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a) \(x^3+3x^2+3x+2=0\)
<=> \(x^3+x^2+x+2x^2+2x+2=0\)
<=> \(x\left(x^2+x+1\right)+2\left(x^2+x+1\right)=0\)
<=> \(\left(x+2\right)\left(x^2+x+1\right)=0\)
tự làm
b) \(x^4-2x^3+2x-1=0\)
<=> \(\left(x^4-3x^3+3x^2-x\right)+\left(x^3-3x^2+3x-1\right)=0\)
<=> \(x\left(x^3-3x^2+3x-1\right)+\left(x^3-3x^2+3x-1\right)=0\)
<=> \(\left(x^3-3x^2+3x-1\right)\left(x+1\right)=0\)
<=> \(\left(x-1\right)^3\left(x+1\right)=0\)
tự làm
c) \(x^4-3x^3-6x^2+8x=0\)
<=> \(x\left(x^3-3x^2-6x+8\right)=0\)
<=> \(x\left[\left(x^3+x^2-2x\right)-\left(4x^2+4x-8\right)\right]=0\)
<=>\(x\left[x\left(x^2+x-2\right)-4\left(x^2+x-2\right)\right]=0\)
<=> \(x\left(x-4\right)\left(x^2+x-2\right)=0\)
<=> \(x\left(x-4\right)\left(x-1\right)\left(x+2\right)=0\)
tự làm
a/ => x3 = 64 => x3 = 43 => x = 4
b/ => 4x2 - 12x + 9 - x2 - 10x - 25 = 0
=> 3x2 - 22x - 16 = 0
=> (x - 8)(3x + 2) = 0
=> x - 8 = 0 => x = 8
hoặc 3x + 2 = 0 => 3x = -2 => x = -2/3
Vậy x = 8 ; x = -2/3
c/ => x3 - x2 - 4x2 + 8x - 4 = 0
=> x3 - 5x2 + 8x - 4 = 0
=> (x - 2)2 (x - 1) = 0
=> (x - 2)2 = 0 => x - 2 = 0 => x = 2
hoặc x - 1 = 0 => x = 1
Vậy x = 2 ; x = 1
a) 2x2+3x-5=0
=> 2x2+5x-2x-5=0
=> x(2x+5)-(2x-5)=0
=> (2x-5)(x-1)=0
=> 2x-5=0, x-1=0
=> x=5/2; 1
\(2x^2+3x-5=0< =>2x^2-2+3x-3=0\)
\(< =>2\left(x+1\right)\left(x-1\right)-3\left(x-1\right)=0\)
\(< =>\left(x-1\right)\left(2x-1\right)=0< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
1,
<=> \(\left(x-1\right)\left(x-2\right)^2=0\)
=> x=1 hoặc x=2
2,
<=>\(\left(x+1\right)\left(2x^2-3x+6\right)\)=0
=> x=-1
1.
<=> ( x -1 ) ( x - 2 ) 2 = 0
=> x = 1 hoặc x = 2
2.
<=> ( x + 1 ) ( 2x2 - 3x + 6 ) = 0
=> x = -1
\(2x^3+x^2-8x-4=0\)
\(\Leftrightarrow\)\(x^2\left(2x+1\right)-4\left(2x+1\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(x-2\right)\left(x+2\right)=0\)
đến đây bạn làm tiếp nha
\(2x^3+x^2-8x-4=0\)
\(x^2\left(2x+1\right)-4\left(2x+1\right)=0\)
\(\left(x^2-4\right)\left(2x+1\right)=0\)
\(1.x^2-4=0\)
\(\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow x=\pm2\)
\(2.2x+1=0\)
\(x=-\frac{1}{2}\)
\(2x^3+x^2-8x-4=0\)
\(\Leftrightarrow x^2.\left(2x+1\right)-4\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\)2x + 1 = 0 => x = -1/2
Hoặc x - 2 = 0 => x = 2
Hoặc x + 2 = 0 => x = -2
Vậy x = -1/2 hoặc x = 2 hoặc x = -2
2x3 + x2 - 8x - 4 = 0
<=> (2x3 - 8x) + (x2 - 4) = 0
<=> 2x(x2 - 4) + (x2 - 4) = 0
<=> (2x + 1) (x2 - 4) = 0
\(< =>\orbr{\orbr{\begin{cases}2x+1=0\\x^2-4=0\end{cases}< =>\orbr{\begin{cases}x=\frac{-1}{2}\\\orbr{\begin{cases}x=2\\x=-2\end{cases}}\end{cases}}}}\)\(< =>\orbr{\begin{cases}2x+1=0\\x^2-4\end{cases}}\)\(< =>\orbr{\begin{cases}x=\frac{-1}{2}\\\orbr{\begin{cases}x=2\\x=-2\end{cases}}\end{cases}}\)<=> x = \(\frac{-1}{2}\)hoặc -2 hoặc 2