Tìm nghiệm
D(x)=3x^3+x
E(x)=x^2-3x+2
F(x)=4x^2-4x+1
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4 tháng 2 2021
b) Ta có: \(x^2-4x+6\)
\(=x^2-4x+4+2\)
\(=\left(x-2\right)^2+2\)
Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x-2\right)^2+2\ge2>0\forall x\)
hay \(x^2-4x+6>0\forall x\)
Vậy: phương trình \(x^2-4x+6=0\) vô nghiệm
c) Ta có: \(\left|x-2\right|=-1\)
mà \(\left|x-2\right|>0>-1\forall x\)
nên phương trình \(\left|x-2\right|=-1\) vô nghiệm(đpcm)
d) Ta có: \(\left|x\right|=x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=x\\x=-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-x=0\\x+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=0\left(luônđúng\right)\\2x=0\end{matrix}\right.\Leftrightarrow x\in R\)
Vậy: S={x|\(x\in R\)}
23 tháng 10 2021
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
15 tháng 12 2021
\(a,=x\left(x-2\right)^2\\ b,=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\\ c,=x^2\left(2x-1\right)-4\left(2x-1\right)=\left(x-2\right)\left(x+2\right)\left(2x-1\right)\\ d,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ e,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x\left[\left(x-2\right)^2-y^2\right]=x\left(x-y-2\right)\left(x+y-2\right)\\ g,=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\\ h,=x^3-x-2x+2=x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\\ =\left(x-1\right)\left(x^2+x-2\right)=\left(x-1\right)^2\left(x+2\right)\\ i,=3x^2+3x-10x-10=\left(x+1\right)\left(3x-10\right)\)
31 tháng 10 2021
\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
31 tháng 10 2021
a.
4x(x-5) - (x-1)(4x-3)-5=0
4x^2-20x-4x^2+3x+4x+3=0
(4x^2-4x^2)+(-20x+3x+4x)+3=0
13x+3 = 0
13x=-3
x=-3/13
b,
(3x-4)(x-2)-3x(x-9)+3=0
3x^2-6x-4x+8 - 3x^2+27x+3=0
(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0
17x+11=0
17x=-11
x=-11/17
c, 2(x+3)-x^2-3x=0
2(x+3) - x(x+3)=0
(x+3)(2-x)=0
TH1: x+3 = 0; x=-3
TH2: 2-x=0;x=2
14 tháng 10 2021
1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)
\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)
\(\Leftrightarrow x=2\)
3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)
\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)
\(\Leftrightarrow6x=6\)
hay x=1
`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 \(x\in\left\{\dfrac{1}{2}\right\}\)