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a) x3-x2-21x+45=0
<=> x3+5x2-6x2-30x+9x+45=0
<=> (x+5)(x2-6x+9)=0
<=> (x+5)(x2-3x-3x+9)=0
<=> (x+5)(x-3)2=0
Vậy S={-5;3}
b) X3+3X2+4X+2=0
<=> X3+X2+2X2+2X+2X+2=0
<=> (X+1)(X2+2X+2)=0
VÌ X2+2X+2 >=0
NÊN S={-1}
C) X4+7X-8=0
<=> X4-X3+X3-X2+X2-X+8X-8=0
<=> (X-1)(X3+X2+X+8)=0
VÌ X3+X2+X+8>=0
NÊN S={1}
D) 6X4-X3-7X2+X+1=0
<=> 6X4-6X3+5X3-5X2-2X2+2X-X+1=0
<=> (X-1)(6X3+5X2-2X-1)=0
<=> (X-1)(6X3-3X2+8X2-4X+2X-1)=0
<=> (X-1)(2X-1)(3X2_4X+1)=0
<=> (X-1)(2X-1)(3X2-3x-x+1)=0
<=> (X-1)2(2X-1)(3x-1)=0
vậy S={1/3;1/2;1}
(x+1)(6x2+2x)+(x-1)(6x2+2x)
<=> (6x2+2x)(x+1+x-1)
<=> 2x(3x+1)2x
<=> 4x2(3x+1)
<=> x2=0
3x+1=0
<=> x=0
x= -1/3 (-1 phần 3)
c) \(x^3-9x^2+6x+16=x^3-8x^2-x^2+8x-2x+16\)
\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)=\left(x-8\right)\left(x^2-x-2\right)=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)
d) \(2x^3+3x^2+3x+1=\left(2x+1\right)\left(x^2+x+1\right)\)
e) \(2x^3-5x^2+5x-3=\left(2x-3\right)\left(x^2-x+1\right)\)
( 3x - 1 )( x + 3 ) + 9x2 - 1 = 0
<=> 3x2 + 9x - x - 3 + 9x2 - 1 = 0
<=> 12x2 + 8x - 4 = 0
<=> 4( 3x2 + 2x - 1 ) = 0
<=> 3x2 + 2x - 1 = 0
<=> 3x2 + 3x - x - 1 = 0
<=> ( 3x2 + 3x ) - ( x + 1 ) = 0
<=> 3x( x + 1 ) - 1( x + 1 ) = 0
<=> ( 3x - 1 )( x + 1 ) = 0
<=> \(\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)
Vậy S = { 1/3 ; -1 }
\(\frac{x+1}{3}>\frac{3x-2}{5}\)
\(\Leftrightarrow\frac{5\left(x+1\right)}{15}>\frac{3\left(3x-2\right)}{15}\)
\(\Leftrightarrow5x+5>9x-6\)
\(\Leftrightarrow5x-9x>-6-5\)
\(\Leftrightarrow-4x>-11\)
\(\Leftrightarrow x< \frac{11}{4}\)
Bài làm:
a) \(\left(3x-1\right)\left(x+3\right)+9x^2-1=0\)
\(\Leftrightarrow3x^2+8x-3+9x^2-1=0\)
\(\Leftrightarrow12x^2+8x-4=0\)
\(\Leftrightarrow3x^2+2x-1=0\)
\(\Leftrightarrow\left(3x^2+3x\right)-\left(x+1\right)=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)
Vậy tập nghiệm của PT \(S=\left\{-1;\frac{1}{3}\right\}\)
b) \(\frac{x+1}{3}>\frac{3x-2}{5}\Leftrightarrow\frac{5\left(x+1\right)}{15}>\frac{3\left(3x-2\right)}{15}\)
\(\Rightarrow5x+5>9x-6\)
\(\Leftrightarrow4x< 11\)
\(\Rightarrow x< \frac{11}{4}\)
a) (x + 3)2 - (x - 2)2 = 2x
=> (x + 3 - x + 2)(x + 3 + x - 2) = 2x
=> 5(2x + 1) = 2x
=> 10x + 5 = 2x
=> 10x - 2x = -5
=> 8x = -5
=> x = -5/8
b) 7x(x - 2) = x - 2
=> 7x(x - 2) - (x - 2) = 0
=> (7x - 1)(x - 2) = 0
=> \(\orbr{\begin{cases}7x-1=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{7}\\x=2\end{cases}}\)
c) 8x3 - 12x2 + 6x - 1 = 0
=> (2x - 1)3 = 0
=> 2x - 1 = 0
=> 2x = 1
=> x = 1/2
b) \(x^3-6x^2+9x=0\)
\(\Leftrightarrow x.\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow x.\left(x-3\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Vậy \(x=0\)hoặc \(x=3\)
a. ( x - 1 )3 + 1 + 3x ( x - 4 ) = 0
<=> x3 - 3x2 + 3x - 1 + 1 + 3x2 - 12x = 0
<=> x3 - 9x = 0
<=> x ( x2 - 9 ) = 0
<=> \(\orbr{\begin{cases}x=0\\x^2-9=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}\)
b. x3 - 6x2 + 9x = 0
<=> x ( x2 - 6x + 9 ) = 0
<=> x ( x - 3 )2 = 0
<=> \(\orbr{\begin{cases}x=0\\\left(x-3\right)^2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=0\\x=3\end{cases}}\)