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a) (2x2 - x) + 4x - 2 = 0
x(2x - 1) + 2(2x - 1) = 0
(2x - 1)(x + 2) = 0
2x - 1 = 0 hoặc x + 2 = 0
* 2x - 1 = 0
2x = 1
x = \(\frac{1}{2}\)
* x + 2 = 0
x = -2
Vậy x = -2; x = \(\frac{1}{2}\)
b) x2 - 6x + 8 = 0
x2 - 2x - 4x + 8 = 0
(x2 - 2x) + (-4x + 8) = 0
x(x - 2) - 4(x - 2) = 0
(x - 2)(x - 4) = 0
x - 2 = 0 hoặc x - 4 = 0
* x - 2 = 0
x = 2
* x - 4 = 0
x = 4
Vậy x = 2; x = 4
c) x4 - 8x2 - 9 = 0
x4 + x2 - 9x2 - 9 = 0
(x4 - 9x2) + (x2 - 9) = 0
x2(x2 - 9) + (x2 - 9) = 0
(x2 - 9)(x2 + 1) = 0
x2 - 9 = 0 (vì x2 + 1 > 0 với mọi x)
x2 = 9
x = 3 hoặc x = -3
Vậy x = 3; x = -3
\(f\left(x\right)=x^3-x^2+3x-3\)
\(=x^2\left(x-1\right)+3\left(x-1\right)\)
\(=\left(x^2+3\right)\left(x-1\right)\)
Để \(f\left(x\right)>0\Leftrightarrow\left(x^2+3\right)\left(x-1\right)>0\)
Mà \(x^2\ge0\forall x\Leftrightarrow x^2+3>0\)
\(\Rightarrow x-1>0\Leftrightarrow x=1\)
\(h\left(x\right)=4x^3-14x^2+6x-21< 0\)
\(\Leftrightarrow0\left(x-\frac{7}{2}\right)\left(4x^2+6\right)< 0\)
Mà \(4x^2+6>0\forall x\Leftrightarrow h\left(x\right)< 0\Leftrightarrow x-\frac{7}{2}< 0\Leftrightarrow x< \frac{7}{2}\)
f(x)=x3−x2+3x−3f(x)=x3−x2+3x−3
=x2(x−1)+3(x−1)=x2(x−1)+3(x−1)
=(x2+3)(x−1)=(x2+3)(x−1)
Để f(x)>0⇔(x2+3)(x−1)>0f(x)>0⇔(x2+3)(x−1)>0
Mà x2≥0∀x⇔x2+3>0x2≥0∀x⇔x2+3>0
⇒x−1>0⇔x=1⇒x−1>0⇔x=1
h(x)=4x3−14x2+6x−21<0h(x)=4x3−14x2+6x−21<0
⇔0(x−72)(4x2+6)<0⇔0(x−72)(4x2+6)<0
Mà 4x2+6>0∀x⇔h(x)<0⇔x−72<0⇔x<72
\(\left(x-4\right)^2=\left(2x+1\right)^2\)
\(\Leftrightarrow\left(x-4\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(x-4-2x-1\right)\left(x-4+2x+1\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\3\left(x-1\right)=0\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}}\)
a)4x3y-6xy2
=2xy(2x2-3y)
b)4x2-4x+1
=(2x)2-2*2x*1+12
=(2x-1)2
c)x2-2xy-3x+6y
=x(x-2y)-3(x-2y)
=(x-3)(x-2y)
d)x3-2x2+x-xy2
=x(x2-2x+1-y2)
=x[(x-1)2-y2]
=x(x-y-1)(x+y-1)
e)x2-x+y2-y-x2y2+xy
=xy2-x+y2-y-x2y2+x2-xy2+xy
=(xy2-x+y2-y)-x(xy2-x+y2-y)
=(1-x)(xy2-x+y2-y)
=(1-x)[xy2+xy+y2-(xy+y+x)]
=(1-x)[y(xy+y+x)-(xy+y+x)]
=(1-x)(y-1)(xy+y+x)
Bài 2:
a)x(x-y)+y(y-x)
=x2-xy+y2-xy
=(x-y)2.Tại x=53 và y=3 ta có:
N=(53-3)2=502=2500
b) x2013-53x2012+103x2011-51x2010
=x2010(x3-53x2+103x-51)
=x2010[x3-2x2+x-51x2+102x-51]
=x2010[x(x2-2x+1)-51(x2-2x+1)]
=x2010(x-51)(x2-2x+1).Tại x=51 ta có:
M=512010(51-51)(512-2*51+1)=0
\(\left(x^3-x^2-5x+21\right):\left(x^2-4x+7\right)\)
\(=\left(x^3-4x^2+3x^2+7x-12x+21\right):\left(x^2-4x+7\right)\)
\(=\left[\left(x^3-4x^2+7x\right)+\left(3x^2-12x+21\right)\right]:\left(x^2-4x+7\right)\)
\(=\left[x\left(x^2-4x+7\right)+3\left(x^2-4x+7\right)\right]:\left(x^2-4x+7\right)\)
\(=\left[\left(x^2-4x+7\right)\left(x+3\right)\right]:\left(x^2-4x+7\right)\)
\(=x+3\)
a ) -36a2 + x2 + 4y2 - 4xy
= ( x2 - 4xy + 4y2 ) - (6a)2
= ( x -2y )2 - (6a)2
= ( x - 2y - 6a ).(x - 2y + 6a )
b ) 10ax - 5ay +2x - y
= ( 10ax - 5ay ) + ( 2x - y )
= 5a ( 2x - y ) + ( 2x - y )
= ( 2x - y ) . (5a + 1 )
c ) 2a2b(x + y) - 4ab2(-x - y )
= 2a2b( x+ y ) + 4ab2(x + y )
= 2ab(x + y ) ( a + 2b )
a, \(-36a^2+x^2+4y^2-4xy=\left(x+2y\right)^2-\left(6a\right)^2=\left(x+2y-6a\right)\left(x+2y+6a\right)\)
b, \(10ax-5ay+2x-y=5a\left(2x-y\right)+2x-y=\left(5a+1\right)\left(2x-y\right)\)
c, \(2a^2b\left(x+y\right)-4ab^2\left(-x-y\right)=2a^2b\left(x+y\right)+4ab^2\left(x+y\right)\)
\(=\left(2a^2b+4ab^2\right)\left(x+y\right)=2ab\left(a+2b\right)\left(x+y\right)\)
ta có: \(\left(3x-5\right)^2+\left(2-x\right)^3+\left(3-2x\right)^3=0\)
<=>\(\left(5-3x\right)^2+\left(2-x+3-2x\right)\left[\left(2-x\right)^2+\left(2-x\right)\left(3-2x\right)+\left(3-2x\right)^2\right]=0\)
<=>\(\left(5-3x\right)^2+\left(5-3x\right)\left(4-4x+x^2-6+7x-2x^2+9-12x+4x^2\right)=0\)
<=>\(\left(5-3x\right)^{^2}+\left(5-3x\right)\left(7-9x-3x^2\right)=0\)
<=>\(\left(5-3x\right)\left(5-3x+7-9x-3x^2\right)=0\)
<=>\(3.\left(5-3x\right)\left(4-4x-x^2\right)=0\)
Mà 4-4x-x^2>0 nên 5-3x=0 <=>x=5/3