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\(y=-4\\ \Rightarrow-4=3x^2-7\\ \Rightarrow3x^2=3\\ \Rightarrow x^2=1\\ \Rightarrow x=\pm1\)
\(y=5\\ \Rightarrow5=3x^2-7\\ \Rightarrow3x^2=12\\ \Rightarrow x^2=4\\ \Rightarrow x=\pm2\)
\(y=-6\dfrac{2}{3}\\ \Rightarrow-6\dfrac{2}{3}=3x^2-7\\ \Rightarrow3x^2=\dfrac{1}{3}\\ \Rightarrow x^2=\dfrac{1}{9}\\ \Rightarrow x=\pm\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(\frac{1}{2}x+y\right)\left(\frac{1}{2}x+y\right)\)
\(=\frac{1}{2}x.\left(\frac{1}{2}x+y\right)+y.\left(\frac{1}{2}x+y\right)\)
\(=\frac{1}{4}x^2+\frac{1}{2}xy+\frac{1}{2}xy+y^2\)
\(=\frac{1}{4}x^2+xy+y^2\)
b) \(\left(x^3-2x^2+x\right).\left(5-x\right)\)
\(=x^3.\left(5-x\right)-2x^2.\left(5-x\right)+x.\left(5-x\right)\)
\(=5x^3-x^4-10x^2+2x^3+5x-x^2\)
\(=7x^3-x^4-11x^2+5x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
F(x)=62+5x+8+3x-3x2+3x3
=(36+8)+(5x+3x)-3x2+3x3
=3x3-3x2+8x+44
G(x)=12x2-6-9x2+3x3
=3x3+(12x2-9x2)-6
=3x3+3x2-6
F(x)+G(x)=3x3-3x2+8x+44+3x3+3x2-6
=(3x3+3x3)+(-3x2+3x2)+8x+(44-6)
=6x3+8x+38
\(F\left(x\right)=G\left(x\right)\\ \Rightarrow6^2-5x+8+3x-3x^2+3x^3=12x^2-6-9x^2+3x^3\\ \Leftrightarrow-3x^2-2x+44=3x^2-6\\ \Leftrightarrow6x^2+2x-50=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{301}}{6}\\x=\dfrac{-1-\sqrt{301}}{6}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
c) \(2x\left(3x-1\right)-3x\left(5+2x\right)=0\)
\(\Rightarrow x\left[2\left(3x-1\right)-3\left(5+2x\right)\right]=0\)
\(\Rightarrow x\left(6x-2-15-6x\right)\)
\(\Rightarrow-16x=0\)
\(\Rightarrow x=0\)
d) \(\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\)
\(\Rightarrow9x^2-4-4x+4=0\)
\(\Rightarrow9x^2-4x=0\)
\(\Rightarrow x\left(9x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)
\(a,\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt `3(x+2)-1/3(6-3x)=0`
`<=>3(x+2)-(2-x)=0`
`<=>3x+2+x-2=0`
`<=>4x=0`
`<=>x=0`
Vậy nghiệm của đa thức là 0
`3x(x-5)-(x+3x)=0`
`<=>3x(x-5)-4x=0`
`<=>x(3x-15-4)=0`
`<=>x(3x-19)=0`
`<=>[(x=0),(3x-19=0):}`
`<=>[(x=0),(x=19/3):}`
Vậy nghiệm đa thức là 0 và `19/3`.
a) Đặt \(3\left(x+2\right)-\dfrac{1}{3}\left(6-3x\right)=0\)
\(\Leftrightarrow3x+6-2+x=0\)
\(\Leftrightarrow4x=-4\)
hay x=-1
b) Đặt 3x(x-5)-(x+3x)=0
\(\Leftrightarrow3x^2-15x-4x=0\)
\(\Leftrightarrow x\left(3x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`D(x)=3x^3+x=0`
`\Leftrightarrow 3x^2*x+x=0`
`\Leftrightarrow x(3x^2+1)=0`
`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\3x^2+1=0\end{matrix}\right.\)
`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\3x^2=-1\text{(loại)}\end{matrix}\right.\)
Vậy, nghiệm của đa thức là `x=0`
`E(x)=x^2-3x+2=0`
`\Leftrightarrow x^2-2x-x+2=0`
`\Leftrightarrow (x^2-2x)-(x-2)=0`
`\Leftrightarrow x(x-2)-(x-2)=0`
`\Leftrightarrow (x-2)(x-1)=0`
`\Leftrightarrow `\(\left[{}\begin{matrix}x-2=0\\x-1=0\end{matrix}\right.\)
`\Leftrightarrow `\(\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
Vậy, nghiệm của đa thức là `x= {2 ; 1}`
`F(x)=4x^2-4x+1=0`
`\Leftrightarrow (2x+1)^2=0`
`\Leftrightarrow 2x-1=0`
`\Leftrightarrow 2x=1`
`\Leftrightarrow x=1/2`
Vậy, nghiệm của đa thức là `x=1/2`
`D(x)=3x^3+x`
`-> 3x^3 +x=0`
`=> x(3x^2 +1)=0`
\(\Rightarrow\left[{}\begin{matrix}x=0\\3x^2+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\3x^2=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x\in\left\{0\right\}\)
__
`E(x)=x^2-3x+2`
`-> x^2-3x+2=0`
`=> x^2 -2x-x+2=0`
`=> (x^2-2x) -(x-2)=0`
`=> x(x-2)-(x-2)=0`
`=>(x-2)(x-1)=0`
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{2;1\right\}\)
__
`F(x)=4x^2-4x+1`
`-> 4x^2-4x+1=0`
`=> 4x^2 -2x-2x+1=0`
`=> (4x^2-2x)-(2x-1)=0`
`=> 2x(2x-1)-(2x-1)=0`
`=> (2x-1)(2x-1)=0`
`=>(2x-1)^2=0`
`=>2x-1=0`
`=>2x=1`
`=>x=1/2`
Vậy \(x\in\left\{\dfrac{1}{2}\right\}\)
Hoặc
`->4x^2-4x+1=0`
`=> (2x-1)^2=0`
`=> 2x-1=0`
`=>2x=1`
`=>x=1/2`
Vậy \(x\in\left\{\dfrac{1}{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)Tìm x
a) (x+1)(x-2)<0
=>Có 2TH:
TH1:
x+1<0=>x< -1
x-2>0=>x>2
=>Vô lí
TH2:
x+1>0=>x> -1
x-2<0=>x<2
=> -1<x<2
Vậy x thuộc {0;1}
b) Tương tự a thôi ạ.
c) (x-2)(3x+2)
=> Có hai TH:
TH1:
x-2<0=>x<2
3x+2<0=>3x< -2=>x< -2/3
=>x< -2/3
TH2:
x-2>0=>x>2
3x+2>0=>3x> -2=>x> -2/3
=>x>2
Vậy x< -2/3 hoặc x>2
2)Tìm x
x.x=x
<=>x²-x=0
<=>x(x-1)=0
<=>x=0 hoặc x=1
Ta có: \(x^2< 3x\)
\(\Leftrightarrow x^2-3x< 0\)
\(\Leftrightarrow x\left(x-3\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-3>0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x< 3\end{matrix}\right.\\\left[{}\begin{matrix}x< 0\\x>3\end{matrix}\right.\left(loại\right)\end{matrix}\right.\Leftrightarrow0< x< 3\)
Vậy: S={x|0<x<3}