Phân tích đa thức \(P\left( x \right) = \left( {x + 1} \right)\left( {2x - 1} \right) + \left( {x + 1} \right)x\) thành nhân tử.
phân tích đa thức thành nhân tử :
a, \( \left(x-5\right)^2-4\left(x-3\right)^2+2\left(2x-1\right)\left(x-5\right)+\left(2x-1\right)^2\)
(x - 5)2 - 4(x - 3)2 + 2(2x - 1)(x - 5) + (2x - 1)2
= [(x - 5)2 + 2(2x - 1)(x - 5) + (2x - 1)2) - [2(x - 3)]2
= (x - 5 + 2x - 1)2 - (2x - 6)2
= (3x - 6)2 - (2x - 6)2
= (3x - 6 - 2x + 6)(3x - 6 + 2x - 6) = x(5x - 12)
( x - 5 )2 - 4( x - 3 )2 + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )2
= [ ( x - 5 )2 + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )2 ] - 22( x - 3 )2
= ( x - 5 + 2x - 1 )2 - ( 2x - 6 )2
= ( 3x - 6 )2 - ( 2x - 6 )2
= ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 )
= x( 5x - 12 )
\(\left(x-5\right)^2-4\left(x-3\right)^2+2\left(2x-1\right)\left(x-5\right)+\left(2x-1\right)^2\)
\(=\left(x-5\right)^2+2\left(2x-1\right)\left(x-5\right)+\left(2x-1\right)^2-4\left(x-3\right)^2\)
\(=\left(x-5+2x-1\right)^2-\left(2x-6\right)^2\)
\(=\left(3x-6\right)^2-\left(2x-6\right)^2\)
\(=\left[\left(3x-6\right)-\left(2x-6\right)\right].\left[\left(3x-6\right)+\left(2x-6\right)\right]\)
\(=\left(3x-6-2x+6\right)\left(3x-6+2x-6\right)\)
\(=\left(5x-12\right)x\)
Phân tích đa thức thành nhân tử
\(2x\left(y-1\right)-z\left(1-y\right)\)
\(a\left(x-y\right)-b\left(x+y\right)+x-y\)
\(a\left(x-y\right)-b\left(y-x\right)+c\left(x-y\right)\)
\(a^m-a^{m+2}\)
a: \(a\left(x-y\right)-b\left(y-x\right)+c\left(x-y\right)\)
\(=a\left(x-y\right)+b\left(x-y\right)+c\left(x-y\right)\)
\(=\left(x-y\right)\left(a+b+c\right)\)
b: \(a^m-a^{m+2}\)
\(=a^m-a^m\cdot a^2\)
\(=a^m\left(1-a^2\right)\)
\(=a^m\left(1-a\right)\left(1+a\right)\)
Phân tích đa thức thành nhân tử: \(2\left(x^2+x+1\right)^2-\left(2x+1\right)^2-\left(x^2+2x\right)^2\)
\(2\left(x^2+x+1\right)^2-\left(2x+1\right)^2-\left(x^2+2x\right)^2\)
\(=2.\left[x^4+x^2+1+2x^3+2x+2x^2\right]-\left(4x^2+4x+1\right)-\left(x^4+4x^3+4x^2\right)\)
\(=x^4-2x^2+1=\left(x^2-1\right)^2=\left(x-1\right)^2\left(x+1\right)^2\)
Chúc bạn học tốt.
Phân tích đa thức thành nhân tử:
a) \(x^2\left(1-x^2\right)-4-4x^2\)
b) \(\left(1+2x\right)\left(1-2x\right)-x\left(x+2\right)\left(x-2\right)\)
Phân tích đa thức thành đa nhân tử :
\(3x\left(x-1\right)-4x\left(x-1\right)\)
\(=-x\left(x-1\right)=x\left(1-x\right)\)
phân tích đa thức thành nhân tử
a.\(\left(1+2x\right)\left(1-2x\right)-x\left(x+2\right)\left(x-2\right)\)
b.\(x^2+y^2-x^2y^2+xy-x-y\)
Phân tích các đa thức sau thành nhân tử:
\(A=4x^2+6x\). \(B=\left(2x+3\right)^2-x\left(2x+3\right)\). \(C=\left(9x^2-1\right)-\left(3x-1\right)^2\).
