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a) (15x2-1+9x4-6x3+2x) :( 5 + 3x2-2x)
b) ( -19x+ 10+ 3x4- 5x2+11x3) : ( 3x+ x2-2)
c) (x4-14-x) : (x-2)
c: \(\dfrac{x^4-x-14}{x-2}\)
\(=\dfrac{x^4-2x^3+2x^3-4x^2+4x^2-8x+7x-14}{x-2}\)
\(=x^3+2x^2+4x+7\)
`1,(4x^3+3x^3):x^3+(15x^2+6x):(-3x)=0`
`<=> 4 + 3 + (-5x) + (-2)=0`
`<=> -5x+5=0`
`<=>-5x=-5`
`<=>x=1`
`2,(25x^2-10x):5x +3(x-2)=4`
`<=> 5x - 2 + 3x-6=4`
`<=> 8x -8=4`
`<=> 8x=12`
`<=>x=12/8`
`<=>x=3/2`
`3,(3x+1)^2-(2x+1/2)^2=0`
`<=> [(3x+1)-(2x+1/2)][(3x+1)+(2x+1/2)]=0`
`<=>( 3x+1-2x-1/2)(3x+1+2x+1/2)=0`
`<=>( x+1/2) (5x+3/2)=0`
`@ TH1`
`x+1/2=0`
`<=>x=0-1/2`
`<=>x=-1/2`
` @TH2`
`5x+3/2=0`
`<=> 5x=-3/2`
`<=>x=-3/2 : 5`
`<=>x=-15/2`
`4, x^2+8x+16=0`
`<=>(x+4)^2=0`
`<=>x+4=0`
`<=>x=-4`
`5, 25-10x+x^2=0`
`<=> (5-x)^2=0`
`<=>5-x=0`
`<=>x=5`
1: Sửa đề: 3x-5
\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)
2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
=5x^2+14x^2+12x+8
3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)
5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)
\(2x-x^2=2\\ \Leftrightarrow x^2-2x+2=0\\ \Leftrightarrow\left(x^2-2x+1\right)+1=0\\ \Leftrightarrow\left(x-1\right)^2+1=0\\ Mà:\left(x-1\right)^2\ge0\forall x\in R\\ \Rightarrow\left(x-1\right)^2+1\ge1\forall x\in R\\ Vậy:Pt.vô.nghiệm\\ x^3+15x^2+75x+125=0\\ x^3+3.x^2.5+3.x.5^2+5^3=0\\ \left(x+5\right)^3=0\\ \Leftrightarrow x+5=0\\ \Leftrightarrow x=-5\\ x^3+48x=12x^2+64\\ \Leftrightarrow x^3-12x^2+48x-64=0\\ \Leftrightarrow x^3-3.x^2.4+3.x.4^2-4^2=0\\ \Leftrightarrow\left(x-4\right)^3=0\\ \Leftrightarrow x-4=0\\ \Leftrightarrow x=4\)
\(=5x^4-15x^3+10x^2\)
Vậy không đáp án nào đúng
Kì thế ạ :vv