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a) \(x^3+4x^2-21x\)
\(=x\left(x^2+4x-21\right)\)
\(=x\left(x^2-3x+7x-21\right)\)
\(=x\left[x\left(x-3\right)+7\left(x-3\right)\right]\)
\(=x\left(x-3\right)\left(x+7\right)\)
b) \(5x^3+6x^2+x\)
\(=x\left(5x^2+6x+1\right)\)
\(=x\left(5x^2+5x+x+1\right)\)
\(=x\left[5x\left(x+1\right)+\left(x+1\right)\right]\)
\(=x\left(x+1\right)\left(5x+1\right)\)
c) \(x^3-7x+6\)
\(=x^3+2x^2-3x-2x^2-4x+6\)
\(=x\left(x^2+2x-3\right)-2\left(x^2+2x-3\right)\)
\(=\left(x-2\right)\left(x^2+2x-3\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(x+3\right)\)
d) \(3x^3+2x-5\)
\(=3x^3+3x^2+5x-3x^2-3x-5\)
\(=x\left(3x^2+3x+5\right)-\left(3x^2+3x+5\right)\)
\(=\left(x-1\right)\left(3x^2+3x+5\right)\)
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: \(5x^2-5x-10=5\left(x^2-x-2\right)=5\left(x-2\right)\left(x+1\right)\)
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/ \(x^2+5x-6=\left(x^2-x\right)+\left(6x-6\right)=x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x+6\right)\)
b: \(\Leftrightarrow\dfrac{x-2}{A}=\dfrac{\left(5x-1\right)\left(x-2\right)}{x^2\left(5x-1\right)+3\left(5x-1\right)}=\dfrac{x-2}{x^2+3}\)
hay \(A=x^2+3\)
a: \(A\left(x\right)=2x^4-x^3+3x^2+9x-2\)
\(B\left(x\right)=2x^4-5x^3-x+9\)
\(C\left(x\right)=x^4+4x^2+5\)
A(x): bậc 4; hệ số cao nhất là 2; hệ số tự do là -2
B(x): bậc 4; hệ số cao nhất là 4; hệ số tự do là 9
b: M(x)=A(x)+B(x)=4x^4-6x^3+3x^2+8x+7
N(x)=B(x)-A(x)=-4x^3-3x^2-10x+11
c: Q(x)=-N(x)=4x^3+3x^2+10x-11
a) Ta có : x2 + 7x + 12
= x2 + 3x + 4x + 12
= (x2 + 3x) + (4x + 12)
= x(x + 3) + 4(x + 3)
= (x + 4)(x + 3)
Bạn ơi mk nhầm đề rồi số 30 thay bằng số 60 còn 36 thay bằng 72 và 39 thay bằng 75 nha
2: =(2x+1)^2-y^2
=(2x+1+y)(2x+1-y)
3: =x^2(x^2+2x+1)
=x^2(x+1)^2
4: =x^2+6x-x-6
=(x+6)(x-1)
5: =-6x^2+3x+4x-2
=-3x(2x-1)+2(2x-1)
=(2x-1)(-3x+2)
6: =5x(x+y)-(x+y)
=(x+y)(5x-1)
7: =2x^2+5x-2x-5
=(2x+5)(x-1)
8: =(x^2-1)*(x^2-4)
=(x-1)(x+1)(x-2)(x+2)
9: =x^2(x-5)-9(x-5)
=(x-5)(x-3)(x+3)
\(1,\\ a,=-35x^5y^4z\\ b,=6x^2-30x-6x^2-3x=-33x\\ c,=x^3-9x^2-2x^2+18x-x+9=x^3-11x^2+17x+9\\ 2,\\ A\left(x\right)+B\left(x\right)=10-2x+4x^3-5x^2-10x^3-5x+6x^2-20\\ =-6x^3+x^2-7x-10\\ A\left(x\right)-B\left(x\right)=10-2x+4x^3-5x^2+10x^3+5x-6x^2+20\\ =14x^3-11x^2+3x+30\\ 3,\\ a,M\left(x\right)=5x+20=0\\ \Leftrightarrow x=-4\\ b,N\left(x\right)=100x^2-49=0\\ \Leftrightarrow\left(10x-7\right)\left(10x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\\ c,P\left(x\right)=3x-15=0\\ \Leftrightarrow x=5\)
Làm 1 cách là đủ rồi mà (: 6 cách thì đến bao giờ :v
a) x2 + x - 6 = x2 - 2x + 3x - 6 = x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( x + 3 )
b) x2 - 4x + 3 = x2 - x - 3x + 3 = x( x - 1 ) - 3( x - 1 ) = ( x - 1 )( x - 3 )
c) x2 + 5x + 4 = x2 + x + 4x + 4 = x( x + 1 ) + 4( x + 1 ) = ( x + 1 )( x + 4 )
d) x2 - x - 6 = x2 + 2x - 3x - 6 = x( x + 2 ) - 3( x + 2 ) = ( x + 2 )( x - 3 )
e) 2x2 + 5x + 3 = 2x2 + 2x + 3x + 3 = 2x( x + 1 ) + 3( x + 1 ) = ( x + 1 )( 2x + 3 )
g) 2x2 - 7x + 3 = 2x2 - 6x - x + 3 = 2x( x - 3 ) - ( x - 3 ) = ( x - 3 )( 2x - 1 )
h) 3x2 + 10x - 8 = 3x2 + 12x - 2x - 8 = 3x( x + 4 ) - 2( x + 4 ) = ( x + 4 )( 3x - 2 )
k) \(\frac{1}{2}x^2-\frac{19}{6}x+1=\frac{1}{2}x^2-\frac{1}{6}x-3x+1=\frac{1}{2}x\left(x-\frac{1}{3}\right)-3\left(x-\frac{1}{3}\right)=\left(x-\frac{1}{3}\right)\left(\frac{1}{2}x-3\right)\)
bị đòi làm 6 cách :')