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Ta có: \(x^4-30x^2+31x-30=0\) \(\Rightarrow x^4+x-30x^2+30x-30=0\)
\(\Rightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\Rightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)
Xét \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)
\(\Rightarrow x^2+x-30=0\Rightarrow x^2-5x+6x-30=0\)
\(\Rightarrow\left(x-5\right)\left(x+6\right)=0\Rightarrow\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}}\)
Vậy x=5 hoặc x = -6
\(\Leftrightarrow x^4-5x^3+5x^3-25x^3-5x^3+25x+6x-30=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^3+5x^2-5x+6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\cdot\left(x^3+6x^2-x^2-6x+x+6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+6\right)\left(x^2-x+1\right)=0\)
hay \(x\in\left\{5;-6\right\}\)
\(1,\\ a,=3x^2+2x\\ b,=x^2+13x+40\\ c,=x^3+6x^2+8x^2+48x-x-6=x^3+14x^2+47x-6\\ 2,\\ a,=x^2+4x+4\\ b,=x^2-16y^2\\ c,=4x^2-12xy+9y^2\\ d,=x^3-27\\ 3,\\ a,=3x\left(x+2\right)\\ b,=\left(x+y\right)\left(4x+5\right)\\ c,=6x\left(2x^2-x+3\right)\)
\(=\left(x-y\right)\left(x+y\right)+11\left(x-y\right)=\left(x+y+11\right)\left(x-y\right)\)
\(x+y+z=0\Leftrightarrow x^2+y^2+z^2=-2\left(xy+xz+yz\right)\)
Mẫu số nhân ra : \(2\left(x^2+y^2+z^2\right)-2\left(xy+xz+yz\right)=3\left(x^2+y^2+z^2\right)\)
\(A=\dfrac{18\left(x^2+y^2+z^2\right)}{3\left(x^2+y^2+z^2\right)}=6\)
3. A = x3 - 64 - ( x3 - x2 + x - 1 ) = x3 - 64 - x3 + x2 - x + 1 = x2 - x - 63
B = x3 + 8 - ( x3 - 8 ) = x3 + 8 - x3 + 8 = 16
C = x3 - 3x2 + 3x - 1 - ( 4x2 - 1 ) = x3 - 3x2 + 3x - 1 - 4x2 + 1 = x3 - 7x2 + 3x
D = x( x2 - 25 ) - ( x3 + 1 ) = x3 - 25x - x3 - 1 = -25x - 1
4. a) x2 - 4x + 1 = 0 <=> ( x2 - 4x + 4 ) - 3 = 0 <=> ( x - 2 )2 - (√3)2 = 0
<=> ( x - 2 - √3 )( x - 2 + √3 ) = 0 <=> x = 2 ± √3
b) 9x2 - 6x - 8 = 0 <=> ( 9x2 - 6x + 1 ) - 9 = 0 <=> ( 3x - 1 )2 - 32 = 0
<=> ( 3x - 4 )( 3x + 2 ) = 0 <=> x = 4/3 hoặc x = -2/3
c) x3 - 3x2 + 3x + 7 = 0 <=> ( x3 - 3x2 + 3x - 1 ) + 8 = 0
<=> ( x - 1 )3 + 23 = 0 <=> ( x + 1 )( x2 - 4x + 7 ) = 0
<=> x + 1 = 0 <=> x = -1 ( vì x2 - 4x + 7 = ( x - 2 )2 + 3 > 0 )