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a) \(x^2-xy+4x-2y+4\)
\(=\left(x^2+4x+4\right)-\left(xy+2y\right)\\ =\left(x+2\right)^2-y.\left(x+2\right)\)
\(=\left(x+2\right).\left(x+2-y\right)\)
b) \(2x^2-5x-3\)
\(=2x^2+x-6x-3\)
\(=\left(2x^2+x\right)-\left(6x+3\right)=x\left(2x+1\right)-3\left(2x+1\right)\)
\(=\left(2x+1\right).\left(x-3\right)\)
c)\(\)
c);d);e) tạm thời tớ chưa nghĩ ra-.-"
tham khả tạm 2 câu ạ, chúc học tốt'.'
1) (3x+4)(x+1) = 3x2+7x+4 đặt là a
(6x+7)2= 36x2+84x+49 = 12a+1
=> a(12a+1)- 6 = 12a2 -a -6 = (3a+2)(4a-3) = (9x2+21x+14)(12x2+28x+13)
2) (x-2)2=x2-4x+4 đặt là a
(2x-5)(2x-3)= 4x2-16x+15 =4a-1
=> a(4a-1)-5 = 4a2-a-5 = (4a-5)(a+1) = ( 4x2-16x+11)(x2-4x+5)
3) đặt (x+3)2 =a ta làm tương tự
4) (x-2)(x-10)(x-4)(x-5) = (x2-12x+20)(x2-9x+20)
đặt x2+20=a => (a-12x)(a-9x)-54x2 = a2-21ax+54x2 = (a-18x)(a-3x) = (x2-18x+20)(x2-3x+20)
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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a) \(x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
a/\(\left(x^2-x\right)^2+4\left(x^2-x\right)-12.\)
cho \(\left(x^2-x\right)=a\)
\(\Rightarrow a^2+4a-12\)
\(=a^2+6a-2a-12\)
\(=\left(a^2+6a\right)-\left(2a+12\right)\)
\(=a\left(a+6\right)-2\left(a+6\right)\)
\(=\left(a+6\right)\left(a-2\right)\)
\(=\left(x^2-x+6\right)\left(x^2-x-2\right)\)
b/ \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24\)
\(=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
Gọi \(x^2+5x+5=a\)
\(\Rightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=\left(a-1\right)\left(a+1\right)-24\)
\(=a^2-1-24\)
\(=a^2-25\)
\(=\left(a-5\right)\left(a+5\right)\)
\(\Rightarrow\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
a) \(x^2-6x+8=x^2-4x-2x+8=x\left(x-4\right)-2\left(x-4\right)=\left(x-2\right)\left(x-4\right)\)
b) \(x^2-8xy+12y^2=x^2-6xy-2xy+12y^2=x\left(x-6y\right)-2y\left(x-6y\right)=\left(x-6y\right)\left(x-2y\right)\)c) \(x^4+4x^2-5=x^4-x^3+x^3-x^2+5x^2-5\)
\(=x^3\left(x-1\right)+x^2\left(x-1\right)+5\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x^3+x^2+5x+5\right)\)
\(=\left(x-1\right)\left[x^2\left(x+1\right)+5\left(x+1\right)\right]=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
d) \(x^4+x^2+1=x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\)
\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
a, \(x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
b, \(x^2-2.4xy+16y^2-4y^2=\left(x-4y\right)^2-4y^2=\left(x-6y\right)\left(x-2y\right)\)
c, \(x^4+4x^2+4-9=\left(x^2+2\right)^2-9=\left(x^2-1\right)\left(x^2+5\right)=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
d, \(x^4-2x^22y^2+4y^4+4x^2y^2=\left(x^2-2y^2\right)^2+4x^2y^2\)bạn xem lại đề nhé
e, \(x^4+x^2+\dfrac{1}{4}-\dfrac{1}{4}+1=\left(x^2+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)bạn xem lại đề nhé