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a) \(x^2-5x+6=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\)
b)\(3x^2+9x-30=3x^2-6x+15x-30=3\left(x-2\right)\left(x+5\right)\)
c)\(x^2-7x+12=x^2-3x-4x+12=\left(x-3\right)\left(x-4\right)\)
d)\(x^2-7x+10=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\)
a) \(x^2-5x+6=x^2-2x-3x+6=\left(x^2-2x\right)-\left(3x-6\right)\)
\(=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
b) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)
\(=3\left[\left(x^2-2x\right)+\left(5x-10\right)\right]=3\left[x\left(x-2\right)+5\left(x-2\right)\right]\)
\(=3\left(x-2\right)\left(x+5\right)\)
c) \(x^2-7x+12=x^2-3x-4x+12=\left(x^2-3x\right)-\left(4x-12\right)\)
\(=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)
d) \(x^2-7x+10=x^2-2x-5x+10=\left(x^2-2x\right)-\left(5x-10\right)\)
\(=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)
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) \(x^2-9x+18=x^2-3x-6x+18\)
\(=x\left(x-3\right)-6\left(x-3\right)=\left(x-3\right)\left(x-6\right)\)
c) \(x^2-6x+5=x^2-x-5x+5\)
\(=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)
d) \(3x^2+5x-30=3\left(x^2+\dfrac{5x}{3}-10\right)=3\left(x^2+2.x.\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{5347}{500}\right)\)
Câu này bạn xem lại đề nha
e) \(3x^2-5x-2=3x^2-6x+x-2\)
\(3x\left(x-2\right)+x-2=\left(x-2\right)\left(3x+1\right)\)
Bài 1 : Phân tích các đa thức sau thành nhân tử : ( tách một hạn tử thành nhiều hạng tử )
a, 3x2 + 9x - 30
= 3(x2 + 3x - 10)
= 3(x2 + 5x - 2x - 10)
= 3[x(x + 5) - 2(x + 5)]
= 3(x + 5)(x - 2)
b, x2 - 3x + 2
= x2 - x - 2x + 2
= x(x - 1) - 2(x - 1)
= (x - 1)(x - 2)
c, x2 - 9x + 18
= x2 - 6x - 3x + 18
= x(x - 6) - 3(x - 6)
= (x - 6)(x - 3)
d, x2 - 6x + 8
= x2 - 4x - 2x + 8
= x(x - 4) - 2(x - 4)
= (x - 4)(x - 2)
e, x2 - 5x - 14
= x2 + 2x - 7x - 14
= x(x + 2) - 7(x + 2)
= (x + 2)(x - 7)
f, x2 + 6x + 5
= x2 + x + 5x + 5
= x(x + 1) + 5(x + 1)
= (x + 1)(x + 5)
h, x2 - 7x + 12
= x2 - 3x - 4x + 12
= x(x - 3) - 4(x - 3)
= (x - 3)(x - 4)
i, x2 - 7x + 10
= x2 - 2x - 5x + 10
= x(x - 2) - 5(x - 2)
= (x - 2)(x - 5)
#Học tốt!