\(D=x^3-16x\). \(E=4x^2-25y^2\). \(G=\left(2x+3\right)^2-\left(2x-3\right)^2\).
\(A=4x^2+6x=2x\left(2x+3\right)\)
\(B=\left(2x+3\right)^2-x\left(2x+3\right)=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\)
\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)
\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)
\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)
\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)
1.Phân tích đa thức thành nhân tử
a.\(2x^3+3x^2-2x\) b.\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
2.Cho A=\(\dfrac{2x+1}{\left(x-4\right)\left(x-3\right)}-\dfrac{x+3}{x-4}+\dfrac{2x-1}{x-3}\)
a.Rút gọn biểu thức A
b.tính giá trị của A biết \(x^2+20=9x\)
3.Tìm đa thức thương và đa thức dư trong phép chia:\(\left(2x^2-7x^2:13x:2\right):\left(2x-1\right)\)
Bài 1:
a. $2x^3+3x^2-2x=2x(x^2+3x-2)=2x[(x^2-2x)+(x-2)]$
$=2x[x(x-2)+(x-2)]=2x(x-2)(x+1)$
b.
$(x+1)(x+2)(x+3)(x+4)-24$
$=[(x+1)(x+4)][(x+2)(x+3)]-24$
$=(x^2+5x+4)(x^2+5x+6)-24$
$=a(a+2)-24$ (đặt $x^2+5x+4=a$)
$=a^2+2a-24=(a^2-4a)+(6a-24)$
$=a(a-4)+6(a-4)=(a-4)(a+6)=(x^2+5x)(x^2+5x+10)$
$=x(x+5)(x^2+5x+10)$
Bài 2:
a. ĐKXĐ: $x\neq 3; 4$
\(A=\frac{2x+1-(x+3)(x-3)+(2x-1)(x-4)}{(x-3)(x-4)}\\ =\frac{2x+1-(x^2-9)+(2x^2-9x+4)}{(x-3)(x-4)}\\ =\frac{x^2-7x+14}{(x-3)(x-4)}\)
b. $x^2+20=9x$
$\Leftrightarrow x^2-9x+20=0$
$\Leftrightarrow (x-4)(x-5)=0$
$\Rightarrow x=5$ (do $x\neq 4$)
Khi đó: $A=\frac{5^2-7.5+14}{(5-4)(5-3)}=2$
Bài 3:
$(2x^2-7x^2:13x:2):(2x-1)=(2x^2-\frac{7}{26}x):(2x-1)$
$=[x(2x-1)+\frac{19}{52}(2x-1)+\frac{19}{52}]:(2x-1)$
$=[(2x-1)(x+\frac{19}{52})+\frac{19}{52}]: (2x-1)$
$\Rightarrow$ thương là $x+\frac{19}{52}$ và thương là $\frac{19}{52}$
phân tích đa thức thành nhân tử
\(x\left(x+2\right)\left(x^2+2x+2\right)+1\)
Phân tích đa thức thành nhân tử:
\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
(x+1)(x+3)(x+5)(x+8)+15
=[(x+1)(x+7)][(x+3)(x+5)]+15
=(x2+8x+7)(x2+8x+15)+15
Đặt t=x2+8x+7
=>x2+8x+15=t+8
=>(x2 +8x+7)(x2+8x+15)+15
=t(t+8)+15
=t2+8t+15
=t2+3t+5t+15
=t(t+3)+5(t+3)
=(t+3)(t+5)
=(x2+8x+10)(x2+8x+12)
Đặt \(A=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(\Rightarrow A=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+11=t\)
\(\Rightarrow A=\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1=\left(t+1\right)\left(t-1\right)\)
\(=\left(x^2+8x+11+1\right)\left(x^2+8x+11-1\right)=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)
\(=\left(x^2+2x+6x+12\right)\left(x^2+8x+10\right)\)\(=\left[x\left(x+2\right)+6\left(x+2\right)\right]\left(x^2+8x+10\right)\)
\(=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)