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Bài làm:
a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)
Đặt \(x^2+5x+5=t\)\(\Rightarrow\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2\)
\(=\left(x^2+5x+5\right)^2\)
b) Tương tự như a phân tích và đặt ra được: \(t^2-1-24=t^2-25=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)=x\left(x+5\right)\left(x^2+5x+10\right)\)
c) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+11=t\)\(\Rightarrow\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1\)
\(=\left(t-1\right)\left(t+1\right)=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
\(=\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)
d) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+11=t\)\(\Rightarrow\left(t-1\right)\left(t+1\right)-24=t^2-1-24=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
Làm mẫu cho 1 vd:
a, (x+1)(x+2)(x+3)(x+4)+1
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)(1)
Đặt \(y=x^2+5x+5\)
Khi đó ::
(1) = \(\left(y-1\right)\left(y+1\right)+1\)
\(=y^2-1+1=y^2\)
Thay vào ta được: \(\left(x^2+5x+5\right)^2\)
a) (x-1)(2x+5)
b) (x+1)(x-5)
c) [(x+1)^2](x^2+x+1)
d) (x-1)(x^3-x-1)
e) (x+y)(x-y-1)
a) 2x2 + 3x - 5 = 2x2 + 5x - 2x - 5 = x(2x + 5) - (2x + 5) = (x - 1)(2x + 5)
b) x2 - 4x - 5 = x2 - 5x + x - 5 = x(x - 5) + (x - 5) = (x + 1)(x - 5)
c) x4 + x3 + x + 1 = x3(x + 1) + (x + 1) = (x + 1)(x3 + 1) = (x + 1)2(x2 - x + 1)
d) x4 - x3 - x2 + 1 = x3(x - 1) - (x - 1)(x + 1) = (x - 1)(x3 - x - 1)
e) -x - y2 + x2 - y = -(x + y) + (x - y)(x + y) = (-1 + x - y)(x + y)
a) Ta có: \(a^3y^3+125\)
\(=\left(ay+5\right)\left(a^2y^2-5ay+25\right)\)
b) Ta có: \(8x^3-y^3-6xy\cdot\left(2x-y\right)\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)-6xy\left(2x-y\right)\)
\(=\left(2x-y\right)\left(4x^2+2xy-6xy+y^2\right)\)
\(=\left(2x-y\right)^3\)
a) \(x^3\left(x^2-7\right)^2-36x=x\left[\left(x^3-7x\right)^2-6^2\right]\)
\(=x\left(x^3-7x-6\right)\left(x^3-7x+6\right)\)
\(x\left[\left(x-3\right)\left(x+1\right)\left(x+2\right)\right].\left[\left(x+3\right)\left(x-2\right)\left(x-1\right)\right]\)
\(=\left(x-3\right)\left(x-2\right)\left(x-1\right).x.\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
b) Không pt được.
c) Không pt được.
Bài làm:
a) \(2x^2+7x+5=\left(2x^2+2x\right)+\left(5x+5\right)=2x\left(x+1\right)+5\left(x+1\right)\)
\(=\left(2x+5\right)\left(x+1\right)\)
b) \(x^3-2x-4=\left(x^3-2x^2\right)+\left(2x^2-4x\right)+\left(2x-4\right)\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)=\left(x-2\right)\left(x^2+2x+2\right)\)
c) \(x^2+4x+3=\left(x^2+x\right)+\left(3x+3\right)=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
2x2 + 7x + 5 = 2x2 + 2x + 5x + 5 = ( 2x2 + 2x ) + ( 5x + 5 ) = 2x( x + 1 ) + 5( x + 1 ) = ( 2x + 5 )( x + 1 )
x2 + 4x + 3 = x2 + x + 3x + 3 = ( x2 + x ) + ( 3x + 3 ) = x( x + 1 ) + 3( x + 1 ) = ( x + 3 )( x + 1 )
a. \(x^5+x+1\)
\(=\left(x^5-x^2\right)+x^2+x+1\)
\(=x^2\left(x^3-1\right)+x^2+x+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)\)\(+x^2+x+1\)
\(=\left[x^2\left(x-1\right)+1\right]\left(x^2+x+1\right)\)
\(=\left(x^3-x^2+1\right)\left(x^2+x+1\right)\)
b.\(x^3+x^2+4\)
=\(x^3+2x^2-x^2-2x+2x+4\)
\(=x^2\left(x+2\right)-x\left(x+2\right)+2\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-x+2\right)\)
c.\(x^4+2x^2-24\)
\(=x^4+2x^3-2x^3-4x^2+6x^2+12x-12x-24\)
\(=x^3\left(x+2\right)-2x^2\left(x+2\right)+6x\left(x+2\right)-12\left(x+2\right)\)
\(=\left(x^3-2x^2+6x-12\right)\left(x+2\right)\)
\(=\left[x^2\left(x-2\right)+6\left(x-2\right)\right]\left(x+2\right)\)
\(=\left(x^2+6\right)\left(x-2\right)\left(x+2\right)\)
a, x^5 + x + 1 = x ^ 5 - x^2 + (x ^2 + x + 1) = x^2 ( x-1) ( x^2+x+1) + ( x^2+x+1) = ( x^2+x+1 ) ( x^3-x^2+1)
c, x^4 + 2x^2 -24 = (x^4 +6x^2) - ( 4x^2+24) = x^2( x^2+6) - 4(x^2+6) = (x^2-4)(x^2 +6 ) = (x-2)(x+2)(x^2+6)