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1/ \(2x^2+3x-5=\left(2x^2+2x\right)-\left(5x+5\right)=2x\left(x+1\right)-5\left(x+1\right)=\left(x+1\right)\left(2x-5\right)\)
2/ \(16x-5x^2-3=\left(15x-5x^2\right)+\left(x-3\right)=5x\left(3-x\right)-\left(3-x\right)=\left(3-x\right)\left(5x-1\right)\)
3/ \(7x-6x^2-2=\left(3x-6x^2\right)-\left(2-4x\right)=3x\left(1-2x\right)-2\left(1-2x\right)=\left(1-2x\right)\left(3x-2\right)\)
4/ \(x^2+5x-6=\left(x^2-x\right)+\left(6x-6\right)=x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x+6\right)\)
1/ \(x^2+x-90=\left(x^2-10x\right)+\left(9x-90\right)=x\left(x-10\right)+9\left(x-10\right)=\left(x-10\right)\left(x+9\right)\)
2/ \(2x^2+4xy+2y^2=\left(2x^2+2xy\right)+\left(2xy+2y^2\right)=2x\left(x+y\right)+2y\left(x+y\right)=\left(x+y\right)\left(2x+2y\right)\)
3/ \(2y^2-14y+24=2\left(y^2-7y+12\right)=2\left[\left(y^2-4y\right)+\left(12-3y\right)\right]=2\left[y\left(y-4\right)-3\left(y-4\right)\right]\)
\(=2\left(y-4\right)\left(y-3\right)\)
4/ \(x^8+x^4+1=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x+1\right)\)
\(=\left(x^2+x+1\right)\left[\left(x^6-x^5+x^4\right)-\left(x^4-x^3+x^2\right)+\left(x^2-x+1\right)\right]\)
\(=\left(x^2+x+1\right)\left[x^4\left(x^2-x+1\right)\right]-x^2\left(x^2-x+1\right)+\left(x^2-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4-x^2+1\right)\)
1)\(x^4+2x^3+x^2\)
=\(\left(x^4+x^3\right)+\left(x^3+x^2\right)\)đật nhân tử chung ra
=\(x^2\left(x+1\right)^2\)
2) pt => \(\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
=\(\left(x+y\right)^3-\left(x+y\right)\)
=\(\left(x+y\right)\left(\left(x+y\right)^2+1\right)\)
3)chia tất cả cho 5 pt => \(x^2-2xy+y^2-4x^2\)
=\(\left(x+y\right)^2-4z^2\)
=\(\left(x+y+2z\right)\left(x+y-2z\right)\)
4)pt => \(2\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
=\(2\left(x-y\right)-\left(x-y\right)^2\)
=\(\left(x-y\right)\left(2-x+y\right)\)
k chi nha
a) \(x^2\left(x-3\right)+12-4x=x^2\left(x-3\right)-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x^2-2^2\right)\)
\(=\left(x+3\right)\left(x-2\right)\left(x+2\right)\)
b)\(x^2-4+\left(x-2\right)^2=x^2-2^2+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2+x-2\right)\)
\(=\left(x-2\right)2x\)
c)\(x^3-4x^2-12x+27=x^3+3x^2-7x^2-21x+9x+27\)
\(=x^2\left(x+3\right)-7x\left(x+3\right)+9\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
a) => x2.(x-3)-4(x-3)=(x-3)(x2-4)=(x-3)(x-2)(x+2)
b) => (x+2)(x-2)+(x-2)2=(x-2)(x+2+x-2)=2x(x-2)
c) => x3+27-(4x2+12x)=(x+3)(x2-3x+3)-4x(x+3)=(x+3)(x2-3x+3-4x)=(x-3)(x2-7x+3)
a) x2 - 3x + 2 = x2 - x - 2x + 2 = x( x - 1 ) - 2( x - 1 ) = ( x - 1 )( x - 2 )
