Giải pt:\(x^4+x^3+4x^2+3x+3\) =0
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a: Ta có: \(x^2+3x+4=0\)
\(\text{Δ}=3^2-4\cdot1\cdot4=9-16=-7< 0\)
Do đó: Phương trình vô nghiệm
1: 5(2-3x)(x-2)=3(1-3x)
=>5(3x-2)(x-2)=3(3x-1)
=>\(5\cdot\left(3x^2-6x-2x+4\right)=9x-3\)
=>\(15x^2-40x+20-9x+3=0\)
=>\(15x^2-49x+23=0\)
\(\Delta=49^2-4\cdot15\cdot23=1021>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left[\begin{array}{l}x=\frac{49-\sqrt{1021}}{2\cdot15}=\frac{49-\sqrt{1021}}{30}\\ x=\frac{49+\sqrt{1021}}{2\cdot15}=\frac{49+\sqrt{1021}}{30}\end{array}\right.\)
2: \(4x^2+4x+1=0\)
=>\(\left(2x+1\right)^2=0\)
=>2x+1=0
=>2x=-1
=>\(x=-\frac12\)
3: \(4x^2-9=0\)
=>\(4x^2=9\)
=>\(x^2=\frac94\)
=>\(\left[\begin{array}{l}x=\frac32\\ x=-\frac32\end{array}\right.\)
4: \(5x^2-10x=0\)
=>5x(x-2)=0
=>x(x-2)=0
=>\(\left[\begin{array}{l}x=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=2\end{array}\right.\)
5: \(x^2-3x=-2\)
=>\(x^2-3x+2=0\)
=>(x-1)(x-2)=0
=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)
6: |x-5|-3=0
=>|x-5|=3
=>\(\left[\begin{array}{l}x-5=3\\ x-5=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=2\end{array}\right.\)
\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)
\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)
\(\Leftrightarrow x^2-9-x^2+3x=0\)
\(\Leftrightarrow3x-9=0\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\left(n\right)\)
Vậy \(S=\left\{3\right\}\)
\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)
\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)
\(\Leftrightarrow12x-9-12x+20+2x-7>0\)
\(\Leftrightarrow2x+4>0\)
\(\Leftrightarrow2x>-4\)
\(\Leftrightarrow x>-2\)
1:
a: =>3x=6
=>x=2
b: =>4x=16
=>x=4
c: =>4x-6=9-x
=>5x=15
=>x=3
d: =>7x-12=x+6
=>6x=18
=>x=3
2:
a: =>2x<=-8
=>x<=-4
b: =>x+5<0
=>x<-5
c: =>2x>8
=>x>4
Ta có: \(x^4+x^3+4x^2+3x+3=0\)
=>\(x^4+x^3+x^2+3x^2+3x+3=0\)
=>\(\left(x^2+x+1\right)\left(x^2+3\right)=0\)
mà \(x^2+3\ge3>0\forall x;x^2+x+1=\left(x+\frac12\right)^2+\frac34\ge\frac34>0\forall x\)
nên x∈∅
x=1