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a) ( x - 1 )( 2x + 1 ) + 3( x - 1 )( x + 2 )( 2x + 1 )
= ( x - 1 )( 2x + 1 )[ 1 + 3( x + 2 ) ]
= ( x - 1 )( 2x + 1 )( 1 + 3x + 6 )
= ( x - 1 )( 2x + 1 )( 3x + 7 )
b) ( 6x + 3 ) - ( 2x - 5 )( 2x + 1 )
= 3( 2x + 1 ) - ( 2x - 5 )( 2x + 1 )
= ( 2x + 1 )[ 3 - ( 2x - 5 ) ]
= ( 2x + 1 )( 3 - 2x + 5 )
= ( 2x + 1 )( 8 - 2x )
= 2( 2x + 1 )( 4 - x )
c) ( x - 5 )2 + ( x + 5 )( x - 5 ) - ( 5 - x )( 2x + 1 )
= ( x - 5 )2 + ( x + 5 )( x - 5 ) + ( x - 5 )( 2x + 1 )
= ( x - 5 )[ ( x - 5 ) + ( x + 5 ) + ( 2x + 1 ) ]
= ( x - 5 )( x - 5 + x + 5 + 2x + 1 )
= ( x - 5 )( 4x + 1 )
d) ( 3x - 2 )( 4x - 3 ) - ( 2 - 3x )( x - 1 ) - 2( 3x - 2 )( x + 1 )
= ( 3x - 2 )( 4x - 3 ) + ( 3x - 2 )( x - 1 ) - 2( 3x - 2 )( x + 1 )
= ( 3x - 2 )[ ( 4x - 3 ) + ( x - 1 ) - 2( x + 1 ) ]
= ( 3x - 2 )( 4x - 3 + x - 1 - 2x - 2 )
= ( 3x - 2 )( 3x - 6 )
= 3( 3x - 2 )( x - 2 )
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1.a) 2x4-4x3+2x2
=2x2(x2-2x+1)
=2x2(x-1)2
b) 2x2-2xy+5x-5y
=2x(x-y)+5(x-y)
=(2x+5)(x-y)
2.
a) 4x(x-3)-x+3=0
=>4x(x-3)-(x-3)=0
=>(4x-1)(x-3)=0
=> 2 TH:
*4x-1=0 *x-3=0
=>4x=0+1 =>x=0+3
=>4x=1 =>x=3
=>x=1/4
vậy x=1/4 hoặc x=3
b) (2x-3)^2-(x+1)^2=0
=> (2x-3-x-1).(2x-3+x+1)=0
=>(x-4).(3x-2)=0
=> 2 TH
*x-4=0
=> x=0+4
=> x=4
*3x-2=0
=>3x=0-2
=>3x=-2
=>x=-2/3
vậy x=4 hoặc x=-2/3
a)(ab−1)2+(a+b)2
=a2b2−2ab+1+a2+2ab+b2
=a2b2+1+a2+b2=a2(b2+1)+(b2+1) = (a2+1)(b2+1)
c)x3−4x2+12x−27
=x3−27+(−4x2+12x)
=(x−3)(x2+3x+9)−4x(x−3)
=(x−3)(x2+3x+9−4x)
=(x−3)(x2−x+9)
b)x3+2x2+2x+1
=x3+2x2+x+x+1
=x(x2+2x+1)+(x+1)
=x(x+1)2+(x+1)
=(x+1)(x(x+1)+1)
=(x+1)(x2+x+1)
d)x4−2x3+2x−1
=x4−2x3+x2−x2+2x−1
=x2(x2−2x+1)−(x2−2x+1)
=(x2−2x+1)(x2−1)
=(x−1)2(x−1)(x+1)
=(x−1)3(x+1)
e)x4+2x3+2x2+2x+1
=x4+2x3+x2+x2+2x+1
=x2(x2+2x+1)+(x2+2x+1)
=(x2+2x+1)(x2+1)
=(x+1)2(x2+1)
1) \(\left(3x+7\right)^2-\left(2x-3\right)^2=0\)
\(\Leftrightarrow\left(3x+7-2x+3\right)\left(3x+7+2x-3\right)=0\)
\(\Leftrightarrow\left(x+10\right)\left(5x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+10=0\\5x+4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-10\\x=\frac{-4}{5}\end{cases}}\)
Vạy ...
phần 2 tương tự áp dụng \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\((4x-1)^2-(5-3x)^2=0\)
\(\Leftrightarrow(4x-1-5-3x)(4x+1+5-3x)=0\)
\(\Leftrightarrow(x-6)(x+6)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
Vậy : ...
a, \(x^3-2x^2+3x-6=x\left(x^2+3\right)-2\left(x^2+3\right)=\left(x-2\right)\left(x^2+3\right)\)
b, \(x^2+2x+1-4y^2=\left(x+1\right)^2-\left(2y\right)^2=\left(x+1-2y\right)\left(x+1+2y\right)\)
help mình với ad ưi :/
\(a,\left(2x+3\right)\left(2x-3\right)-\left(2x+1\right)^2\)
\(=4x^2-9-4x^2-4x-1\)
\(=-4x-10\)
\(=-2\left(2x+5\right)\)
b,Tương tự