b) 2x2 - x - 6 = 2x2 - 4x + 3x - 6 = 2x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( 2x + 3 )
c) x2 - 5x - 6 = x2 + x - 6x - 6 = x( x + 1 ) - 6( x + 1 ) = ( x + 1 )( x - 6 )
d) x2 + 8x + 7 = x2 + x + 7x + 7 = x( x + 1 ) + 7( x + 1 ) = ( x + 1 )( x + 7 )
e) 3x2 + 2x - 5 = 3x2 - 3x + 5x - 5 = 3x( x - 1 ) + 5( x - 1 ) = ( x - 1 )( 3x + 5 )
f) 4x2 - 3x - 1 = 4x2 - 4x + x - 1 = 4x( x - 1 ) + ( x - 1 ) = ( x - 1 )( 4x + 1 )
a \(x^2-3x+2=x^2-x-2x+2=\left(x-1\right)\left(x-2\right)\)
b, \(2x^2-x-6=2x^2-4x+3x-6=\left(x-2\right)\left(2x+3\right)\)
c, \(x^2-5x-6=x^2+x-6x-6=\left(x+1\right)\left(x-6\right)\)
d, \(x^2+8x+7=x^2+x+7x+7=\left(x+1\right)\left(x+7\right)\)
e, \(3x^2+2x-5=3x^2-3x+5x-5=\left(x-1\right)\left(3x+5\right)\)
f, \(4x^2-3x-1=4x^2-4x+x-1=\left(x-1\right)\left(4x+1\right)\)
a) \(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
b) \(x^2+x-12=x^2+4x-3x-12=x.\left(x+4\right)-3.\left(x+4\right)=\left(x+4\right)\left(x-3\right)\)
a) \(x^3-\frac{1}{4}x=x\left(x^2-\frac{1}{4}\right)=x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\)
b) \(\left(2x-1\right)^2-\left(x+3\right)^2=\left(2x-1-x-3\right)\left(2x-1+x+3\right)=\left(x-4\right)\left(3x+2\right)\)
c) \(x^2-y^2-2y-1=x^2-\left(y^2+2y+1\right)=x^2-\left(y+1\right)^2=\left(x-y-1\right)\left(x+y+1\right)\)
d) \(x^2\left(x-3\right)+12-4x=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-2^2\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
Phép tính b):
Đặt 2x - 1 = a ; x + 3 = b. Từ đầu bài suy ra:
\(\left(2x-1\right)^2-\left(x+3\right)^2\Rightarrow a^2-b^2\)
\(\Rightarrow a^2-b^2-\left(ab-ab\right)\Rightarrow\left(a^2-ab\right)-\left(b^2-ab\right)\)
\(\Rightarrow a\left(a-b\right)-b\left(b-a\right)\Rightarrow a\left(a-b\right)+b\left(a-b\right)\)
\(\Rightarrow\left(a+b\right)\left(a-b\right)\)
Thế lại vào ta có:
\(\orbr{\begin{cases}a+b=\left(2x-1\right)+\left(x+3\right)=\left(2x+x\right)-\left(1-3\right)=3x+2\\a-b=\left(2x-1\right)-\left(x-3\right)=\left(2x-x\right)-\left(1-3\right)=x+2\end{cases}}\)
\(\Rightarrow\left(a+b\right)\left(a-b\right)=\left(3x+2\right)\left(x+2\right)\)
1) x2 - 4x + 3
= x2 - x - 3x + 3
= (x2 - x) - (3x - 3)
= x.(x - 1) - 3.(x - 1)
= (x - 1).(x - 3)
2) x2 - x - 6
= x2 + 2x - 3x - 6
= (x2 + 2x) - (3x + 6)
= x.(x + 2) - 3.(x + 2)
= (x + 2).(x - 3)
3) x2 + 5x + 4
= x2 + x + 4x + x
= (x2 + x) + (4x + x)
= x.(x + 1) + 4.(x + 1)
= (x + 1).(x + 4)
4) x2 + 5x + 6
= x2 + 2x + 3x + 6
= (x2 + 2x) + (3x + 6)
= x.(x + 2) + 3.(x + 2)
= (x + 2).(x + 3)
a,=x^2+x+3x+3
=x(x+1)+3(x+1)
=(x+3)(x+1)
b,x^2-3x+2x-6
=x(x-3)+2(x-3)
=(x+2)(x-3)
2 câu còn lại từ lm nha.........