x^2+3x-18 tách hạng tử
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\(x^2+3x-18=x^2-3x+6x-18=x\left(x-3\right)+6\left(x-3\right)=\left(x+6\right)\left(x-3\right)\)
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 )
a, 3x^2 + 13x + 10
= 3x^2 + 3x + 10x + 10
= 3x(x + 1) + 10(x + 1)
= (3x + 10)(x + 1)
b, x^2 - 10x + 21
= x^2 - 3x - 7x + 21
= x(x - 3) - 7(x - 3)
= (x - 7)(x - 3)
c, 6x^2 - 5x + 1
= 6x^2 - 3x - 2x + 1
= 3x(2x - 1) - (2x - 1)
= (3x - 1)(2x - 1)
Bạn đăng 1 lần nhiều bài như vậy làm người khác nản lắm đấy =) đơn giản bài rất dài mà mik cx ko chắc là bản thân mik có đc k hay ko nên phải nản vậy thôi :)
1a)\(3x^2+13x+10=3x^2+3x+10x+10\)
\(3x\left(x+1\right)+10\left(x+1\right)=\left(3x+10\right)\left(x+1\right)\)
b)\(x^2-10x+21=x^2-3x-7x+21\)
\(=x\left(x-3\right)-7\left(x-3\right)=\left(x-7\right)\left(x-3\right)\)
c)\(6x^2-5x+1=6x^2-3x-2x+1\)
\(=3x\left(2x-1\right)-\left(2x-1\right)=\left(3x-1\right)\left(2x-1\right)\)
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) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x+5\right)\left(x-2\right)\)
a, x2-5x+6=(x2-2x)-(3x-6)=x(x-2)-3(x-2)=(x-2)(x-3)
b, 3x2+9x-30=(3x2-6x)+(15x-30)=3x(x-2)+15(x-2)=3(x-2)(x+5)
c, x2-3x+2=(x2-x)-(2x-2)=x(x-1)-2(x-1)=(x-1)(x-2)
a, x^2-5x+6=x^2-2x-3x+6=(x^2-2x)-(3x-6)=x(x-2)-3(x-2)=(x-3)(x-2)
b, 3x^2+9x-30=3x^2-6x+15x-30=(3x^2-6x)+(15x-30)=3x(x-2)+3(x-2)=(3x+3)(x-2)
c, x^2-3x+2=x^2-x-2x+2=(x^2-x)-(2x-2)=x(x-1)-2(x-1)=(x-2)(x-1)
Phân tích đa thức thành nhân tử(tách hạng tử)
1)x^2+2x-3=x^2-x+3x-3=x(x-1)+3(x-1)=(x-1)(x+3)
2)x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)
3)x^2+7x+12=(x+3)(x+4)
4)x^2-x-12=(x-4)(x+3)
5)3x^2+3x-36=3[(x-3)(x+4)]
6)5x^2-5x-10=5[(x-2)(x+1) ]
7)3x^2-7x-6=(x-3)(3x+2)
8)4x^2+4x-3=4x^2+6x-2x-3=(2x-1)(2x+3)
9)8x^2-2x-3=8x^2+4x-6x-3=(4x-3)(2x+1)
1: \(x^2+2x-3=\left(x+3\right)\left(x-1\right)\)
2: \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)
3: \(x^2+7x^2+12x=4x\left(2x+3\right)\)
4: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)
5: \(3x^2+3x-36=3\left(x^2+x-12\right)=3\left(x+4\right)\left(x-3\right)\)
6: \(5x^2-5x-10=5\left(x^2-x-2\right)=5\left(x-2\right)\left(x+1\right)\)
\(x^3+3x^2-4\)
\(=\left(x^3+4x^2\right)-\left(x^2+4\right)\)
\(=\left(x^2+4\right)\left(x-1\right)\)
Mình nhìn nhầm đề
\(x^3+3x^2-4\)
\(=\left(x^3+2x^2\right)+\left(x^2-4\right)\)
\(=x^2\left(x+2\right)+\left(x-2\right)\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+x-2\right)\)
\(=\left(x+2\right)\left[\left(x^2+x\right)-\left(2x+2\right)\right]\)
\(=\left(x+2\right)\left(x+2\right)\left(x-1\right)\)
\(=\left(x+2\right)^2\left(x-1\right)\)
x2 + 3x -18
= x2 + 6x - 3x -18
= x(x+6) - 3(x+6)
= (x+6)(x-3)
x2 + 3x - 18
= x2 - 3x + 6x - 18
= x( x - 3 ) + 6( x - 3 )
= ( x - 3 )( x + 6 )