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a) (3x-5)2 - (x+1)2 =0
\(\Leftrightarrow\left(3x-5+x+1\right)\left[\left(3x-5\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(4x-4\right)\left(2x-6\right)=0\)
\(\Leftrightarrow8\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-3=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=3\end{array}\right.\)
b) 4x3 - 36x =0
\(\Leftrightarrow4x\left(x^2-9\right)=0\)
\(\Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}4x=0\\x-3=0\\x+3=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=3\\x=-3\end{array}\right.\)
Phân tích đa thức thành nhân tử
a) 4x2-36x+56 = 4\(x^2\)-36x+81-25 = \(\left(2x\right)^2\)-2.2x.9+\(9^2\) - 25 = \(\left(2x-9\right)^2\)-\(5^2\) = ( 2x - 9 - 5 ).( 2x - 9 + 5 ) = ( 2x - 14 ).( 2x - 4 )
b ) x4+4x2 = \(x^2.\left(x^2+4\right)\)
c ) x4 + x2 +1 = \(x^2\left(x^2+2\right)\)
d) 64\(x^4+1\)=\(64\left(x^4+1\right)\)
e ) 81x4+1=\(81\left(x^4+1\right)\)
\(4x^3-36x=0\)
\(x.\left[\left(2x\right)^2-6^2\right]=0\)
\(x.\left(2x-6\right)\left(2x+6\right)=0\)
\(\Rightarrow\)\(\orbr{\begin{cases}x=0\\2x-6=0\end{cases}}\)hoặc \(2x+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)hoặc \(x=-3\)
KL:...............................................
a) a3 - a2c + a2b - abc
= a( a2 - ac + ab - bc )
= a[ a( a - c ) + b( a - c ) ]
= a( a - c )( a + b )
b) ( x2 + 1 )2 - 4x2
= ( x2 + 1 )2 - ( 2x )2
= ( x2 - 2x + 1 )( x2 + 2x + 1 )
= ( x - 1 )2( x + 1 )2
c) x2 - 10x - 9y2 + 25
= ( x2 - 10x + 25 ) - 9y2
= ( x - 5 )2 - ( 3y )2
= ( x - 3y - 5 )( x + 3y - 5 )
d) 4x2 - 36x + 56
= 4( x2 - 9x + 14 )
= 4( x2 - 7x - 2x + 14 )
= 4[ x( x - 7 ) - 2( x - 7 ) ]
= 4( x - 7 )( x - 2 )
a,\(a^3-a^2c+a^2b-abc\)
\(=a\left(a^2-ac+ab-bc\right)\)
\(=a\left[a\left(a-c\right)+b\left(a-c\right)\right]\)
\(=a\left(a-b\right)\left(a-c\right)\)
b,\(\left(x^2+1\right)^2-4x^2\)
\(=\left(x^2+1-2x\right)\left(x^2+1+2x\right)\)
\(=\left(x-1\right)^2\left(x+1\right)^2\)
c,\(x^2-10x-9y^2+25\)
\(=\left(x^2-10x+25\right)-9y^2\)
\(=\left(x-5\right)^2-\left(3y\right)^2\)
\(=\left(x-5-3y\right)\left(x-5+3y\right)\)
d,\(4x^2-36x+56\)
\(=4\left(x^2-9x+14\right)\)
\(=4\left(x^2-7x-2x+14\right)\)
\(=4\left(x-7\right)\left(x-2\right)\)
4x2 - 36x + 56
= 4(x2 - 9x + 14)
= 4(x2 - 2x - 7x + 14)
= 4[x(x - 2) - 7(x - 2)]
= 4(x - 2)(x - 7)
\(4x^2\)−36x+56=04x2−36x+56
⇒4(x2−9x+14)=0⇒4(x2−9x+14)
⇒4(x2−7x−2x+14)=0⇒4(x2−7x−2x+14)
⇒4x(x−2)−7(x−2)=0⇒4x(x−2)−7(x−2)
⇒4(x−7)(x−2)=0⇒4(x−7)(x−2)
⇒(x−7)(x−2)=0⇒(x−7)(x−2)
⇒[x−7=0x−2=0⇒[x−7=0x−2=0
⇒x=7;x=2⇒x=7;x=2.
b) x(x - 3)+ 4( 3 - x) =0
=> x(x - 3) - 4( x - 3) = 0
=> (x - 3)( x - 4) =0
<=> x - 3 = 0 hoặc x - 4= 0
=> x= 3 hoặc => x= 4
Vậy x= 3 hoặc 4
a) 7x2 - 2x3 + 56 - 16x = 0
=> x2 ( 7 - 2x) + 8 ( 7 - 2x) = 0
=> ( 7 - 2x) ( x2 +8) =0
<=> 7 - 2x = 0 hoặc x2 + 8 =0
=> x= 7/2 hoặc x2 = -8 ( loại vì x2 \(\ge\) 0 )
Vậy x= 7/2
\(4x^2-36x+56=0\)
\(\Rightarrow4\left(x^2-9x+14\right)=0\)
\(\Rightarrow4\left(x^2-7x-2x+14\right)=0\)
\(\Rightarrow4x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Rightarrow4\left(x-7\right)\left(x-2\right)=0\)
\(\Rightarrow\left(x-7\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-7=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow x=7;x=2\).
\(4x^2-36x+56=0\)
\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot9+9^2-25=0\)
\(\Leftrightarrow\left(2x-9\right)^2-25=0\)
\(\Leftrightarrow\left(2x-9\right)^2=25\Leftrightarrow\left[{}\begin{matrix}2x-9=5\\2x-9=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=2\end{matrix}\right.\)
Vậy....................