Phân tích đa thức thành nhân tử (phương pháp tách hạng tử)
1) x2 + x - 90
2) 2x2 + 4xy + 2y2
3) 2y2 - 14y + 24
4*) x8 + x4 + 1
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(x - 2)2 + (x + 3)2 - 2x2 = 0
<=> x2 - 4x + 4 + x2 + 6x + 9 - 2x2 = 0
<=> 2x + 13 =0
<=> 2x = -13
<=> x = -6,5
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/ \(x^2+5x-6=\left(x^2-x\right)+\left(6x-6\right)=x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x+6\right)\)
\(6^2.6^4-4^3\left(3^6-1\right)=6^6-2^6\left(3^6-1\right)\)
\(=6^6-6^6+2^6\)
\(=64\)
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.........
ta có A+B=180 ( 2 góc kề phụ, AD//BC) ; A-B=20 (GT)
=> A+B+A-B=200 => 2A=200 => A= 100
=> B=80
có A+C=150 mà A=100 => C=50
có D+C=180 ( 2gocs kề phụ, AD//BC) mà C=50 =>D=130
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/ \(x^8+x^4+1=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x+1\right)\)
\(=\left(x^2+x+1\right)\left[\left(x^6-x^5+x^4\right)-\left(x^4-x^3+x^2\right)+\left(x^2-x+1\right)\right]\)
\(=\left(x^2+x+1\right)\left[x^4\left(x^2-x+1\right)\right]-x^2\left(x^2-x+1\right)+\left(x^2-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4-x^2+1\right)\)