay x=-2 vào A, ta được:
\(A=\dfrac{-2+3}{12\cdot\left(-2\right)^2}=\dfrac{1}{48}\)
c: Để \(A=\dfrac{2x+1}{x^2}\) thì \(\dfrac{x+3}{12x^2}=\dfrac{2x+1}{x^2}\)
=>x+3=24x+12
=>24x+12=x+3
=>23x=-9
hay x=-9/23
d: Để A<0 thì x+3<0
hay x<-3
Ta có: \(\dfrac{x+1}{2018}+\dfrac{x+1}{2019}+\dfrac{x+1}{2020}+\dfrac{x+1}{2021}=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
Ta có: \(\left|y+3\right|\ge0\forall y\)
\(\left|2x+y\right|\ge0\forall x,y\)
Do đó: \(\left|y+3\right|+\left|2x+y\right|\ge0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}y+3=0\\2x+y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-3\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-3\end{matrix}\right.\)
\(x-\dfrac{3}{6}=\dfrac{28}{6}-\dfrac{1}{2}\\ x-\dfrac{3}{6}=\dfrac{28}{6}-\dfrac{3}{6}\\ x-\dfrac{3}{6}=\dfrac{25}{6}\\ =>x=\dfrac{25}{6}+\dfrac{3}{6}=\dfrac{28}{6}=\dfrac{14}{3}\)
\(x-\dfrac{3}{6}=\dfrac{28}{6}-\dfrac{1}{2}\)
\(x-\dfrac{3}{6}=\dfrac{28-3}{6}=\dfrac{25}{6}\)
\(=>x=\dfrac{25}{6}+\dfrac{3}{6}=\dfrac{25+3}{6}=\dfrac{28}{6}=\dfrac{4}{1}\)
Câu 1:Tính hợp lí
a) 28.72+28.28
=28.(72+28)
=28.100
= 2800
b)50.5.2
=50.2.5
=100.5
= 500
Câu 2:Tìm x
a) 3x+6=15
3x =15-6
3x = 9
x = 9:3
x = 3
b)100-(4x+8):10=8
100-(4x+8) =8.10
100-(4x+8) = 80
4x+8 = 100-80
4x+8 = 20
4x = 20-8
4x = 12
x = 12:4
x = 3
K cho mk nhé!!!!!Thanks>_<
câu 1
a )
28x72+28x28
=28+72
=100
b )
50x5x2
=50x10
=500
câu 2
a )
3x+6=15
3x=15-6
3x=9
x=9:3
x=3
b)
100-(4x+8):10=8
100-(4x+8=)8x10
100-(4x+8)=80
4x+8=100-80
4x+8=20
4x=20-8
4x=12
x=12:4
x=3
Ta có: \(\left(\dfrac{2}{3}x-\dfrac{4}{9}\right)\left(\dfrac{1}{2}-\dfrac{3}{7}x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{9}\\\dfrac{3}{7}x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{7}{6}\end{matrix}\right.\)
x3 + 3x2 + 1 + 28 = 0
x3 + 3x2 + 29 = 0
Mà: 3x2 > 0
29 > 0
nên không có x thỏa mãn.
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