a) x2 - 5x + 6
= x2 - 2x - 3x + 6
=(x2 - 2x) - (3x + 6)
=x.(x - 2) - 3.(x - 2)
=(x-2).(x-3)
b) 3x2+9x-30
=3x2+15x-6x-30
=(3x2+15x) - (6x+30)
= 3x(x+5) - 6(x+5)
=(x+5).(3x-6)
c) x2-3x+2
=x2-2x-x+2
=(x2-2x) - (x-2)
=x(x-2)-(x-2)
=(x-2)(x-1)
a)x2 - 5x + 6
= x2 - 2x - 3x + 6
=x.(x - 2) - 3(x - 2)
=(x - 2).(x - 3)
b)3x2 +9x -30
=3x2 +15x - 6x -30
=3x.(x+5) - 6.(x + 5)
=(x+5).(3x - 6)
c)x2 - 3x +2
=x2 - 2x - x +2
=x.(x- 2) - 1.(x-2)
=(x-2).(x - 1)
d)x2 - 9x +18
=x2 - 6x -3x +18
=x.(x - 6) -3.(x - 6)
=(x - 6).(x - 3)
e)x2 - 6x +8
=x2 - 2x - 4x +8
=x.(x - 2)- 4.(x - 2)
=(x - 2).(x - 4)
f)x2 - 5x -14
=x2 + 2x - 7x - 14
=x.(x + 2) -7.(x + 2)
=(x + 2).(x - 7)
\(a,x^2-5x+6\\=x^2-3x-2x+6\\=x(x-3)-2(x-3)\\=(x-3)(x-2)\\---\\b,3x^2+9x-30\\=3x^2-6x+15x-30\\=3x(x-2)+15(x-2)\\=(x-2)(3x+15)\\=3(x-2)(x+5)\\---\)
\(c,x^2-3x+2\\=x^2-x-2x+2\\=x(x-1)-2(x-1)\\=(x-1)(x-2)\\---\\d,3x^2-5x-2\\=3x^2-6x+x-2\\=3x(x-2)+(x-2)\\=(x-2)(3x+1)\\Toru\)
a) \(x^2-3x+2\)
\(\Leftrightarrow x^2-2x-x+2\)
\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\)
b) \(x^2-6x+8\)
\(\Leftrightarrow x^2-4x-2x+8\)
\(\Leftrightarrow x\left(x-4\right)-2\left(x-4\right)\)
\(\Leftrightarrow\left(x-4\right)\left(x-2\right)\)
c) \(3x^2+9x-30\)
\(\Leftrightarrow3\left(x^2+3x-10\right)\)
\(\Leftrightarrow3\left[\left(x^2+2\cdot\frac{3x}{2}+\frac{9}{4}\right)-\frac{49}{4}\right]\)
\(\Leftrightarrow3\left[\left(x+\frac{3}{2}\right)^2-\left(\frac{7}{2}\right)^2\right]\)
\(\Leftrightarrow3\left(x+\frac{3}{2}+\frac{7}{2}\right)\left(x+\frac{3}{2}-\frac{7}{2}\right)\)
\(\Leftrightarrow3\left(x-2\right)\left(x+5\right)\)
d) \(x^2-9x+18\)
\(\Leftrightarrow x^2-3x-6x+18\)
\(\Leftrightarrow x\left(x-3\right)-6\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x-6\right)\)
TK MK NKA !!!! TH@NK !!!
a) x2 - 3x + 2 ( như này mới phân tích được ạ :) )
= x2 - x - 2x + 2
= x( x - 1 ) - 2( x - 1 )
= ( x - 2 )( x - 1 )
b) x2 - 6x + 8
= x2 - 2x - 4x + 8
= x( x - 2 ) - 4( x - 2 )
= ( x - 4 )( x - 2 )
c) 3x2 + 9x - 30
= 3( x2 + 3x - 10 )
= 3( x2 - 2x + 5x - 10 )
= 3[ x( x - 2 ) + 5( x - 2 )]
= 3( x + 5 )( x - 2 )
d) x2 - 9x + 18
= x2 - 3x - 6x + 18
= x( x - 3 ) - 6( x - 3 )
= ( x - 6 )( x - 3 )
1) \(x^2-6x+3\)
\(=x^2-6x+9-6\)
\(=\left(x-3\right)^2-6\)
\(=\left(x-3+\sqrt{6}\right)\left(x-3-\sqrt{6}\right)\)
2) \(2m^2+10m+8\)
\(=2m^2+2m+8m+8\)
\(=2m\left(m+1\right)+8\left(m+1\right)\)
\(=\left(2m+8\right)\left(m+1\right)\)
\(=2\left(m+4\right)\left(m+1\right)\)
3) \(9x^2+6x-8\)
\(=\left(9x^2+6x+1\right)-9\)
\(=\left(3x+1\right)^2-9\)
\(=\left(3x+4\right)\left(3x-2\right)\)
4) \(x^3-5x^2-14x\)
\(=x\left(x^2-5x-14\right)\)
\(=x\left(x^2-2x+7x-14\right)\)
\(=x\left[x\left(x-2\right)+7\left(x-2\right)\right]\)
\(=x\left(x+7\right)\left(x-2\right)\)
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/
\(3x^2+9x-30=3\left(x^2+3x-10\right)\)
c/
\(x^2-3x+2=x^2-x-2x+2=x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(x-2\right)\)
d/\(x^2-9x+18=x^2-3x-6x+18=x\left(x-3\right)-6\left(x-3\right)=\left(x-3\right)\left(x-6\right)\)e/
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-2\right)\left(x-4\right)\)f/\(x^2-5x-14=x^2+2x-7x-14=x\left(x+2\right)-7\left(x+2\right)=\left(x+2\right)\left(x-7\right)\)
g/
\(x^2-6x+5=x^2-x-5x+5=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)
h/
\(x^2-7x+12=x^2-4x-3x+12=x\left(x-4\right)-3\left(x-4\right)=\left(x-4\right)\left(x-3\right)\)i/\(x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)
a) Ta có: \(x^2-5x+6\)
\(=x^2-2x-3x+6\)
\(=x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x-3\right)\)
b) Ta có: \(3x^2+9x-30\)
\(=3\left(x^2+3x-10\right)\)
\(=3\left(x^2+5x-2x-10\right)\)
\(=3\left[x\left(x+5\right)-2\left(x+5\right)\right]\)
\(=3\left(x+5\right)\left(x-2\right)\)
c) Ta có: \(x^2-3x+2\)
\(=x^2-x-2x+2\)
\(=x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(x-2\right)\)
d) Ta có: \(x^2-9x+18\)
\(=x^2-3x-6x+18\)
\(=x\left(x-3\right)-6\left(x-3\right)\)
\(=\left(x-3\right)\left(x-6\right)\)
e) Ta có: \(x^2-6x+8\)
\(=x^2-4x-2x+8\)
\(=x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
f) Ta có: \(x^2-5x-14\)
\(=x^2-7x+2x-14\)
\(=x\left(x-7\right)+2\left(x-7\right)\)
\(=\left(x-7\right)\left(x+2\right)\)
g) Ta có: \(x^2-6x+5\)
\(=x^2-x-5x+5\)
\(=x\left(x-1\right)-5\left(x-1\right)\)
\(=\left(x-1\right)\left(x-5\right)\)
h) Ta có: \(x^2-7x+12\)
\(=x^2-3x-4x+12\)
\(=x\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x-4\right)\)
i) Ta có: \(x^2-7x+10\)
\(=x^2-2x-5x+10\)
\(=x\left(x-2\right)-5\left(x-2\right)\)
\(=\left(x-2\right)\left(x-5\right)\)
a) x2 + 5x + 6
= x2 + 2x + 3x + 6
= (x2 + 2x) + (3x + 6)
= x(x + 2) + 3 (x + 2)
= (x + 2) (x + 3)
b) x2 + 6x + 8
= x2 + 2x + 4x + 8
= (x2 + 2x) + (4x + 8)
= x(x + 2) + 4(x + 2)
= (x + 2)(x + 4)
c) x2 - 5x - 14
= x2 + 2x - 7x - 14
= (x2 + 2x) - (7x + 14)
= x(x + 2) - 7(x + 2)
= (x + 2)(x - 7)
d) x2 - 9x + 18
= x2 - 3x - 6x + 18
= (x2 - 3x) - (6x + 18)
= x(x - 3) - 6 (x - 3)
= (x - 3)(x - 6)
e) x2 - 7x + 12
= x2 -3x - 4x + 12
= (x2 - 3x) - (4x + 12)
= x(x - 3) - 4(x - 3)
= (x - 3)(x - 4)
f) 3x2 + 9x - 30
= 3(x2 + 3x - 10)
= 3\(\left[\left(x^2+5x-2x-10\right)\right]\)
= 3\(\left[\left(x^2+5x\right)-\left(2x-10\right)\right]\)
= 3\(\left[x\left(x+5\right)-2\left(x+5\right)\right]\)
= 3(x + 5)(x - 2)
Chuc ban hoc tot
a) \(x^2+5x+6=\left(x+2\right)\left(x+3\right)\)
b) \(x^2+6x+8=\left(x+2\right)\left(x+4\right)\)
c) \(x^2-5x-14=\left(x-7\right)\left(x+2\right)\)
d) \(x^2-9x+18=\left(x-3\right)\left(x-6\right)\)
e) \(x^2-7x+12=\left(x-3\right)\left(x-4\right)\)
f) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x+5\right)\left(x-2\right)\)